# Permutations and Combinations Worksheet - Page 11

Permutations and Combinations Worksheet
• Page 11
101.
A 12-member club is choosing its President, Vice president, Secretary and Treasurer. Find the number of ways in which they can be chosen.
 a. 11880 b. 132 c. 990 d. 1320

#### Solution:

President is to be chosen from 12 members, Vice president is to be chosen from 11 members, Secretary is to be chosen from 10 members and Treasurer is to be chosen from 9 members.

From the counting principle, the four people can be chosen in 12 x 11 x 10 x 9 ways.

= 11880
[Multiply.]

There are 11880 different ways in which they can be chosen.

102.
In how many ways can you choose 2 even numbers and 3 odd numbers from the numbers, 1, 2, 3, 4, 5, 6 and 7?
 a. 10 b. 14 c. 12 d. 7

#### Solution:

2, 4 and 6 are the even numbers, which are 3 in number.

1, 3, 5 and 7 are the odd numbers, which are 4 in number.

List all the possible combinations of choosing 2 even numbers as follows:
(2, 4), (2, 6)
(4, 6)

List all the possible combinations of choosing 3 odd numbers as follows:
(1, 3, 5), (1, 3, 7)
(3, 5, 7), (1, 5, 7)

List all the possible combinations of choosing 2 even numbers and 3 odd numbers as follows:
(2, 4, 1, 3, 5), (2, 4, 1, 3, 7), (2, 4, 3, 5, 7), (2, 4, 1, 5, 7)
(2, 6, 1, 3, 5), (2, 6, 1, 3, 7), (2, 6, 3, 5, 7), (2, 6, 1, 5, 7)
(4, 6, 1, 3, 5), (4, 6, 1, 3, 7), (4, 6, 3, 5, 7), (4, 6, 1, 5, 7)

So, 2 even numbers and 3 odd numbers can be chosen in 12 different ways.

103.
State whether the arrangement of 2-lettered words from the alphabets C, H, E, M, I, S, T, R, Y is a permutation or a combination.
 a. Permutation b. Combination

#### Solution:

A permutation is an arrangement in which order is important and a combination is an arrangement in which order does not matter.

The ordering of alphabets is important here and different orders of same letters form different words.

So, the arrangement of 2-lettered words from the alphabets is a permutation.

104.
Henry has to answer 3 questions out of 6 questions in an examination. State whether his selection of 3 questions is a permutation or a combination.
 a. Permutation b. Combination

#### Solution:

A permutation is an arrangement in which order is important and a combination is an arrangement in which order does not matter.

The sequence of selecting 3 questions is not important here and different sequence of orders of same set of questions form the same selection.

So, the selection of questions is a combination.

105.
Tommy plans to study, watch TV and play football on a Sunday. In how many ways can he choose and arrange the activities?
 a. 3 b. 2 c. 6 d. 1

#### Solution:

Total number of activities = 3

First activity can be chosen in 3 ways; second activity can be chosen in 2 ways and third activity can be chosen in 1 way.

By counting principle,
1st Activity   2nd Activity   3rd Activity
3        x         2        x         1    =     6

Tommy can choose and arrange the 3 activities in 6 different ways.

106.
The combination to open your lock has three pairs of numbers 13, 37 and 27. 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. 12 b. 3 c. 9 d. 6

#### Solution:

Total number of pairs of the numbers = 3

Number of combinations to open the lock with the 3 given pairs of numbers 13, 37, and 27 = 3!

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

The list of all combinations to open the lock is
(i) 27, 13, 37
(ii) 27, 37, 13
(iii) 13, 27, 37
(iv) 13, 37, 27
(v) 37, 27, 13
(vi) 37, 13, 27

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

107.
Find the number of combinations of choosing 2 cubes from the 5 cubes.

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

#### 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.

108.
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.

109.
What is the value of $\frac{3!}{\left(2-2\right)!}$?
 a. 3 b. 9 c. 4 d. 6

#### Solution:

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

(2 - 2)! = (0)! = 1
[0! is always 1.]

3!(2-2)! = 3!0! = 61 = 6
[Substitute the values of 3! and 0!.]

The value of 3! / (2-2)! is 6.

110.
Find: ($\frac{5!}{4!}$) × ($\frac{1}{2!}$)
 a. 3 b. $\frac{2}{5}$ c. $\frac{5}{2}$ d. $\frac{5}{3}$

#### Solution:

5! = 5 ×4 ×3 ×2 ×1
[Expand 5!.]

4! = 4 ×3 ×2 ×1
[Expand 4!.]

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

(5!4!) = 5
[Simplify the fraction.]

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

(5! / 4!) × (1 / 2!) = 5 × (1 / 2)

= 52
[Simplify.]

The value of (5! / 4!) × (1 / 2!) is 5 / 2.