Aim: Permutations Course: Math Lit.. Aim: Is Spiderman a permutation or what? A = {a, b, c, d, e}...

Aim: Permutations Course: Math Lit..

Aim: Is Spiderman a permutation or what?

A = {a, b, c, d, e}

What could be a two element subset of A?

{a, b}Is = {b, a} ?

Yes order of elements in set is not important


Subsets & Arrangements

If order were important

{a, b}is = {b, a} ? No

A = {a, b, c, d, e}

If the two elements a and b are selected from A, then

there is one subset (order not important): {a, b}

there are two arrangements (order important): (a, b) and (b, a)

note: both are selected without repetitions (without replacement)


Model Problem

Consider selecting the elements a, b, and c from set A = {a, b, c, d, e}. List all possible subsets of these elements, as well as all possible arrangements.

1 subset: {a, b, c}

6 arrangements: (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a)

a

b

c

b c

c ba c

c aa b

b a


Counting Principle at Work

6 • 5 • 4 • 3 • 2 • 1 = 720

How many different arrangements can be made using all six names?

5 ways to

choose 2nd

name

4 way to

choose 3rd

name

6 ways to

choose 1st

name

2 ways to

choose 5th

name

1 way to

choose 6th

name

3 ways to

choose 4th

name

How many different arrangements can be made using 9 names?

9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880

w/o repetition

Tom, Dick, Harry, Mary Larry & Joe

Factorial: n! = n•(n - 1)(n - 2)(n - 3). . . . 1. When arranging all objects in the set


Counting Principle at Work

6 • 5 • 4 = 120

How many different arrangements using only 3 names can be made from six names?

5 ways to

choose 2nd

name

4 ways to

choose 3rd

name

6 ways to

choose 1st

name

How many different arrangements using 6 names from a selection of 9 names?

9 • 8 • 7 • 6 • 5 • 4 = 60480

Since our selection is limited

to only 3 names we stop after the

3rd selection

Since our selection is limited

to only 3 names we stop after the

3rd selection

How many three letter ‘words’ can be made from the word ‘STUDY’? 5 • 4 • 3 = 60

Tom, Dick, Harry, Mary Larry & Joe


Permutations

A permutation is an arrangement of objects in a specific order.

The number of permutation of n things taken n at a time is

nPn = n! = n(n – 1)(n – 2)(n – 3) . . . 3, 2, 1

The number of permutation of n things taken r at a time is

( 1)( 2)n rP n n n L

r factors

!

( )!

n

n r

Factorial: n! = n•(n - 1)(n - 2)(n - 3). . . . 1. When arranging all objects in the set:

Definition - 0! = 1.


Model Problems

How many three letter ‘words’ can be made from the letters of

CAT

GOAT

CLONE

5P5

8P6

6P1

= n! = 5 • 4 • 3 • 2 • 1 = 120

n! (n - r)!

= 8! (8 - 6)!

=

8 • 7 • 6 • 5 • 4 • 3 • 2 • 1

2 • 1

8! 2!

=

= 20160

6! 5!

= = 6

3P3 = 6

4P3 = 24

5P3 = 60


Model Problems

How many arrangements can be made from the word ANGLE?

5P5 = 120

How many arrangements can be made from the word ANGLE if repetition of letters is allowed?

5 • 5 • 5 • 5 • 5 = 55 = 3125

Repetition is key issueRepetition is key issue

How many 5-letter arrangements can be made from E, Q, B, X, R, T, L, A, and V if the last two letters must be vowels. no repetition

7 • 6 • 5 • 2 • 1 = 420


Repetitions

How many 3-letter arrangements can you make from the word TEA?

How many 3-letter arrangements can you make from the word TEE?

TEA EATATETAEETAAET

TEE EET ETE TEE ETE EET

Duplicates

6

Repeated letters alter number of arrangements possible because they lead to duplicated ‘words’.

Repeated letters alter number of arrangements possible because they lead to duplicated ‘words’.

3


Permutations with Repetitions

The number of permutations of n things, taken n at a time with r of these things identical, is given by

How many 3-letter arrangements can you make from the word TEE?

n = 3 letters to arrange

r = 2, the number of times E is repeated

3!

2!= 3

!

!

n

r !n nP

r


Permutations with Repetitions

The number of permutations of n things, taken n at a time with r things identical, s things identical, and t things identical, is given by

Find the number of permutations of the letters in MISSISSIPPI.

n = 11 letters to arrange

r = 4, the number of times I is repeated

n!

r!s! t!

s = 4, the number of times S is repeated

s = 2, the number of times P is repeated

11!

4!4!2! = 34,650


Model Problems

How many different 5-letter permutations are there is the letters in

APPLE

ADDED

VIVID

How many different 5-digit numerals can be written using all 5 digits listed?

1, 2, 3, 4, 5

1, 1, 2, 2, 2

n!

r!s! t!

5!

2! = 60

= 205!

3!

= 305!

2!2!

5! = 120

= 105!

3!2!


Model Problems

How many 5-letter arrangements can be made from E, Q, B, X, R, T, L, A, and V if the last two letters must be vowels. no repetition

Counting Principle E1 E2 E3 E4 E5

Permutations

9 letters altogether; A and E are only vowels; 7 consonants for 1st three events

7 6 5 2 1· · · · = 420

)!(

!

rn

nPrn

)!(

!

rn

nPrn

7 3P

210

2 2P

2

7!

(7 3)!2! = 420

·

·

·


Model Problems

Sea-going vessels use different flags or arrangements of flags to send messages to other vessels. If 10 different flags are available, and if every message consists of three different flags, how many different messages are possible? 10P3

How messages are possible if the top flag must be one of three specific flags? 3 • 9 • 8 = 216

How many different ways can John place a math book, a history book, a science book and an art book on a shelf? 4P4 = 4! = 24

How many different ways can John place the 4 books if the math book must be first?

1 • 3 • 2 = 6


Model Problems

Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation.

A. There are no restrictions

B. The first letter must be A

ONLY 1 CHOICE FOR FIRST LETTER - A

)!(

!

rn

nPrn

)!(

!

rn

nPrn

7 P4 7!

(7 4)!

7!

3!

7 6 5 4 3 2 1

3 2 1

7 6 5 4 840

AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3 PLACES OR 6P3

1 • = 1206P3


Model Problems

Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation.C. The third letter must be A

6 5 1 4• • • = 120

ONLY 1 CHOICE

FOR THIRD LETTER - A

AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3

PLACES OR 6P3

1 • 6P3 = 120


Model Problems

Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation.D.. The last two letters must be R or T in either order 5 4 2 1• • • = 40

ONLY 2 CHOICES FOR THIRD LETTER & FOURTH LETTER

AFTER THE R AND T ARE PLACED, THERE ARE 5 LETTERS

LEFT FOR 2 PLACES OR 5P2

5P2 = 402 1• •


Model Problem

Using the word SQUARE, find how many 6-letter arrangements with no repetitions are possible if the: A. First letter is S

1 5 4 3 2 1• • • • •

ONLY 1 CHOICE FOR FIRST LETTER - S

5P5

= 120AFTER THE S IS PLACED, THERE ARE 5 LETTERS LEFT FOR 5 PLACES OR 5P5


Model Problem

Using the word SQUARE, find how many 6-letter arrangements with no repetitions are possible if the: B. Vowels and consonants alternate beginning with a vowel

3 2 13 2 1• • • • • = 36

VOWELS - U, A, E CONSON. - S, Q R

ONLY 3 CHOICES

U,A,E

ONLY 3 CHOICES

S,Q,R

• = 363P33P3


Model Problems

AMDKDLK3EW

AMDKDLK3EW

AMDKDLK3EW

AMDKDLK3EW

AMDKDLK3EW

AMDKDLK3EW

AMDKDLK3EW

Four different biology books and three different chemistry books are to be placed on a shelf with the biology books together and to the right of the chemistry books. In how many ways can this be done?

2 1 4 3 2 1• • • • • = 1443 •

First three books are chemistry

Last four books are

biology

3P3 4P4

3P3 • 4P4 = 144


Model Problem

Frances has 8 tulip bulbs, 10 daffodil bulbs, and 28 crocus bulbs. In how many different ways can Frances plant these bulbs in a row in her garden?

n!

r!s! t!n = # bulbs = 46r = # of tulips = 8s = # of daffodils = 10

t = # of crocus = 28

46!

8!10!28!= 1.233. . . x 1017


Model Problems

A combination lock has 60 numbers around its dial. How many different arrangments are possible if every combination has three numbers?

60 60 60• • = 216,000

Question: can a number be used more than once?

A three digit number is formed by selecting from the digits 4, 5, 6, 7, 8, and 9 with no repetitions. How many of these 3-digit numbers will be greater than 700?

3 • = 60ONLY 3 CHOICES FOR FIRST DIGIT 7,8,9

5 4•

5P2


Model Problem

Find the total number of different twelve-letter arrangements that can be formed using the letters in the word PENNSYLVANIA.

Aim: Permutations Course: Math Lit.. Aim: Is Spiderman a permutation or what? A = {a, b, c, d, e}...

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