Counting License Plateselyse/220/2016/3Counting.pdf · 3 Counting 13.1 Counting Lists 13.2...
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3 Counting 13.1 Counting Lists 13.2 Factorials 13.3 Counting Subsets 13.4 Pascal’s Triangle and the Binomial Theorem Counting License Plates License plates in BC generally have the form of three numbers, followed by three letters.
Transcript of Counting License Plateselyse/220/2016/3Counting.pdf · 3 Counting 13.1 Counting Lists 13.2...
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
First, an easier problem: how many license plates are possible if the first threenumbers are all zeroes and ones, and the letters are all vowels?
Digit 1: 2 ways
Digit 2: 2× 2 ways
Digit 3: 2× 2× 2 ways
0
00
000 001
01
010 011
1
10
100 101
11
110 111
Vowel 1: 5 ways
Vowel 2: 5× 5 ways
Vowel 3: 5× 5× 5 ways
a
•••••
e
•••••
i
•••••
o
•••••
u
•••••
Total: 103 plates.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
How many such plates are possible?
103263 = 17 576 000
Population of British Columbia: about 4.6 million(according to Google)
Number of cars in BC: 2,859,463(also according to Google)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
How many such plates are possible?
103263 = 17 576 000
Population of British Columbia: about 4.6 million(according to Google)
Number of cars in BC: 2,859,463(also according to Google)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
How many such plates are possible?
103263 = 17 576 000
Population of British Columbia: about 4.6 million(according to Google)
Number of cars in BC: 2,859,463(also according to Google)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting License Plates
License plates in BC generally have the form of three numbers, followed bythree letters.
How many such plates are possible?
103263 = 17 576 000
Population of British Columbia: about 4.6 million(according to Google)
Number of cars in BC: 2,859,463(also according to Google)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists
Definition
A list is an ordered sequence of objects.
The objects are called entries .
The number of entries in a list is the length of the list.
Example: the list (0, 1, a, 8, z , z) has third entry a; it has length 6.
Example: the list () is the empty list . It has no entries and its length is0.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists
Definition
A list is an ordered sequence of objects.
The objects are called entries .
The number of entries in a list is the length of the list.
Example: the list (0, 1, a, 8, z , z) has third entry a; it has length 6.
Example: the list () is the empty list . It has no entries and its length is0.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists
Definition
A list is an ordered sequence of objects.
The objects are called entries .
The number of entries in a list is the length of the list.
Example: the list (0, 1, a, 8, z , z) has third entry a; it has length 6.
Example: the list () is the empty list . It has no entries and its length is0.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
A password is a list of characters.
If you only use lower-case letters in the English alphabet, how manypasswords of length n are there?
If you can use upper-case letters and lower-case letters, how manypasswords of length n are there?
If you can use upper- and lower-case letters, as well as numbers, howmany passwords of length n are there?
n 26n 52n = 2n26n 62n ≈ 2.38n26n
0 1 1 11 26 52 622 766 2704 38443 17 576 140 608 238 3284 456 976 7 311 616 14 776 3368 208 827 064 576 53 459 728 531 456 218 340 105 584 896
12 95 quadrillion 390 quintillion 3 sextillion
Google: most English speakers have an active vocabulary of about 20,000words.The Oxford English dictionary has more like 180,000.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
A password is a list of characters.
If you only use lower-case letters in the English alphabet, how manypasswords of length n are there?
If you can use upper-case letters and lower-case letters, how manypasswords of length n are there?
If you can use upper- and lower-case letters, as well as numbers, howmany passwords of length n are there?
n 26n 52n = 2n26n 62n ≈ 2.38n26n
0 1 1 11 26 52 622 766 2704 38443 17 576 140 608 238 3284 456 976 7 311 616 14 776 3368 208 827 064 576 53 459 728 531 456 218 340 105 584 896
12 95 quadrillion 390 quintillion 3 sextillion
Google: most English speakers have an active vocabulary of about 20,000words.The Oxford English dictionary has more like 180,000.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
A password is a list of characters.
If you only use lower-case letters in the English alphabet, how manypasswords of length n are there?
If you can use upper-case letters and lower-case letters, how manypasswords of length n are there?
If you can use upper- and lower-case letters, as well as numbers, howmany passwords of length n are there?
n 26n 52n = 2n26n 62n ≈ 2.38n26n
0 1 1 11 26 52 622 766 2704 38443 17 576 140 608 238 3284 456 976 7 311 616 14 776 3368 208 827 064 576 53 459 728 531 456 218 340 105 584 896
12 95 quadrillion 390 quintillion 3 sextillion
Google: most English speakers have an active vocabulary of about 20,000words.The Oxford English dictionary has more like 180,000.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
Google: most English speakers have an active vocabulary of about 20,000words.How many phrases are there that contain three of these 20,000 words?
20 0003 = 8 000 000 000 000
More than the number of 8-character, lower-case-only passwords; less than thenumber of 8-character, letter-only passwords.
How many passwords have the following pattern:
1st entry: number
2nd entry: lower-case letter
3rd entry: number
10× 26× 10 = 2600
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
Google: most English speakers have an active vocabulary of about 20,000words.How many phrases are there that contain three of these 20,000 words?
20 0003 = 8 000 000 000 000
More than the number of 8-character, lower-case-only passwords; less than thenumber of 8-character, letter-only passwords.
How many passwords have the following pattern:
1st entry: number
2nd entry: lower-case letter
3rd entry: number
10× 26× 10 = 2600
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
Google: most English speakers have an active vocabulary of about 20,000words.How many phrases are there that contain three of these 20,000 words?
20 0003 = 8 000 000 000 000
More than the number of 8-character, lower-case-only passwords; less than thenumber of 8-character, letter-only passwords.
How many passwords have the following pattern:
1st entry: number
2nd entry: lower-case letter
3rd entry: number
10× 26× 10 = 2600
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Passwords
Google: most English speakers have an active vocabulary of about 20,000words.How many phrases are there that contain three of these 20,000 words?
20 0003 = 8 000 000 000 000
More than the number of 8-character, lower-case-only passwords; less than thenumber of 8-character, letter-only passwords.
How many passwords have the following pattern:
1st entry: number
2nd entry: lower-case letter
3rd entry: number
10× 26× 10 = 2600
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
3 ways for Rami to get a medal; 4 choices for one of the remaining medals; 3choices for the last remaining medal:3× 4× 3 = 36 ways.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Alternate solution: Rami wins gold, silver, or bronze.Case 1 : Rami wins gold.
4 choices for silver, 3 choices for bronze: 12
Case 2 : Rami wins silver.
4 choices for gold, 3 choices for bronze: 12
Case 3 : Rami wins bronze.
4 choices for gold, 3 choices for silver: 12
All together:
3(12) = 36 options.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Alternate solution: Rami wins gold, silver, or bronze.Case 1 : Rami wins gold. 4 choices for silver, 3 choices for bronze: 12
Case 2 : Rami wins silver.
4 choices for gold, 3 choices for bronze: 12
Case 3 : Rami wins bronze.
4 choices for gold, 3 choices for silver: 12
All together:
3(12) = 36 options.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Alternate solution: Rami wins gold, silver, or bronze.Case 1 : Rami wins gold. 4 choices for silver, 3 choices for bronze: 12
Case 2 : Rami wins silver. 4 choices for gold, 3 choices for bronze: 12
Case 3 : Rami wins bronze.
4 choices for gold, 3 choices for silver: 12
All together:
3(12) = 36 options.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Alternate solution: Rami wins gold, silver, or bronze.Case 1 : Rami wins gold. 4 choices for silver, 3 choices for bronze: 12
Case 2 : Rami wins silver. 4 choices for gold, 3 choices for bronze: 12
Case 3 : Rami wins bronze. 4 choices for gold, 3 choices for silver: 12
All together:
3(12) = 36 options.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Alternate solution: Rami wins gold, silver, or bronze.Case 1 : Rami wins gold. 4 choices for silver, 3 choices for bronze: 12
Case 2 : Rami wins silver. 4 choices for gold, 3 choices for bronze: 12
Case 3 : Rami wins bronze. 4 choices for gold, 3 choices for silver: 12
All together: 3(12) = 36 options.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists Without Repetition
Suppose there are 5 contestants in a proof-writing contest. You will give onegold medal, one silver medal, and one bronze medal to three differentcontestants. How many possible outcomes are there?
1
5 ways
2
4 ways
3
3 ways
5× 4× 3 = 60 possible outcomes.
Suppose Rami is one of the five contestants, and for sure he will get amedal.
Another alternate solution: there are 5× 4× 3 = 60 total ways for all 5contestants to win medals, and there are 4× 3× 2 = 26 ways that Rami doesnot win a medal. So there are 60− 24 = 36 ways that Rami does win amedal.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists With More Stipulations
Suppose your friend has a 7-digit phone number. You’re sure the phonenumber contains the number 3, but other than that, there are no restrictions.(For instance, the number might start with 0, or it might have lots of 3s.)How many 7-digit lists contain the number 3?
There are 107 total phone numbers, and 97 of them do not contain any 3s.So, the number that do contain at least one 3 is 107 − 97.
Suppose you’re sure your friend has a 7-digit phone number that contains thenumber 3 at least twice. How many 7-digit lists contain the number 3 atleast twice?
We need to subtract from the last answer the number of lists that containexactly one 3. If a list contains exactly one 3, then there are 7 places where itcan be. That leaves 6 places, each of which can be filled by one of 9 digits.So, there are 7× 96 lists that contain exactly one 3.
Therefore, the number of lists that contain at least two 3s is107 − 97 − 7 · 96.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists With More Stipulations
Suppose your friend has a 7-digit phone number. You’re sure the phonenumber contains the number 3, but other than that, there are no restrictions.(For instance, the number might start with 0, or it might have lots of 3s.)How many 7-digit lists contain the number 3?
There are 107 total phone numbers, and 97 of them do not contain any 3s.So, the number that do contain at least one 3 is 107 − 97.
Suppose you’re sure your friend has a 7-digit phone number that contains thenumber 3 at least twice. How many 7-digit lists contain the number 3 atleast twice?
We need to subtract from the last answer the number of lists that containexactly one 3. If a list contains exactly one 3, then there are 7 places where itcan be. That leaves 6 places, each of which can be filled by one of 9 digits.So, there are 7× 96 lists that contain exactly one 3.
Therefore, the number of lists that contain at least two 3s is107 − 97 − 7 · 96.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists With More Stipulations
Suppose your friend has a 7-digit phone number. You’re sure the phonenumber contains the number 3, but other than that, there are no restrictions.(For instance, the number might start with 0, or it might have lots of 3s.)How many 7-digit lists contain the number 3?
There are 107 total phone numbers, and 97 of them do not contain any 3s.So, the number that do contain at least one 3 is 107 − 97.
Suppose you’re sure your friend has a 7-digit phone number that contains thenumber 3 at least twice. How many 7-digit lists contain the number 3 atleast twice?
We need to subtract from the last answer the number of lists that containexactly one 3. If a list contains exactly one 3, then there are 7 places where itcan be. That leaves 6 places, each of which can be filled by one of 9 digits.So, there are 7× 96 lists that contain exactly one 3.
Therefore, the number of lists that contain at least two 3s is107 − 97 − 7 · 96.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Lists With More Stipulations
Suppose your friend has a 7-digit phone number. You’re sure the phonenumber contains the number 3, but other than that, there are no restrictions.(For instance, the number might start with 0, or it might have lots of 3s.)How many 7-digit lists contain the number 3?
There are 107 total phone numbers, and 97 of them do not contain any 3s.So, the number that do contain at least one 3 is 107 − 97.
Suppose you’re sure your friend has a 7-digit phone number that contains thenumber 3 at least twice. How many 7-digit lists contain the number 3 atleast twice?
We need to subtract from the last answer the number of lists that containexactly one 3. If a list contains exactly one 3, then there are 7 places where itcan be. That leaves 6 places, each of which can be filled by one of 9 digits.So, there are 7× 96 lists that contain exactly one 3.
Therefore, the number of lists that contain at least two 3s is107 − 97 − 7 · 96.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Practice!
You choose an outfit that consists of a shirt, a pair of pants, a pair of shoes,and a hat. If you have 5 shirts, 2 pairs of pants, 3 pairs of shoes, and 7 hats,how many outfits can you make?
5× 2× 3× 7 = 210
Postal codes have the form (letter, number, letter, number, letter, number).How many postal codes are possible?
263 × 103 = 17 576 000 (same as license plates!)Area of Canada: about 10 million square km
How many postal codes contain exactly two 0s?
263 × 9× 3 = 474 552
How many postal codes contain two 0s and one 1?
263 × 3 = 52 728
How many postal codes have no repeated letters or numbers at all?
26× 25× 24× 10× 9× 8 = 11 232 000
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
More Practice!
In a certain family, there are 7 possibilities for the first and middle name of ababy, and 3 possibilities for a surname. No baby has the same first and middlename. How many possible names are there?
7× 6× 3 = 126
A more realistic phone number does not start with 0 or 1, and does not startwith 555. How many such 7-digit phone numbers are there?
8 · 106 − 104
Of the phone numbers above, how many contain exactly one 0?
8 · 6 · 95 − 4 · 93
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
More Practice!
In a certain family, there are 7 possibilities for the first and middle name of ababy, and 3 possibilities for a surname. No baby has the same first and middlename. How many possible names are there?
7× 6× 3 = 126
A more realistic phone number does not start with 0 or 1, and does not startwith 555. How many such 7-digit phone numbers are there?
8 · 106 − 104
Of the phone numbers above, how many contain exactly one 0?
8 · 6 · 95 − 4 · 93
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
More Practice!
In a certain family, there are 7 possibilities for the first and middle name of ababy, and 3 possibilities for a surname. No baby has the same first and middlename. How many possible names are there?
7× 6× 3 = 126
A more realistic phone number does not start with 0 or 1, and does not startwith 555. How many such 7-digit phone numbers are there?
8 · 106 − 104
Of the phone numbers above, how many contain exactly one 0?
8 · 6 · 95 − 4 · 93
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
More Practice!
In a certain family, there are 7 possibilities for the first and middle name of ababy, and 3 possibilities for a surname. No baby has the same first and middlename. How many possible names are there?
7× 6× 3 = 126
A more realistic phone number does not start with 0 or 1, and does not startwith 555. How many such 7-digit phone numbers are there?
8 · 106 − 104
Of the phone numbers above, how many contain exactly one 0?
8 · 6 · 95 − 4 · 93
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
Given a finite set A, a permutation of the elements of A is a list that uses
each element of A precisely once.
Example: If A = {1, 2, 3}, the permutations of A are:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
If n ≥ 1, then the number of permutations of n elements isn(n − 1)(n − 2) · · · (1).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
Given a finite set A, a permutation of the elements of A is a list that uses
each element of A precisely once.
Example: If A = {1, 2, 3}, the permutations of A are:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
If n ≥ 1, then the number of permutations of n elements isn(n − 1)(n − 2) · · · (1).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
Given a finite set A, a permutation of the elements of A is a list that uses
each element of A precisely once.
Example: If A = {1, 2, 3}, the permutations of A are:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
If n ≥ 1, then the number of permutations of n elements is
n(n − 1)(n − 2) · · · (1).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
Given a finite set A, a permutation of the elements of A is a list that uses
each element of A precisely once.
Example: If A = {1, 2, 3}, the permutations of A are:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
If n ≥ 1, then the number of permutations of n elements isn(n − 1)(n − 2) · · · (1).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
If n is a non-negative integer, then the factorial of n, denoted n!, is thenumber of permutations of n elements. That is, the factorial of n is thenumber of non-repetitive lists of length n that can be made from nsymbols.
We already saw that, when n ∈ N, n! = n(n − 1)(n − 2) · · · (1).What should 0! be?The only list containing 0 elements is the empty list. So the number of ways
to permute 0 elements is 1, so 0! = 1 .
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
If n is a non-negative integer, then the factorial of n, denoted n!, is thenumber of permutations of n elements. That is, the factorial of n is thenumber of non-repetitive lists of length n that can be made from nsymbols.
We already saw that, when n ∈ N, n! = n(n − 1)(n − 2) · · · (1).
What should 0! be?The only list containing 0 elements is the empty list. So the number of ways
to permute 0 elements is 1, so 0! = 1 .
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
If n is a non-negative integer, then the factorial of n, denoted n!, is thenumber of permutations of n elements. That is, the factorial of n is thenumber of non-repetitive lists of length n that can be made from nsymbols.
We already saw that, when n ∈ N, n! = n(n − 1)(n − 2) · · · (1).What should 0! be?
The only list containing 0 elements is the empty list. So the number of ways
to permute 0 elements is 1, so 0! = 1 .
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Definition
If n is a non-negative integer, then the factorial of n, denoted n!, is thenumber of permutations of n elements. That is, the factorial of n is thenumber of non-repetitive lists of length n that can be made from nsymbols.
We already saw that, when n ∈ N, n! = n(n − 1)(n − 2) · · · (1).What should 0! be?The only list containing 0 elements is the empty list. So the number of ways
to permute 0 elements is 1, so 0! = 1 .
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Repetition Repetition
Suppose you want to make a list of length 7 from the elements{1, 2, 3, 4, 5, 6, 7}.
How many lists are there if repetitions are allowed?
How many lists are there if repetitions are not allowed?
How many lists are there if repetitions are not allowed, and the firstthree entries must be odd?
How many lists are there if repetitions are allowed, and in fact the listdoes contain at least one repeated number?
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Repetition Repetition Repetition
Suppose you want to make a list of length 5 from the elements{1, 2, 3, 4, 5, 6, 7}.
How many lists are there if repetitions are allowed?
How many lists are there if repetitions are not allowed?
How many lists are there if repetitions are not allowed, and the firstthree entries must be odd?
How many lists are there if repetitions are allowed, and in fact the listdoes contain at least one repeated number?
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0?
1 (empty set)
How many subsets of A have size 1?
5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0?
1 (empty set)
How many subsets of A have size 1?
5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1?
5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1?
5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5?
1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5? 1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4?
5 (missing one element each)
How many subsets of A have size 5? 1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4? 5 (missing one element each)
How many subsets of A have size 5? 1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2?
harder question
How many subsets of A have size 3?
harder question
How many subsets of A have size 4? 5 (missing one element each)
How many subsets of A have size 5? 1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Combinations
A = {1, 2, 3, 4, 5}
How many subsets of A have size 0? 1 (empty set)
How many subsets of A have size 1? 5 (single-element sets)
How many subsets of A have size 2? harder question
How many subsets of A have size 3? harder question
How many subsets of A have size 4? 5 (missing one element each)
How many subsets of A have size 5? 1 (A)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have only two elements?
Choice 1: 5 ways
Choice 2: 5× 4 ways
1
{1, 2}
{1, 3}
{1, 4}
{1, 5}
2
{2, 1}
{2, 3}
{2, 4}
{2, 5}
3
{3, 1}
{3, 2}
{3, 4}
{3, 5}
4
{4, 1}
{4, 2}
{4, 3}
{4, 5}
5
{5, 1}
{5, 2}
{5, 3}
{5, 4}
Number of subsets:4× 5
2= 10
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
Let f (X ) = X be a function from subsets of A of size 2 to subsets of A of size3.Since f is a bijection, there are 10 subsets of A with three elements.
{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 2}{1, 3, 4}{1, 3, 5}{1, 4, 2}{1, 4, 3}{1, 4, 5}{1, 5, 2}{1, 5, 3}{1, 5, 4}
{2, 1, 3}{2, 1, 4}{2, 1, 5}{2, 3, 1}{2, 3, 4}{2, 3, 5}{2, 4, 1}{2, 4, 3}{2, 4, 5}{2, 5, 1}{2, 5, 3}{2, 5, 4}
{3, 1, 2}{3, 1, 4}{3, 1, 5}{3, 2, 1}{3, 2, 4}{3, 2, 5}{3, 4, 1}{3, 4, 2}{3, 4, 5}{3, 5, 1}{3, 5, 2}{3, 5, 4}
{4, 1, 2}{4, 1, 3}{4, 1, 5}{4, 2, 1}{4, 2, 3}{4, 2, 5}{4, 3, 1}{4, 3, 2}{4, 3, 5}{4, 5, 1}{4, 5, 2}{4, 5, 3}
{5, 1, 2}{5, 1, 3}{5, 1, 4}{5, 2, 1}{5, 2, 3}{5, 2, 4}{5, 3, 1}{5, 3, 2}{5, 3, 4}{5, 4, 1}{5, 4, 2}{5, 4, 3}
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets
How many subsets of A = {1, 2, 3, 4, 5} have three elements?
There are 5× 4× 3 lists of length 3 with elements in A
Given three elements of A, there are 3! = 6 ways to permute them.
Since every subset makes 6 different lists, the number of subsets of length3 is:
5× 4× 3
3× 2× 1= 10
In general, suppose we want have a finite set of cardinality n, and we want toknow how many subsets of that set have cardinality k (for somek < n).
There are n(n − 1)(n − 2) · · · (n − k + 1) =n!
(n − k)!lists of length k
Given k elements in a list, there are k! ways to permute them.
Since every subset makes k! different lists, the number of subsets oflength k is:
n(n − 1)(n − 2) · · · (n − k + 1)
k!=
n!
(n − k)!k!
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
We write: (n
k
)=
n!
k!(n − k)!
This is sometimes called the binomial coefficient, or “n choose k”. It is thenumber of subsets of size k from a set of size n, when n ≥ k.
A pizza shop offers 10 different toppings. How many ways are there toorder a pizza with between one and three toppings?
(101
)+(102
)+(103
)= 10 + 45 + 120 = 175
A multiple-choice question asks you to select the two correct answersfrom five choices. How many ways are there to choose two of theanswers? What are the odds that you guess correctly? What are the oddsthat you guess one correctly but not the other?
(52
)= 10 ways to choose two answers, so your odds of guessing correctly
are 1 in 10, or 10%.There are two ways to choose a correct answer, and 3 ways to choose anincorrect answer, so your odds of guessing one correctly but not the otherare 6 in 10, or 60%.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
We write: (n
k
)=
n!
k!(n − k)!
This is sometimes called the binomial coefficient, or “n choose k”. It is thenumber of subsets of size k from a set of size n, when n ≥ k.
A pizza shop offers 10 different toppings. How many ways are there toorder a pizza with between one and three toppings?
(101
)+(102
)+(103
)= 10 + 45 + 120 = 175
A multiple-choice question asks you to select the two correct answersfrom five choices. How many ways are there to choose two of theanswers? What are the odds that you guess correctly? What are the oddsthat you guess one correctly but not the other?
(52
)= 10 ways to choose two answers, so your odds of guessing correctly
are 1 in 10, or 10%.There are two ways to choose a correct answer, and 3 ways to choose anincorrect answer, so your odds of guessing one correctly but not the otherare 6 in 10, or 60%.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
We write: (n
k
)=
n!
k!(n − k)!
This is sometimes called the binomial coefficient, or “n choose k”. It is thenumber of subsets of size k from a set of size n, when n ≥ k.
A pizza shop offers 10 different toppings. How many ways are there toorder a pizza with between one and three toppings?(101
)+(102
)+(103
)= 10 + 45 + 120 = 175
A multiple-choice question asks you to select the two correct answersfrom five choices. How many ways are there to choose two of theanswers? What are the odds that you guess correctly? What are the oddsthat you guess one correctly but not the other?
(52
)= 10 ways to choose two answers, so your odds of guessing correctly
are 1 in 10, or 10%.There are two ways to choose a correct answer, and 3 ways to choose anincorrect answer, so your odds of guessing one correctly but not the otherare 6 in 10, or 60%.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
We write: (n
k
)=
n!
k!(n − k)!
This is sometimes called the binomial coefficient, or “n choose k”. It is thenumber of subsets of size k from a set of size n, when n ≥ k.
A pizza shop offers 10 different toppings. How many ways are there toorder a pizza with between one and three toppings?(101
)+(102
)+(103
)= 10 + 45 + 120 = 175
A multiple-choice question asks you to select the two correct answersfrom five choices. How many ways are there to choose two of theanswers? What are the odds that you guess correctly? What are the oddsthat you guess one correctly but not the other?(52
)= 10 ways to choose two answers, so your odds of guessing correctly
are 1 in 10, or 10%.There are two ways to choose a correct answer, and 3 ways to choose anincorrect answer, so your odds of guessing one correctly but not the otherare 6 in 10, or 60%.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways are there to have a hand of 5 cards, out of 52 cardstotal?
(525
)= 2 598 960
A “full house” is a 5-card hand consisting of two cards that have thesame face value, and three cards that have another face value. Forexample: two sevens and three queens. How many ways are there tomake a full house?
13 choices for the double, then 12 choices for the triple. In each the 13choices for the double, there are
(42
)= 6 ways to choose the two cards; in
each of the 12 choices for the triple, there are(43
)= 4 ways to choose the
three cards. All together: 13(42
)+ 12
(43
)= 126 different full houses.
What are your odds of being dealt a full house?
126
2 598 960≈ 1
20 627
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways are there to have a hand of 5 cards, out of 52 cardstotal?(525
)= 2 598 960
A “full house” is a 5-card hand consisting of two cards that have thesame face value, and three cards that have another face value. Forexample: two sevens and three queens. How many ways are there tomake a full house?
13 choices for the double, then 12 choices for the triple. In each the 13choices for the double, there are
(42
)= 6 ways to choose the two cards; in
each of the 12 choices for the triple, there are(43
)= 4 ways to choose the
three cards. All together: 13(42
)+ 12
(43
)= 126 different full houses.
What are your odds of being dealt a full house?
126
2 598 960≈ 1
20 627
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways are there to have a hand of 5 cards, out of 52 cardstotal?(525
)= 2 598 960
A “full house” is a 5-card hand consisting of two cards that have thesame face value, and three cards that have another face value. Forexample: two sevens and three queens. How many ways are there tomake a full house?
13 choices for the double, then 12 choices for the triple. In each the 13choices for the double, there are
(42
)= 6 ways to choose the two cards; in
each of the 12 choices for the triple, there are(43
)= 4 ways to choose the
three cards. All together: 13(42
)+ 12
(43
)= 126 different full houses.
What are your odds of being dealt a full house?
126
2 598 960≈ 1
20 627
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways are there to have a hand of 5 cards, out of 52 cardstotal?(525
)= 2 598 960
A “full house” is a 5-card hand consisting of two cards that have thesame face value, and three cards that have another face value. Forexample: two sevens and three queens. How many ways are there tomake a full house?
13 choices for the double, then 12 choices for the triple. In each the 13choices for the double, there are
(42
)= 6 ways to choose the two cards; in
each of the 12 choices for the triple, there are(43
)= 4 ways to choose the
three cards. All together: 13(42
)+ 12
(43
)= 126 different full houses.
What are your odds of being dealt a full house?
126
2 598 960≈ 1
20 627
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways can 22 people be split into two groups of 11?
It’s a little ambiguous, but either(2211
)or 1
2
(2211
). Once you choose the first
team, the other is automatic. But, we overcount by two if the groups arenot differentiated.
How many ways can 22 people be divided into two teams of 11, one redteam and one blue team, where each team has a captain?
(2211
)· 112: 11 choices for each captain.
Alternately: 22(21)(2010
)
How many 8-digit passwords, with only lower-case letters and no repeats,are in alphabetical order?
(268
): first choose the letters, then put them in order.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways can 22 people be split into two groups of 11?
It’s a little ambiguous, but either(2211
)or 1
2
(2211
). Once you choose the first
team, the other is automatic. But, we overcount by two if the groups arenot differentiated.
How many ways can 22 people be divided into two teams of 11, one redteam and one blue team, where each team has a captain?
(2211
)· 112: 11 choices for each captain.
Alternately: 22(21)(2010
)
How many 8-digit passwords, with only lower-case letters and no repeats,are in alphabetical order?
(268
): first choose the letters, then put them in order.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways can 22 people be split into two groups of 11?
It’s a little ambiguous, but either(2211
)or 1
2
(2211
). Once you choose the first
team, the other is automatic. But, we overcount by two if the groups arenot differentiated.
How many ways can 22 people be divided into two teams of 11, one redteam and one blue team, where each team has a captain?(2211
)· 112: 11 choices for each captain.
Alternately: 22(21)(2010
)How many 8-digit passwords, with only lower-case letters and no repeats,are in alphabetical order?
(268
): first choose the letters, then put them in order.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Coefficient
How many ways can 22 people be split into two groups of 11?
It’s a little ambiguous, but either(2211
)or 1
2
(2211
). Once you choose the first
team, the other is automatic. But, we overcount by two if the groups arenot differentiated.
How many ways can 22 people be divided into two teams of 11, one redteam and one blue team, where each team has a captain?(2211
)· 112: 11 choices for each captain.
Alternately: 22(21)(2010
)How many 8-digit passwords, with only lower-case letters and no repeats,are in alphabetical order?(268
): first choose the letters, then put them in order.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Choose: 5 numbers from 1 to 49; one “grand” number from 1 to 7.
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:
(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers:
1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:
(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers:
1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:
(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:
(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:(5k
)·(
445−k
)
Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)
Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)
Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:
6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)
Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:
6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:
6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 1 to 49:(495
)= 1 906 884.
Number of ways to choose the 5 winning numbers: 1.
Number of ways to choose 5 numbers such that k of them match thewinning numbers:(5k
)·(
445−k
)Odds of choosing k of the 5 winning numbers:
(5k
)·(
445−k
)(495
)Odds of choosing k of the 5 winning numbers,
and also choosing the grand number:
(5k
)·(
445−k
)7 ·(495
)Odds of choosing k of the 5 winning numbers,
and not choosing the grand number:6 ·(5k
)·(
445−k
)7 ·(495
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game
Number of ways to choose 5 numbers from 49 possibilities:(495
)= 1 906 884.
Number of ways to choose 1 number from 7 possibilities:(71
)= 7.
result prize odds5/5+1/1 $7 000 000 1
(495 )· 17
= 113 348 188
5/5+0/1 $50 000 1
(495 )· 67
= 12 224 698
4/5+1/1 $1 000(54)·(
441 )
(495 )· 17≈ 1
60 674
4/5+0/1 $500(54)·(
441 )
(495 )· 67≈ 1
10 112
3/5+1/1 $100(53)·(
442 )
(495 )· 17≈ 1
1411
3/5+0/1 $20(53)·(
442 )
(495 )· 67≈ 1
235
2/5+1/1 $10(52)·(
443 )
(495 )· 17≈ 1
101
1/5+1/1 $4(51)·(
444 )
(495 )· 17≈ 1
20
0/5+1/1 free play(445 )(495 )· 17≈ 1
12
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Daily Grand Game More Info
http://corporate.bclc.com/content/dam/bclc/universal/
universal-documents/rules-and-regulations/lotto/
daily-grand-game-conditions.pdf
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Let A be a set of finite cardinality n.
The number of subsets of A is 2n.
The number of subsets of A is(n0
)+(n1
)+(n2
)+ · · ·+
(nn
).
Theorem:n∑
k=0
(n
k
)= 2n.
Suppose 0 ∈ A. How many subsets of A contain 0?(n−1k−1
): we choose k − 1 elements other than 0, then add in 0.(
nk
)−(n−1k
): we throw out the subsets that do not contain 0.
Theorem:
(n
k
)=
(n
k − 1
)+
(n − 2
k − 1
).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Let A be a set of finite cardinality n.
The number of subsets of A is 2n.
The number of subsets of A is(n0
)+(n1
)+(n2
)+ · · ·+
(nn
).
Theorem:n∑
k=0
(n
k
)= 2n.
Suppose 0 ∈ A. How many subsets of A contain 0?(n−1k−1
): we choose k − 1 elements other than 0, then add in 0.(
nk
)−(n−1k
): we throw out the subsets that do not contain 0.
Theorem:
(n
k
)=
(n
k − 1
)+
(n − 2
k − 1
).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Let A be a set of finite cardinality n.
The number of subsets of A is 2n.
The number of subsets of A is(n0
)+(n1
)+(n2
)+ · · ·+
(nn
).
Theorem:n∑
k=0
(n
k
)= 2n.
Suppose 0 ∈ A. How many subsets of A contain 0?
(n−1k−1
): we choose k − 1 elements other than 0, then add in 0.(
nk
)−(n−1k
): we throw out the subsets that do not contain 0.
Theorem:
(n
k
)=
(n
k − 1
)+
(n − 2
k − 1
).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Let A be a set of finite cardinality n.
The number of subsets of A is 2n.
The number of subsets of A is(n0
)+(n1
)+(n2
)+ · · ·+
(nn
).
Theorem:n∑
k=0
(n
k
)= 2n.
Suppose 0 ∈ A. How many subsets of A contain 0?(n−1k−1
): we choose k − 1 elements other than 0, then add in 0.(
nk
)−(n−1k
): we throw out the subsets that do not contain 0.
Theorem:
(n
k
)=
(n
k − 1
)+
(n − 2
k − 1
).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Let A be a set of finite cardinality n.
The number of subsets of A is 2n.
The number of subsets of A is(n0
)+(n1
)+(n2
)+ · · ·+
(nn
).
Theorem:n∑
k=0
(n
k
)= 2n.
Suppose 0 ∈ A. How many subsets of A contain 0?(n−1k−1
): we choose k − 1 elements other than 0, then add in 0.(
nk
)−(n−1k
): we throw out the subsets that do not contain 0.
Theorem:
(n
k
)=
(n
k − 1
)+
(n − 2
k − 1
).
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Theorem:
(n
k
)=
(n
n − k
).
Combinatorial proof: Let A be a set of size n. Choosing k elements from A isthe same as choosing n − k elements to exclude from A.
Algebraic proof:(n
n − k
)=
n!
(n − (n − k))!(n − k)!=
n!
k!(n − k)!=
(n
k
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Counting Subsets: Combinatorial Proofs
Theorem:
(n
k
)=
(n
n − k
).
Combinatorial proof: Let A be a set of size n. Choosing k elements from A isthe same as choosing n − k elements to exclude from A.
Algebraic proof:(n
n − k
)=
n!
(n − (n − k))!(n − k)!=
n!
k!(n − k)!=
(n
k
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Theorem
Binomial Theorem: (from homework)For any n ∈ N,
(x + y)n =n∑
k=0
(n
k
)xkyn−k
Theorem: (from earlier)(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Binomial Theorem
Binomial Theorem: (from homework)For any n ∈ N,
(x + y)n =n∑
k=0
(n
k
)xkyn−k
Theorem: (from earlier)(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
)
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3
3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4
6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6
4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5
10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10
10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10
5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
5 10 10 5
(n
k
)=
(n − 1
k − 1
)+
(n − 1
k
) (n
0
)=
(n
n
)= 1
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)= 1
(10
)= 1
(11
)= 1
(20
)= 1
(21
)= 2
(22
)= 1
(30
)= 1
(31
)= 3
(32
)= 3
(33
)= 1
(40
)= 1
(41
)= 4
(42
)= 6
(43
)= 4
(44
)= 1
(50
)= 1
(51
)= 5
(52
)= 10
(53
)= 10
(54
)= 5
(55
)= 1
Expand (2x + 5)5:
(2x + 7)5 = (2x)5 + 5(2x)4(7) + 10(2x)3(72) + 10(2x)2(73) + 5(2x)1(74) + 75
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)= 1
(10
)= 1
(11
)= 1
(20
)= 1
(21
)= 2
(22
)= 1
(30
)= 1
(31
)= 3
(32
)= 3
(33
)= 1
(40
)= 1
(41
)= 4
(42
)= 6
(43
)= 4
(44
)= 1
(50
)= 1
(51
)= 5
(52
)= 10
(53
)= 10
(54
)= 5
(55
)= 1
Expand (2x + 5)5:
(2x + 7)5 = (2x)5 + 5(2x)4(7) + 10(2x)3(72) + 10(2x)2(73) + 5(2x)1(74) + 75
3 Counting
13.1CountingLists
13.2Factorials
13.3CountingSubsets
13.4Pascal’sTriangleand theBinomialTheorem
Pascal’s Triangle
(00
)= 1
(10
)= 1
(11
)= 1
(20
)= 1
(21
)= 2
(22
)= 1
(30
)= 1
(31
)= 3
(32
)= 3
(33
)= 1
(40
)= 1
(41
)= 4
(42
)= 6
(43
)= 4
(44
)= 1
(50
)= 1
(51
)= 5
(52
)= 10
(53
)= 10
(54
)= 5
(55
)= 1
Expand (2x + 5)5:
(2x + 7)5 = (2x)5 + 5(2x)4(7) + 10(2x)3(72) + 10(2x)2(73) + 5(2x)1(74) + 75
