Counting methods
SOLUTIONS
Mathematics higher level
August 6, 2020
1. (a) In how many ways can the letters of the word logarithm be arranged?
(b) In how many ways can the letters be arranged taking them two at a time?
Ans.
(a) There are 9 distinct letters, so there are 9 P9 = 9! = 362,880 ways.
(b) Here we fill two bins, so there are 9 P2 = 9 × 8 = 72 ways.
2. (a) How many four–digit even numbers can be made using the digits 0, 1, 2, 3, 4, 5, 6?
(b) How many of these four–digit even numbers are divisible by 5?
(c) How many of these four–digit even numbers have no repeated digits?
Ans.
(a) There are 7 digits, 4 bins and the last digit must be 0, 2, 4 or 6, so there are
6 × 72 × 4 = 1,176 4–digit even numbers.
(b) Even and divisible by 5 means ending in 0.
a
b
c 0
In the above there are 6 choices for a, 7 for b and 7 for c. This means there are
6 × 72 = 294 4–digit even numbers divisible by 5.
(c) Similar to the above, but this time no repetition. So,
a
b
c 0
We have a 6= 0, so 6 choices for a, 5 for b and 4 for c. That means 6 × 5 × 4 = 120
ways. Now, we consider the case where the last digit is even
a
b
c
d = {2, 4, 6}
There are 3 choices for d, 5 for c, 5 for b and 4 for a. That means 3 × 5 × 5 × 4 = 300
ways. Putting it altogether we have 120 + 300 = 420 4–digit even numbers with
distinct digits.
3. There are 7 girls and 8 boys in a class at school. In how many ways can 10 students be
chosen for a club if
(a) there are no restrictions
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(b) there are 5 boys and 5 girls chosen
(c) at least three of each gender are included?
Ans.
(a)
15 C10
= 3003
(b) 7 C5 ×8 C5 = 1176
(c) 3003 −7 C2 = 2982 There is no way to have only 2 boys chosen, since there are only
7 girls to make up the rest of the club.
4. Three Mathematics books, five English books, four Science books and a dictionary are to
be placed on a student’s shelf so that the books of each subject remain together.
(a) In how many different ways can the books be arranged?
(b) In how many of these will the dictionary be next to the Mathematics books?
Ans.
(a) 4! × (3! × 5! × 4! × 1!) = 414,720 Note, there are 4! ways to arrange the four types
of books!
(b) 3!×(3! × 5! × 4! × ×1! × 2) = 207,360 There are 2 positions for the dictionary, which
then becomes “tied” to the Mathematics books. That leaves 3! ways to arrange the
three sets of books.
5. A team of 4 students is to be selected for a math contest. There are 8 boys and 12 girls
to choose from. In how many ways
(a) can the team be selected?
(b) can the team include at least one of each gender?
(c) can the team have more girls than boys.
Ans.
(a)
20 C4
= 4845
(b)
20 C4
−8 C4 −12 C4 = 4280
(c)
12 C4
+12 C3 ×8 C1 = 2255
6. There are 100 people in a room. Each person shakes hands with every other person exactly
once. How many handshakes occur?
Ans.
100 C2
= 4950
7. How many five digit numbers contain at least one three? (Hint: think about the complement of “at least one”)
Ans.
We have a five digit number...
a
b
c
d
e
where a 6= 0. There are 9 × 104 5–digit numbers (no restrictions). There are 8 × 94 5–digit
numbers that do not contain a 3. Hence, the number of 5–digit numbers containing at
least one 3 is 9 × 104 − 8 × 94 = 37,512 .
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8. A committee of four children is chosen from eight children. The two oldest children cannot
both be chosen. Find the number of ways the committee may be chosen.
Ans.
We subtract the number of committees containing both older children away from the
total number of committees possible with no restrictions.
8 C4
−6 C2 = 55
9. How many different arrangements, each consisting of 5 different digits, can be formed from
the digits 1, 2, 3, 4, 5, 6, 7 if
(a) each arrangement begins and ends with an even digit?
(b) in each arrangement odd and even digits alternate?
Ans.
We have a 5–digit number...
a
b
c
d
e
(a) where a and e can be 2, 4 or 6; while b, c and d can be even or odd. So we get
3 × 5 × 4 × 3 × 2 = 360 arrangements.
(b) In this case we have the following possibilities
even
odd
even
odd
even
odd
even
odd
even
odd
So, the calculation becomes
3 × 4 × 2 × 3 × 1 + 4 × 3 × 3 × 2 × 2 = 72 + 144 = 216
10. A multiple choice test in chemistry consists of eight questions each with 4 possible answers.
For each question only one answer is correct.
Assuming every question is answered,
(a) in how many different ways can the test be completed?
(b) in how many different ways can this test be completed so that exactly six questions
are correct?
Ans.
(a) 48 = 65,536
(b) First we recognize there are 8 C2 ways to get exactly two questions wrong, leaving
exactly 6 correct. There are 3 incorrect choices for each of these two wrongly answered
questions. Hence, the number of different ways to complete the test so that exactly
six questions are correct is
2
8 C2 × 3 = 252
11. There are six boys and five girls in a school tennis club. A team of two boys and two girls
will be selected to represent the school in a tennis competition.
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(a) In how many different ways can the team be selected?
(b) Tim is the youngest boy in the club and Anna is the youngest girl. In how many
different ways can the team be selected if it must include both of them?
Ans.
(a) 6 C2 ×5 C2 = 150
(b) 5 C1 ×4 C1 = 20
12. There are 30 students in a class, of which 18 are girls and 12 are boys. Four students are
selected at random to form a committee. Calculate the probability that the committee
contains
(a) two girls and two boys;
(b) students all of the same gender.
(c) at least one girl.
Ans.
×12 C2
153 × 66
10098
374
=
=
≈ 0.368
=
27405
27405
1015
C
30 4
3060 + 495
18 C4 +12 C4
=
(b) P (G = 4 ∪ B = 4) =
≈ 0.130
27405
30 C4
18 C0 ×12 C4
(c) P (G ≥ 1) = 1 − P (G = 0 ∩ B = 4) = 1 −
≈ 0.982
30 C4
(a) P (G = 2 ∩ B = 2) =
18 C2
13. There are 10 seats in a row in a waiting room. There are six people in the room.
(a) In how many different ways can they be seated?
(b) In the group of six people, there are three sisters who must sit next to each other.
In how many different ways can the group be seated?
Ans.
(a)
10 C6
× 6! =10 P6 = 151200
(b) There are 8 ways the three sisters can be seated together
S
S
S
S
S
S
S
S
S
S
S
S
..
.
Of course the three sisters can be permuted 3! = 6 ways. And there remain 3 people
to still take a seat. The 4th person has 7 choices, the 5th has 6 choices and the 6th
has 5 choices. Thus, the total number of ways the six people can take a seat and
have the three sisters together is
8 × 3! × 7 × 6 × 5 = 10,080
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14. Solve:
(a)
n C2
= 45
(b)
n C2
=n−1 P2
Ans.
(a)
n!
= 45
2!(n − 2)!
n(n − 1) = 90
n = 10
(b)
n!
= (n − 1) × n
2!(n − 2)!
n(n − 1)
(n
−2)!
= (n − 1) × (n − 2)
2!
(n
−2)!
n(n − 1) = 2(n − 1)(n − 2)
n = 2(n − 2)
n= 4
15. The pizza place offers pepperoni, mushrooms, sausages, onions, anchovies and peppers as
toppings for their pizza. How many different pizzas can be made?
Ans.
6
X
6 Ck
=6 C0 +6 C1 + · · · +6 C6 = 64
k=0
16. Five boys and five girls stand in a line. How many arrangements are possible if
(a) all the boys stand in succession.
(b) the boys and girls alternate.
Ans.
(a) Like problem 13. there are six positions the five boys can take. There are 5! ways
the boys can be arranged within their successive standing arrangement, and the five
girls can as well be arranged in 5! ways. Thus the total number of arrangements we
seek is
6 × 5! × 5! = 86,400
(b) A straight–forward way of doing this one is by just counting the number of choices
we have to seat a person into a given place. Consider that there are 10 places to
fill. How many choices do we have for the first place? The answer is 10. How many
for the second place? The answer is 5 because if the first place is a boy then there
remain 5 girls to choose from. The following multiplication should now make sense.
10 × 5 × 4 × 4 × 3 × 3 × 2 × 2 × 1 × 1 = 28,800
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