# Permutations and Combinations Worksheet - Page 6

Permutations and Combinations Worksheet
• Page 6
51.
How many three-letter permutations of 5 vowels can be formed?
 a. 60 b. 12 c. 30 d. 20

#### Solution:

Total number of vowels = 5
[The 5 vowels are a, e, i, o, and u.]

First vowel can be selected in 5 ways, second vowel can be selected in 4 ways and third vowel can be selected in 3 ways.

By counting principle,
1st vowel   2nd vowel   3rd vowel
5       ×       4        ×      3        =    60

There are 60 three-letter permutations of 5 vowels.

52.
How many groups of 8 persons can be formed from 9 men and 4 women?
 a. 1317 b. 1287 c. 1307 d. 1283

#### Solution:

Total number of men and women = 9 + 4 = 13

Number of groups that can be formed by selecting 8 persons from 13 persons = 13C8

13C8 = 1 / 8! × 13P8
[Expand 13C8.]

13P8 = 13×12×11×10×9×8&time

8! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1 = 40320
[Expand 8!.]

13C8 = 140320 × 13×12×11×10×9×8&time
[Substitute the values of 8! and 13P8.]

13C8 = 1287
[Simplify the product.]

So, 1287 groups can be formed from 9 men and 4 women.

53.
The combination to open your lock has three pairs of numbers 16, 39, and 29. You do not remember the order of the numbers. How many different arrangements of the numbers need to be tried out to open the lock?
 a. 29 b. 16 c. 39 d. 6 e. 7

#### Solution:

Total number of pairs of the numbers = 3

Number of combinations to open the lock with the 3 given pairs of numbers 16, 39, and 29 = 3!

= 3 × 2 × 1 = 6
[Expand 3! and simplify]

The list of all combinations to open the lock is
(i) 29, 16, 39
(ii) 29, 39, 16
(iii) 16, 29, 39
(iv) 16, 39, 29
(v) 39, 29, 16
(vi) 39, 16, 29

There are 6 different possible arrangements that need to be tried out to open the lock.

54.
What is the value of $\frac{4!}{\left(3-3\right)!}$?
 a. 28 b. 14 c. 24 d. 12

#### Solution:

4! = 4 ×3 ×2 ×1 = 24
[Expand 4! and multiply.]

(3 - 3)! = (0)! = 1
[0! is always 1.]

4!(3-3)! = 4!0! = 241 = 24
[Substitute the values of 4! and 0!.]

The value of 4! / (3-3)! is 24.

55.
A teacher wants 5 of her 15 students to write a skill test. In how many ways can she select 5 students?
 a. 3008 b. 2998 c. 3013 d. 3003

#### Solution:

Number of ways of selecting 5 students out of 15 students = 15C5

15C5 = 15! × 15P5
[Expand 15C5]

15P5 = 15×14×13×12×11
[Expand 15P5]

5! = 5 ×4 ×3 ×2 ×1 = 120
[Expand 5!]

15C5 = 1120 × 15×14×13×12×11
[Substitute the values of 5! and 15P5.]

15C5 = 3003
[Simplify the above product.]

The teacher can select 5 students out of 15 students in 3003 ways.

56.
How many permutations of 10 digits (0, 1 , . . . 9) can be formed, so as to form a 2-digit number?
 a. 10 b. 90 c. 80 d. 20

#### Solution:

Total number of digits = 10

First digit cannot be a zero.

So, the first digit can be selected in 9 ways and the second digit can be selected in 10 ways.

By counting principle,
1st digit   2nd digit
9      ×    10       =    90

There are 90 permutations of the 10 digits to form a 2-digit number.

57.
What is the value of $\frac{\left(6+2\right)!}{\left(6-2\right)!}$ ?
 a. 840 b. 3360 c. 560 d. 1680

#### Solution:

(6 + 2)! = 8!

8! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1
[Expand 8!.]

(6 - 2)! = 4!
[Subtract.]

4! = 4 ×3 ×2 ×1
[Expand 4!.]

(6+2)!(6-2)! = 1680
[Simplify the above fraction.]

The value of (6+2)! / (6-2)! is 1680.

58.
Find the number of ways of choosing two pairs of socks from the list of socks.
 Color Letter Blue B Yellow Y Red R Green G

 a. 5 b. 6 c. 4 d. 2

#### Solution:

Total number of pairs of socks = 4

List all the possible arrangements of choosing 2 pairs of socks from 4 pairs of socks:
(B, Y), (B, R), (B, G)
(Y, B), (Y, R), (Y, G)
(R, B), (R, Y), (R, G)
(G, B), (G, Y), (G, R)

As choosing (B, Y) pair of socks is same as choosing (Y, B) pair of socks, cancel the duplicate arrangements of pairs of socks as shown below.

List all the combinations of 2 pairs of socks without duplicate arrangements:
(B, Y), (B, R), (B, G)
(Y, R), (Y, G)
(R, G)

So, the number of ways of choosing two pairs of socks from the list of socks = 6

59.
Kelsey wants to give 3 out of 12 paintings to her friend. In how many ways can she choose 3 paintings to give her friend?
 a. 1320 b. 1230 c. 220 d. 120

#### Solution:

Number of ways of selecting 3 out of 12 paintings = 12C3

12C3 = 13! × 12P3
[Expand 12C3.]

12P3 = 12 × 11 × 10
[Expand 12P3.]

3! = 3 × 2 × 1
[Expand 3!.]

12C3 = 16 × 1320
[Substitute the values of 12C3 and 3!.]

12C3 = 220
[Simplify the above product.]

Kelsey can choose 3 out of 12 paintings in 220 different ways.

60.
Make a list of all the combinations to find the number of ways of choosing 1 consonant and 2 vowels from the word QUESTION.
 a. 16 b. 32 c. 24 d. 48

#### Solution:

Total number of letters in the word 'QUESTION' = 8

Vowels in the word 'QUESTION' are U, E, O and I, which are 4 in number.

Consonants in the word are Q, S, T and N, which are 4 in number.

List all the possible combinations of 2 vowels chosen out of 4 vowels as shown below.
(U, E), (U, I), (U, O)
(E, I), (E, O)
(I, O)

There are a total of 6 combinations of choosing 2 vowels.

List all the possible combinations of choosing 1 consonant and 2 vowels as follows:
(Q, U, E), (Q, U, I), (Q, U, O), (Q, E, I), (Q, E, O), (Q, I, O)
(S, U, E), (S, U, I), (S, U, O), (S, E, I), (S, E, O), (S, I, O)
(T, U, E), (T, U, I), (T, U, O), (T, E, I), (T, E, O), (T, I, O)
(N, U, E), (N, U, I), (N, U, O), (N, E, I), (N, E, O), (N, I, O)

So, 1 consonant and 2 vowels can be chosen in 24 different ways.