The Basics of Counting: Selected Exercises. Copyright © Peter Cappello2 Sum Rule Example There are...

The Basics of Counting: Selected Exercises

Copyright © Peter Cappello 2

Sum Rule Example

There are 3 sizes of pink shirts & 7 sizes of blue shirts.

How many types of shirts are there, if a shirt type is a

shirt of a particular color in a particular size?

Pink Blue


Sum Rule

A B = |A B| = |A| + |B|.

A B


Sum Rule Generalization

Let S1, S2, …, Sn be a partition of S.

Then, | S | = | S1 | + | S2 | + … + | Sn |.

When using the sum rule,

• Check 1: Have I partitioned S?

– Are the subsets pairwise disjoint?

– Is their union equal to S?

• Check 2: What equivalence relation corresponds to my partition?


Product Rule

Let A be a set of elements constructed in 2 stages.

• Stage 1 has n1 possible outcomes.

• Stage 2 has n2 possible outcomes.

Then, | A | = n1n2.


Product Rule Example

• A store sells pink shirts & blue shirts;

each comes in small, medium, & large.

• How many types of shirts are there?

A shirt type can be described as an ordered pair:

(color, size).


Product Rule: Counting Ordered Pairs

Let A be a set of objects that are constructed (described) in 2 stages.

• Let S be the set of values from stage 1

• Let T be the set of values from stage 2

Then, | A | = | S | x | T |.

An element of A can be described as an ordered pair (a, b),

where a S and b T.


Product Rule Example

How many sequences of 2 distinct letters are there from a, e, i, o, u ?

• There are 5 ways to select the 1st letter in the sequence.

• There are 4 ways to select the 2nd letter in the sequence.

The set of values in stage 2 depends on which letter was selected in stage 1.

The size of the set of values in stage 2 does not depend on which letter was chosen in stage 1.


Product Rule: Counting Ordered Pairs

• Let A be a set of objects constructed (described) in 2 stages.

• Let S be the set of values from stage 1.

• If a S is selected in stage 1, let Ta be the set of values for stage 2.

• Essentially, A = ( a, b ) | a S & b Ta .

• To use the product rule, for all a, b S, | Ta | = | Tb |.

(Illustrate S and Ta with previous examples)

The product rule is a special case of the sum rule: When

• S1, S2, …, Sn is a partition of A

• | Si | = | Sj |, for 1 ≤ i, j ≤ n

The sum rule reduces to the product rule: | S1 | + | S2 | + … + | Sn | = n| S1 |.


Exercise 10

How many bit strings are there of length 8?


Exercise 10

How many bit strings are there of length 8?

Use the product rule: Count the bit strings of length 8 by

decomposing the process into 8 stages:

count the possibilities for:

the 1st bit (2),

the 2nd bit (2),

…,

the 8th bit (2).

The product: 28 = 256 different bit strings.


Exercise 20

How many positive integers < 1000

1. Are divisible by 7?


Exercise 20



999/7 = 142.

2. Are divisible by 7 & 11?

(Use a Venn diagram)


Exercise 20



999/7 = 142.


999/(7.11) = 12.

3. Are divisible by 7 but not by 11?



Exercise 20



999/7 = 142.


999/(7.11) = 12.

3. Are divisible by 7 but not by 11?

1. Count the number divisible by 7;

2. Subtract the number divisible by 7 & 11;

999/7 - 999/(7.11) = 142 – 12 = 130.


Exercise 20 continued

4. Are divisible by 7 or 11?




4. Are divisible by 7 or 11?

We want property A or property B: use inclusion-exclusion:

1. Count the number that are divisible by 7;

2. Add the number that are divisible by 11;

3. Subtract the number that are divisible by both;

999/7 + 999/11 – 999/(77) = 142 + 90 – 12.

7 11



5. Are divisible by exactly one of 7 & 11?



5. Are divisible by exactly one of 7 or 11?

What region of the Venn diagram represents the answer?

Count the symmetric difference:

1. Count the number that are divisible by 7 or 11:

999/7 + 999/11 – 999/(77) = 142 + 90 – 12 = 220.

• Subtract the number that are divisible by both;

999/(7.11) = 12

7 11



6. Are divisible by neither 7 nor 11?




6. Are divisible by neither 7 nor 11?


1. Count the universe (999).

2. Subtract the number that is divisible by 7 or 11 (220)

giving 779.

7 11


Exercise 20 continue7. Have distinct digits?


Exercise 20 continue7. Have distinct digits?

Use the sum rule to decompose the problem into counting

1. The number of 1-digit numbers: 9

2. The number of 2-digit numbers with distinct digits:

Use the product rule:

• Count the number of ways to select the 10s digit: 9

• Count the number of ways to select the unit digit: 9

• There are 9 . 9 = 81 distinct 2-digit numbers.

• The number of 3-digit numbers with distinct digits:


1. Count the number of ways to select the 100s digit: 9

2. Count the number of ways to select the 10s digit: 9

3. Count the number of ways to select the unit digit: 8

4. There are 9 . 9 . 8 = 648 distinct 3-digit numbers.

The overall answer is 9 + 81 + 648 = 738.

Alternate approach

– Make a 3-level tree of 3-digit numbers

• Top level (100s digit): branch: 0 vs. !0

• Middle level (10s digit): branch: 0 vs. !0

• Bottom level (1s digit): branch: 0 vs. !0

– Add the solutions for the branches representing 3-digit numbers with distinct digits.


1. 000: Invalid

2. 00X: 9

3. 0X0: 9

4. 0XY: 9 x 8 = 72

5. X00: Invalid

6. X0Y: 9 x 8 = 72

7. XY0: 9 x 8 = 72

8. XYZ: 9 x 8 x 7 = 504

Sum = 738Copyright © Peter Cappello 25

Sara’s approach

Count complement set; subtract from 999.

Sum rule:• Exactly 2 digits the same:

– 2-digit numbers: 9– 3-digit numbers:

» No “0”: XYY | YXY | YYX: 9 x 8 x 3» 1 “0”: X0X | XX0: 9 x 2» 2 “0”: X00: 9

• Exactly 3 digits the same: XXX: 9

Sum: 9 + 216 + 18 + 9 + 9 = 261; 999 – 261 = 738



Exercise 30

How many strings of 8 English letters are there:

a) If letters can be repeated?


Exercise 30



Use product rule: (26)8

b) If no letter can be repeated?


Exercise 30





Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19

c) That start with X, if letters can be repeated?


Exercise 30





Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19


Use product rule: 1 . (26)7

d) That start with X, if no letter can be repeated?


Exercise 30





Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19




Use product rule: 1 . 25 . 24 . 23 . 22 . 21 . 20 . 19

e) That start & end with X, if letters can be repeated?


Exercise 30





Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19




Use product rule: 1 . 25 . 24 . 23 . 22 . 21 . 20 . 19

e) That start & end with X, if letters can be repeated?

Use product rule: 1 . 1 . (26)6



f) That start with the letters BO, if letters can be repeated?





g) That start & end with the letters BO, if letters can be repeated?





g) That start & end with the letters BO, if letters can be repeated?

Use product rule: 1 . 1 . 1 . 1 . (26)4

h) That start or end with the letters BO, if letters can be repeated?

Copyright © Peter Cappello 2011 36

h) That start or end with the letters BO, if letters can be repeated?

Use inclusion-exclusion:• That start with the letters BO, if letters can be

repeated: (26)6

• That end with the letters BO, if letters can be repeated: (26)6

• Subtract those that start and end with the letters BO, if letters can be repeated: (26)4

Overall answer: 2 . (26)6 - (26)4


Endw/ BO

Startw/ BO


Exercise 40

How many ways can a wedding photographer arrange 6

people in a row from a group of 10, where the bride &

groom are among these 10, if

a) The bride is in the picture?


Exercise 40

How many ways can a wedding photographer arrange 6

people in a row from a group of 10, where the bride &

groom are among these 10, if

a) The bride is in the picture?


a) Pick the position of the bride: 6

b) Place the remaining 5 people from left to right in the

remaining positions (use the product rule to do this):

9 . 8. 7 . 6 . 5

Overall answer is 6 . 9 . 8. 7 . 6 . 5.



b) Both the bride & groom are in the picture?



b) Both the bride & groom are in the picture?


a) Pick the position for the bride: 6

b) Pick the position for the groom: 5

c) Place the remaining 4 people from the remaining 8 in

the remaining 4 slots, from left to right: 8 . 7 . 6 . 5.

The overall answer is 6 . 5 . 8 . 7 . 6 . 5.



c) Exactly 1 of the bride & groom is in the picture?


40 continued


1. Pick either the bride or the groom: 2.

2. Place that person in the picture: 6.

3. Place remaining 5 from remaining 8 people:

P(8, 5) = 8 . 7 . 6 . 5 . 4.

The overall answer is 2 . 6 . 960 = 80,640.


Exercise 50

– A variable name in C can have uppercase & lowercase letters,

digits, or underscores.

– The name’s 1st character is a letter (uppercase or lowercase), or

an underscore.

– The name of a variable is determined by its 1st 8 characters.

How many different variables can be named in C?


Exercise 50

A variable name in C can have uppercase & lowercase letters, digits,

or underscores.

The name’s 1st character is a letter (uppercase or lowercase), or an

underscore.

The name of a variable is determined by its 1st 8 characters.

How many different variables can be named in C?

Use the sum rule to count the number of variable names of i characters,

for i = 1, 2, …, 8.

The overall answer is the sum of these numbers.



Use the product rule to count the # of names of a fixed size.

Let the name have i characters.

1. The # of ways to pick the 1st character is 2 . 26 + 1 = 53.

2. The # of ways to pick subsequent characters is 53 + 10.

The # of ways to pick the name is 53 . (63)i-1.

The overall answer is Σi=[1,8] 53 . (63)i-1 ≈ 2.1 x 1014.


End


Characters

. ≥ ≡ ~

≈

Ω Θ

Σ


Exercise 20 continue8. Have distinct digits and are even?


Exercise 20 continue8. Have distinct digits and are even?

Easier to:

1. count the number that have distinct digits (738)

2. subtract those that are odd:

• 1-digit: 5

• 2-digit: 40

1. 5 ways to pick the unit digit

2. 8 ways to pick the 10s digit (nonzero)

1. 3-digit: 320

1. 5 ways to pick the unit digit

2. 8 ways to pick the 100s digit (nonzero)

3. 8 ways to pick the 10s digit

The total that have distinct digits & are odd is 5 + 40 + 320 = 365.

The overall answer is 738 – 365 = 373.


20 continued

The hard way: Use the sum rule directly:

1. 1-digit: 4

2. 2-digit:

0 is not picked: 40 is picked: 1low-order digit:

high-order digit: 9 8

9 + 32 = 41


20 continued3. 3-digit:

0 is not picked: 40 is picked: 1low-order digit:

high-order digit: 9 8

72 + 256 = 328

88middle digit:

The overall answer is 4 + 41 + 328 = 373.


40 continued


1. There are 6 . 9 . 8 . 7 . 6 . 5 ways for the bride to be in the picture.

2. There are 6 . 5 . 8 . 7 . 6 . 5. ways for the bride and groom to be in the

picture.

3. The number of ways for the bride only to be in the picture is 6 .

9 . 8 . 7 . 6 . 5 - 6 . 5 . 8 . 7 . 6 . 5 = 6 . 8 . 7 . 6 . 5 (9 – 5) = 40,320.

4. There are the same number of ways for the groom only to be in the

picture (a 1-to-1 correspondence between bride-only & groom-only)

The overall answer is 2 . 40,320.


40 alternate answer for part c



1. Pick the bride or groom to be in the picture: 2.

2. Count the number of ways to fill out that picture.


• Place the bride/groom: 6

• Fill in the other positions from left to right: 8 . 7 . 6 . 5 . 4.

The overall answer is 2 . 6 . 8 . 7 . 6 . 5 . 4 = 80,640.

The Basics of Counting: Selected Exercises. Copyright © Peter Cappello2 Sum Rule Example There are...

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