# Permutations and Combinations Worksheet - Page 3

Permutations and Combinations Worksheet
• Page 3
21.
In how many ways can 6 people sit in 6 chairs?
 a. 721 b. 5040 c. 120 d. 720

#### Solution:

Number of ways in which 6 people can sit in 6 chairs = 6P6

= 6! = 6 ×5 ×4 ×3 ×2 ×1 = 720
[Simplify.]

6 people can sit in 6 chairs in 720 ways.

22.
Find: ($\frac{5!}{3!}$ ) × ($\frac{1}{2!}$ )
 a. 15 b. 10 c. 5 d. None of the above

#### Solution:

5! = 5 ×4 ×3 ×2 ×1
[Expand 5!.]

3! = 3 ×2 ×1
[Expand 3!.]

2! = 2 ×1 = 2
[Expand 2! and multiply.]

(5!3!) = 20
[Simplify the fraction.]

(12!) = 12
[Substitute the value of 2!.]

(5! / 3! ) × (1 / 2! ) = 20 × (1 / 2 )

= 10
[Simplify.]

The value of (5! / 3! ) × (1 / 2! ) is 10 .

23.
Paul lists different arrangements of the letters in the word $\mathrm{JON}$ as follows: $\mathrm{JNO}$, $\mathrm{JON}$, $\mathrm{NOJ}$, $\mathrm{NJO}$, and $\mathrm{ONJ}$. Find the number of arrangements not included in the list.
 a. 2 b. 1 c. 3 d. 8

#### Solution:

Total number of letters in the word JON = 3

= 3! = 3 x 2 = 6
[Expand 3! and multiply.]

Total number of arrangements formed by all the letters of the word JON = 6

Total number of arrangements listed = 5

Number of arrangements not included in the list = 6 - 5 = 1
[Subtract.]

24.
In how many ways can 9 questions be selected out of 14 questions such that the selection always contains the questions numbered 1, 2 and 7?
 a. 452 b. 462 c. 472 d. 467

#### Solution:

Number of questions given in the test paper = 14

Number of questions remaining after selecting questions numbered 1, 2 and 7 = 14 - 3 = 11

The remaining 6 questions have to be selected from these 11 questions.

Number of ways in which 6 questions can be selected out of 11 questions = 11C6

11C6 = 1 / 6! × 11P6

6! = 6 ×5 ×4 ×3 ×2 ×1 = 720
[Expand 6! and multiply.]

11P6 = 11×10×9×8×7 = 332640
[Expand 11P6.]

11C6 = 332640 / 720 = 462
[Substitute the values of 6! and 11P6.]

The number of ways in which 9 questions can be selected out of 14 questions, always selecting questions numbered 1, 2 and 7 is 462.

25.
Which tree diagram represents the possible combinations of the spins of the spinners 1 and 2?

 a. Figure A b. Figure B

#### Solution:

The possible outcomes of spinner 1 are A, B, C, or D. So, the number of outcomes of spinner 1 = 4.

The possible outcomes of spinner 2 are 1 or 2. So, the number of outcomes of spinner 2 = 2

For every outcome of spinner 1 there should be a outcome of spinner 2.

So, the number of possible combinations of the spins of the spinners 1 and 2 are (A, 1), (A, 2), (B, 1),(B, 2), (C, 1), (C, 2), (D, 1), and (D, 2). These are represented as a tree diagram as shown below.

The above tree diagram matches with the tree diagram in figure B.

26.
Stephanie has 7 books to arrange in a shelf. In how many ways can she arrange them?
 a. 720 b. 120 c. 5040 d. None of the above

#### Solution:

Number of books to be arranged = 7

There are 7 books to fill the first place in the shelf, 6 books to fill the second place, 5 books to fill the third place in the shelf and so on.

By counting principle,
1st book × 2nd book × 3rd book . . . 7th book
7     ×        6       ×        5       . . .     1      = 5040

So, there are 5040 different ways in which Stephanie can arrange 7 books.

27.
Find the number of combinations of choosing 2 cubes from 5 cubes shown.

 a. 10 b. 72 c. 56 d. 25

#### Solution:

Total number of cubes = 5

The list of all the possible arrangements of 2 cubes is:
(Red, Yellow), (Red, Blue), (Red, Green), (Red, White),
(Yellow, Red), (Yellow, Blue), (Yellow, Green), (Yellow, White),
(Blue, Red), (Blue, Yellow), (Blue, Green), (Blue, White),
(Green, Red), (Green, Yellow), (Green, Blue), (Green, White)
(White, Red), (White, Yellow), (White, Blue), (White, Green).

As choosing (Red, Yellow) is same as (Yellow, Red), cancel the duplicate arrangements as shown below.

The list of all the combinations of 2 cubes after canceling the duplicates is:
(Red, Yellow), (Red, Blue), (Red, Green), (Red, White),
(Yellow, Blue), (Yellow, Green), (Yellow, White),
(Blue, Green), (Blue, White),
(Green, White)

So, the number of combinations of choosing 2 cubes from 5 cubes is 10.

28.
A school has 14 teachers out of which 8 are male. A team of 8 teachers is to be formed with 4 female teachers. In how many ways can the team be formed?
 a. 1060 b. 1050 c. 1070 d. 1040

#### Solution:

Number of female teachers = 14 - 8 = 6
[Subtract.]

Number of ways to select 4 female teachers out of 6 female teachers = 6C4 = 6! / (6-4)!×4! = 6! / 2!×4! = 15

Number of ways to select 4 male teachers out of 8 male teachers = 8C4 = 8! / (8-4)!×4! = 8! / 4!×4! = 70

Number of ways in which the team can be formed = Number of ways to select 4 female teachers × Number of ways to select 4 male teachers

= 15 × 70 = 1050
[Substitute and multiply.]

A team of 8 teachers can be formed in 1050 different ways.

29.
Find the value of $r$, if 7P4 + $r$ x 7C4 = 0.
 a. -24 b. 24 c. -27 d. 21

#### Solution:

7P4 + r x 7C4 = 0
[Original equation.]

7P4 + r x 14! x 7P4 = 0
[Expand 7C4.]

840 + r x 124 x 840 = 0
[Substitute 7P4 = 7×6×5×4 = 840 and 4! = 4 ×3 ×2 ×1 = 24.]

840 + r x 35 = 0
[Cancel 840 with 24.]

r x 35 = -840
[Subtract -840 from each side.]

r = - 24
[Divide each side by 35.]

The value of r is -24.

30.
A number lock has 9 different digits. A combination of two different digits can be set to open the lock. How many combinations are possible?
 a. 81 b. 72 c. 63 d. 18

#### Solution:

Number of ways of selecting the first number out of 9 numbers = 9C1 = 1 / 1! × 9P1 = 9

Number of digits remaining after selecting the first number = 9 - 1 = 8

Number of ways of selecting the second number out of 8 numbers = 8C1 = 1 / 1! × 8P1 = 8

Total number of possible combinations = Number of ways of selecting the first number × Number of ways of selecting the second number = 9 × 8 = 72

72 combinations can be set to open the lock.