Download Counting and Probability Practice Problems

Counting and Probability Practice Problems
1. Devise an alternate proof that the power set of an n-element set has 2n
elements.
2. Flip 20 fair coins. What is the probability that exactly 7 heads come up?
3. A man works in a building located seven blocks east and eight blocks
north of his home. In how many different ways can he go from home
to work, walking only fifteen blocks?
4. A room has six doors. In how many ways is it possible to enter by one
door and leave by another?
5. A tire store carries eight different sizes of tires, each in both tube and
tubeless variety, each with either nylon or rayon cord, and each with
white sidewalls or plain black. How many different kinds of tires does
the store have?
6. How many integers between 100 and 999 inclusive consist of distinct
odd digits?
7. How many integers between 100 and 999 inclusive consist of distinct
digits?
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8. Of the integers in the preceding problem, how many are odd numbers?
9. How many of the first 1000 positive integers have distinct digits?
10. In how many ways can ten people be seated in a row so that a certain
two of them are not next to each other?
11. A certain men’s club has sixty members. Thirty are business men and
thirty are professors. In how many ways can a committee of eight
be selected if at least three must be business men and at least three
professors.
12. In how many ways can a ballot be validly marked if a citizen is to choose
one of three candidates for mayor, one of four for city councilman, and
one of three for district attorney. A citizen is not required to vote for
all three positions, but he is expected to vote for at least one.
13. In how many ways is it possible to seat eight people at a round table?
14. In the preceding question, what would be the answer if a certain two of
the eight people must not sit in adjacent seats?
15. In how many ways can four men and four women be seated at a round
table if no two men are to be in adjacent seats?
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16. How many permutations are there of the letters, taken all at a time, of
the word MATHEMATICS?
17. Of the permutations in the preceding problem, how many contain the
letters MATH next to each other in order.
18. What is the coefficient of u3 v 7 in (u + v)10 ?
19. How many integers between 1 and 6300 inclusive are divisible by none
of 3, 5, and 7?
20. How many integers from 1 to 1,000,000 inclusive either perfect squares
or perfect cubes?
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SOLUTIONS
1. Let S be a set with n elements. In constructing a subset of S, we run
through each element x of S and decide whether to include x in the
subset or not. Since S has n elements and there are two options per
element, it follows that P(S) has 2n elements.
2. As there are two options per coin, the sample space contains 220 events.
Of those events, we want the ones where exactly 7 heads come up. So
we just need to choose which
7 coins out of the 20 total comes up heads.
The probability is thus
20
7
220
.
3. The man’s trip from home to work can be viewed as an arrangement of
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seven Es and eight N s. This is simply 15
=
.
7
8
4. There are six choices for which door to enter. After picking the entry,
there are then five choices for which door to exit. There are thus 6 · 5
possibilities.
5. Simply multiply the number of options per category. There are thus
8 · 2 · 2 · 2 different kinds of tire.
6. This is the number of permutations of the five distinct odd digits 1, 3,
5!
5!
5, 7, 9, taken three at a time. The answer is P (5, 3) =
= .
(5 − 3)!
2!
7. Unlike the previous problem, this is not a permutation, because the first
digit cannot be 0 whereas the last two can be. There are thus 9 choices
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for the first digit, 9 choices for the second digit (after the first one is
chosen), and 8 digits for the last one (after the first two are chosen).
The answer is 9 · 9 · 8.
8. A number is odd if and only if its units digit is odd. So it is best to
begin the argument by picking the units digit first. There are 5 choices
for the units digit. Next, turn to the the hundreds’ place. There are
eight digits from which a selection can be made, namely all the nonzero
digits except the one already used in the units’ place. Finally, there are
eight choices for the digit in the ten’s place, so the answer is 5 · 8 · 8.
When working such problems, take note how the order you make your
choices affects your answer.
9. Setting aside 1000, whose digits are not distinct, the others can be
separated into three types:
• Integers with one digit: 1, 2, 3, . . . , 9;
• Integers with two digits: 10, 11, 12, . . . , 99;
• Integers with three digits: 100, 101, 102, . . . , 999.
As shown previously, the number of three-digit integers with distinct
digits is 9 · 9 · 8. A similar argument shows that there are 9 · 9 distinct
two-digit integers, and, of course, 9 one-digit integers that meet the
specification of distinct digits. Hence the answer is 9 · 9 · 8 + 9 · 9 + 9.
10. The answer is 10! − 2 · 9!. This is because there are 10! unrestricted
arrangements of the ten people, and the number of arrangements with
the two specific people together is 2 · 9! because the two people can be
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regarded as one unit in this case, but in two orders. Now subtract to
get the number where they are separated.
11. There are three cases, namely 3 professors, 4 professors, or 5 professors
in the committee.
• 3 professors. There are thus 5 business men. The total number of
30
ways to choose such a committee is 30
.
3
5
• 4 professors. There are thus 4 business men. The total number of
30
ways to choose such a committee is 30
.
4
4
• 5 professors. There are thus 3 business men. This too is has
30 30
possiblitites.
5
3
Add each case together to get a total of 2
30
3
30
5
+
30 2
.
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12. There are four options for the mayoral part of the ballot, that is, picking
one of the three candidates or abstaining from voting for the mayor.
Similarly, there are five options for the city councilman part and four
options for the district attorney part. However, the ballot cannot be
left unmarked, and so there are 4 · 5 · 4 − 1 valid ballots.
13. Recall that the number of circular arrangements of n objects is (n − 1)!.
Here n = 8, and so the answer is 7!.
14. Start with the solution of the preceding problem and subtract the number of cases where the two people, say person A and person B, are in
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adjacent seats. Taking person A and person B as one unit, we see that
there are 6! circular arrangements of the eight people with person A to
the left of person B, and 6! cases the other way about. Thus the answer
is 7! − 2 · 6!.
15. The women can be seated in alternate seats in 3! ways. Now since the
women act as a separator for the men, the answer is 3! · 4! since 4! is the
number of ways the men can be seated relative to a fixed arrangement
of the women.
16. Since MATHEMATICS is has 11 letters, 2 of which are M, 2 of which
are A, 2 of which are T, one of which is H, one of which is E, one of
which is I, one of which is C, and one of which is S, the total number
11!
11!
=
.
of permutations is
2!2!2!1!1!1!1!1!
2!2!2!
17. In this case, we treat MATH as one ‘‘letter’’. This including E, M, A,
T, I, C, S are all distinct from each other, and so the answer is simply
8!, the number of arrangements of 8 different objects.
18. By the binomial theorem, the corresponding term is
the desired coefficient is 10
.
7
10
7
u3 v 7 , and so
19. Let A be the set of all integers between 1 and 6300 divisible by 3, B
the set of all integers between 1 and 6300 divisible by 5, and C the set
of all integers between 1 and 6300 divisible by 7. By the principle of
inclusion-exclusion, the total is
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6300−N (A)−N (B)−N (C)+N (A∩B)+N (A∩C)+N (B∩C)−N (A∩B∩C).
Now N (A) = 6300 ÷ 3 = 2100, N (B) = 6300 ÷ 5 = 1260, and
N (C) = 6300 ÷ 7 = 900. Note that any number in A ∩ B is divisible by
both 3 and 5, which happens precisely when that number is divisible by
15. Similarly, numbers in A∩C are divisible by 21, and numbers in B∩C
are divisible by 35. Hence N (A ∩ B) = 6300 ÷ 15 = 420, N (A ∩ C) =
6300 ÷ 21 = 300, and N (B ∩ C) = 6300 ÷ 35 = 180. Lastly, numbers in
A ∩ B ∩ C are divisible by 3, 5, and 7, which occurs precisely when the
number is divisible by 105. Hence N (A ∩ B ∩ C) = 6300 ÷ 105 = 60.
The final answer is thus 6300 - 2100 - 900 + 420 + 300 + 180 - 60 = 2880.
20. Let A be the set of perfect squares between 1 and 1,000,000 and let B
be the set of perfect cubes between 1 and 1,000,000. By the principle of
inclusion-exclusion,
N (A ∪ B) = N (A) + N (B) − N (A ∩ B).
We have N (A) =
√
1, 000, 000 = 1000 and N (B) =
√
3
1, 000, 000 = 100.
Lastly, a number is in A ∩ B when it is both a perfect square and a
perfect cube, which occurs precisely when that number is a perfect sixth
√
power. So N (A ∩ B) = 6 1, 000, 000 = 10. The final answer is thus 1000
+ 100 - 10 = 1090.
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