The Basics of Counting:
Selected Exercises
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
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Sum Rule
A  B =   |A  B| = |A| + |B|.
A
B
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Sum Rule Generalization
Let { S1, S2, …, Sn } be a partition of S.
Then, | S | = | S1 | + | S2 | + … + | Sn |.
When using the sum rule,
1. Check 1: Have I partitioned S?
1. Are the subsets pairwise disjoint?
2. Is their union equal to S?
2. Check 2: What equivalence relation corresponds to my partition?
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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.
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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).
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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 & b  T.
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Product Rule Example
How many sequences of 2 distinct letters are there from
{ a, e, i, o, u } ?
1.
There are 5 ways to select the 1st letter in the
sequence.
2.
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.
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Product Rule: Counting Ordered Pairs
The product rule is a special case of the sum rule: When
1. { S1, S2, …, Sn } is a partition of A
2. | Si | = | Sj | = m, for 1 ≤ i, j ≤ n
Thus, | S1 | + | S2 | + … + | Sn | = n| S1 |.
The sum rule reduces to the product rule:
1.
Pick the subset (n);
2.
Pick the element in the subset (m)
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Exercise 10
How many bit strings are there of length 8?
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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.
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Exercise 20
How many positive integers < 1000
1. Are divisible by 7?
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Exercise 20
How many positive integers < 1000
1. Are divisible by 7?
└
999/7 ┘ = 142.
2. Are divisible by 7 & 11?
(Use a Venn diagram)
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Exercise 20
How many positive integers < 1000
1. Are divisible by 7?
└
999/7 ┘ = 142.
2. Are divisible by 7 & 11?
└
999/(7.11) ┘ = 12.
3. Are divisible by 7 but not by 11?
(Use a Venn diagram)
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Exercise 20
How many positive integers < 1000
1. Are divisible by 7?
└
999/7 ┘ = 142.
2. Are divisible by 7 & 11?
└
999/(7.11) ┘ = 12.
3. Are divisible by 7 but not by 11?
1. Count the # divisible by 7;
2. Subtract the # divisible by 7 & 11;
.11)
999/7
999/(7
└
┘ └
┘ = 142 – 12 = 130.
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Exercise 20 continued
4. Are divisible by 7 or 11?
(Use a Venn diagram)
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Exercise 20 continued
4. Are divisible by 7 or 11?
We want property A or property B: use inclusion-exclusion:
1. Count the # that are divisible by 7;
2. Add the # that are divisible by 11;
3. Subtract the # that are divisible by both;
└
999/7 ┘ + └ 999/11 ┘ – └ 999/(77) ┘ = 142 + 90 – 12.
7
11
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Exercise 20 continued
5. Are divisible by exactly one of 7 & 11?
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Exercise 20 continued
5. Are divisible by exactly one of 7 & 11?
What region of the Venn diagram represents the answer?
Count the symmetric difference: (union – intersection)
1. Count the # that are divisible by 7 or 11;
└
999/7 ┘ + └ 999/11 ┘ – └ 999/(7.11) ┘ = 142 + 90 – 12 =220.
2. Subtract the # that are divisible by both;
└
999/(7.11) ┘ = 12.
7
11
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Exercise 20 continued
6. Are divisible by neither 7 nor 11?
What region of the Venn diagram represents the answer?
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Exercise 20 continued
6. Are divisible by neither 7 nor 11?
What region of the Venn diagram represents the answer?
1. Count the universe (999).
2. Subtract the # that is divisible by 7 or 11 (220) giving
779.
7
11
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Generalize this heuristic
• Given a universe U and property, P(e).
• Let S = { e | e in U, P(e) }
• What is |S|?
• Always ask the question
“Is it easier to count S = U – S?
• If yes, then |S| = |U| - |S|.
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Exercise 20 continue
7. Have distinct digits?
Omit leading 0s.
For example 0 < 9 < 1000 and is composed of distinct digits.
That is 9 is NOT 009.
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Exercise 20 continue
7. Have distinct digits?
Use the sum rule to decompose the problem into counting
1. The # of 1-digit numbers: 9
2. The # of 2-digit numbers with distinct digits:
Use the product rule:
1. Count the # of ways to select the 10s digit: 9
2. Count the # of ways to select the unit digit: 9
3. There are 9 . 9 = 81 distinct 2-digit numbers.
3. The # of 3-digit numbers with distinct digits:
Use the product rule:
1. Count the # of ways to select the 100s digit: 9
2. Count the # of ways to select the 10s digit: 9
3. Count the # 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.
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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.
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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 = 738
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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
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
Use product rule: (26)8
b) If no letter can be repeated?
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
Use product rule: (26)8
b) If no letter can be repeated?
Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19
c) That start with X, if letters can be repeated?
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
Use product rule: (26)8
b) If no letter can be repeated?
Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19
c) That start with X, if letters can be repeated?
Use product rule: 1 . (26)7
d) That start with X, if no letter can be repeated?
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
Use product rule: (26)8
b) If no letter can be repeated?
Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19
c) That start with X, if letters can be repeated?
Use product rule: 1 . (26)7
d) That start with X, if no letter can be repeated?
Use product rule: 1 . 25 . 24 . 23 . 22 . 21 . 20 . 19
e) That start & end with X, if letters can be repeated?
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Exercise 30
How many strings of 8 English letters are there:
a) If letters can be repeated?
Use product rule: (26)8
b) If no letter can be repeated?
Use product rule: 26 . 25 . 24 . 23 . 22 . 21 . 20 . 19
c) That start with X, if letters can be repeated?
Use product rule: 1 . (26)7
d) That start with X, if no letter can be repeated?
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
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Exercise 30 continued
f) That start with the letters BO, if letters can be repeated?
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Exercise 30 continued
f) That start with the letters BO, if letters can be repeated?
Use product rule: 1 . 1 . (26)6
g) That start & end with the letters BO, if letters can be
repeated?
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Exercise 30 continued
f) That start with the letters BO, if letters can be repeated?
Use product rule: 1 . 1 . (26)6
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?
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Exercise 30 continued
h) That start or end with letters BO, if letters can be
repeated?
Use inclusion-exclusion:
a) That start with the letters BO, if letters can be
repeated: (26)6
b) That end with the letters BO, if letters can be repeated:
(26)6
c) Subtract those that start and end with the letters BO, if
letters can be repeated: (26)4
Overall answer: 2 . (26)6 - (26)4
Start
w/ BO
End
w/ BO
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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?
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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?
Use the product rule:
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.
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Exercise 40 continued
b) Both the bride & groom are in the picture?
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Exercise 40 continued
b) Both the bride & groom are in the picture?
Use the product rule:
a) Pick the bride’s position: 6
b) Pick the groom’s position: 5
c) Place 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.
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Exercise 40 continued
c) Exactly 1 of the bride & groom is in the picture?
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40 continued
c) Exactly 1 of the bride & groom is in the picture?
(symmetric difference of what?)
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: 2 . 6 . ( 8 . 7 . 6 . 5 . 4) = 80,640.
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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?
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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 # of variable names of i characters,
for i = 1, 2, …, 8.
The overall answer is the sum of these numbers.
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Exercise 50 continued
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.
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End
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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 and 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
1.
{ S1, S2, …, Sn } is a partition of A
2.
| Si | = | Sj |, for 1 ≤ i, j ≤ n
The sum rule reduces to the product rule: | S1 | + | S2 | + … + | Sn | = n| S1 |.
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Characters
• .≥≡~┌
┐
└ ┘
• ≈
• 
•  Ω Θ
• Σ
• 
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Exercise 20 continue
8. Have distinct digits and are even?
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Exercise 20 continue
8. Have distinct digits and are even?
Easier to:
1.
count the number that have distinct digits (738)
2.
subtract those that are odd:
1. 1-digit: 5
2. 2-digit: 40
1. 5 ways to pick the unit digit
2. 8 ways to pick the 10s digit (nonzero)
3. 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.
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20 continued
The hard way: Use the sum rule directly:
1. 1-digit: 4
2. 2-digit:
low-order digit: 0 is picked: 1
high-order digit:
0 is not picked: 4
8
9
9
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+
32 = 41
52
20 continued
3. 3-digit:
low-order digit:
0 is picked: 1
0 is not picked: 4
high-order digit:
9
8
middle digit:
8
8
72 + 256 = 328
The overall answer is 4 + 41 + 328 = 373.
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40 continued
c) Exactly 1 of the bride & groom is in the picture?
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.
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40 alternate answer for part c
c) Exactly 1 of the bride & groom is in the picture?
Use the product rule:
1. Pick the bride or groom to be in the picture: 2.
2. Count the number of ways to fill out that picture.
Use the product rule:
1. Place the bride/groom: 6
2. 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.
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