﻿ Permutations and Combinations Worksheet - Page 10 | Problems & Solutions

# Permutations and Combinations Worksheet - Page 10

Permutations and Combinations Worksheet
• Page 10
91.
What is the value of?
 a. 20 b. 40 c. 25 d. 23

#### Solution:

5! = 5 × 4 × 3!
[Expand 5!.]

(5 - 2)! = 3!
[Simplify.]

5!(5 - 2)! = 5 × 4 × 3!3!
[Substitute the values of 5! and (5 - 2)!.]

= 5 × 4 = 20
[Simplify.]

The value of 5!(5 - 2)! is 20.

92.
What is the value of $\frac{5!}{\left(5-2\right)!}$?
 a. 25 b. 20 c. 40 d. 23

#### Solution:

5! = 5 x 4 x 3!
[Expand 5!.]

(5 - 2)! = 3!
[Simplify.]

5!(5-2)! = 5x4x3!3!
[Substitute the values of 5! and (5 - 2)!.]

= 5 x 4 = 20
[Simplify.]

The value of 5! / (5-2)! is 20.

93.
In how many ways can you arrange the letters in the word RANDOM?
 a. 720 b. 320 c. 420 d. 520

#### Solution:

Number of possible outcomes for choosing the first letter = 6
[There are 6 letters in the word.]

Number of possible outcomes for choosing the second letter = 5
[There are 5 letters remaining to choose.]

Number of possible outcomes for choosing the third letter = 4
[There are 4 letters remaining to choose after choosing 2 letters.]

Number of possible outcomes for choosing the fourth letter = 3
[There are 3 letters remaining to choose after choosing 3 letters.]

Number of possible outcomes for choosing the fifth letter = 2
[There are 2 letters remaining to choose after choosing 4 letters.]

Number of possible outcomes for choosing the sixth letter = 1
[There is only 1 letter remaining to choose after choosing 5 letters.]

1st letter       2nd letter       3rd letter       4th letter       5th letter       6th letter
6         x        5        x        4        x          3        x         2        x        1        =   720
[Multiply the number of possible outcomes of choosing each letter in the word.]

The letters in the word can be arranged in 720 ways.

94.
In how many ways can you arrange the letters in the word RANDOM?
 a. 520 b. 420 c. 720 d. 320

#### Solution:

Number of possible outcomes for choosing the first letter = 6
[There are 6 letters in the word.]

Number of possible outcomes for choosing the second letter = 5
[There are 5 letters remaining to choose.]

Number of possible outcomes for choosing the third letter = 4
[There are 4 letters remaining to choose after choosing 2 letters.]

Number of possible outcomes for choosing the fourth letter = 3
[There are 3 letters remaining to choose after choosing 3 letters.]

Number of possible outcomes for choosing the fifth letter = 2
[There are 2 letters remaining to choose after choosing 4 letters.]

Number of possible outcomes for choosing the sixth letter = 1
[There is only 1 letter remaining to choose after choosing 5 letters.]

 1st letter 2nd letter 3rd letter 4th letter 5th letter 6th letter 6 x 5 x 4 x 3 x 2 x 1 = 720

[Multiply the number of possible outcomes of choosing each letter in the word.]

The letters in the word can be arranged in 720 ways.

95.
How many outcomes are possible, when three coins are tossed?
 a. 8 b. 3 c. 10 d. 6

#### Solution:

Total number of coins = 3

Each coin has the outcome of either a Head (H) or Tail (T).

The possible outcomes of tossing three coins is HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

There are 8 possible outcomes when the three coins are tossed.

96.
How many outcomes are possible, when the three coins are tossed?

 a. 3 b. 10 c. 8 d. 6

#### Solution:

Total number of coins = 3

Each coin has the outcome of either a Head (H) or Tail (T).

The possible outcomes of tossing three coins is HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

There are 8 possible outcomes when the three coins are tossed.

97.
Find the number of ways of forming a password with 4 letters and 4 digits to create a mail id (repetition not allowed).
 a. 26C4 x 10C4 x 8! b. 26C4 x 10C4 c. 26P4 x 10P4 x 8! d. 36C8

#### Solution:

Three letters out of 26 letters can be chosen in 26C4 ways.

Two digits out of 10 digits can be chosen in 10C4 ways.

The number of ways of arranging each of the selections = 8!

So, the total number of ways of forming the password = 26C4 x 10C4 x 8!

98.
In how many ways can you choose 2 even numbers and 3 odd numbers from the numbers from 3, 4, 5, 6, 7, 8 and 9?
 a. 12 b. 7 c. 10 d. 14

#### Solution:

4, 6 and 8 are the even numbers, which are 3 in number.

3, 5, 7 and 9 are the odd numbers, which are 4 in number.

List all the possible combinations of choosing 2 even numbers as follows:
(4, 6), (4, 8)
(6, 8)

List all the possible combinations of choosing 3 odd numbers as follows:
(3, 5, 7), (3, 5, 9)
(5, 7, 9), (3, 7, 9)

List all the possible combinations of choosing 2 even numbers and 3 odd numbers as follows:
(4, 6, 3, 5, 7), (4, 6, 3, 5, 9), (4, 6, 5, 7, 9), (4, 6, 3, 7, 9)
(4, 8, 3, 5, 7), (4, 8, 3, 5, 9), (4, 8, 5, 7, 9), (4, 8, 3, 7, 9)
(6, 8, 3, 5, 7), (6, 8, 3, 5, 9), (6, 8, 5, 7, 9), (6, 8, 3, 7, 9)

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

99.
William lists different arrangements of the letters in the word $\mathrm{NAD}$ as follows: $\mathrm{NDA}$, $\mathrm{NAD}$, $\mathrm{DAN}$, $\mathrm{DNA}$, and $\mathrm{ADN}$. Find the number of arrangements not included in the list.
 a. 3 b. 2 c. 1 d. 8

#### Solution:

Total number of letters in the word NAD = 3

= 3! = 3 x 2 x 1 = 6
Total number of arrangements formed by all the letters of the word NAD
[Expand 3! and multiply]

Total number of arrangements listed = 5

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

100.
Find the number of three letter words that can be formed with the letters of the word FCP .
 a. 12 b. 6 c. 9 d. 3

#### Solution:

The possible arrangements of the letters in the word FCP can be organized as shown below:
FCP, FPC, CPF, CFP, PFC and PCF

So, the number of words that can be formed with the letters of the word FCP is 6.