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 |

[The 5 vowels are

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,

1

5 × 4 × 3 = 60

There are 60 three-letter permutations of 5 vowels.

Correct answer : (1)

52.

How many groups of 8 persons can be formed from 9 men and 4 women?

a. | 1317 | ||

b. | 1287 | ||

c. | 1307 | ||

d. | 1283 |

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

[Expand

8! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1 = 40320

[Expand 8!.]

[Substitute the values of 8! and

[Simplify the product.]

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

Correct answer : (2)

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 |

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.

Correct answer : (4)

54.

What is the value of $\frac{\mathrm{4!}}{\mathrm{(3-3)!}}$?

a. | 28 | ||

b. | 14 | ||

c. | 24 | ||

d. | 12 |

[Expand 4! and multiply.]

(3 - 3)! = (0)! = 1

[0! is always 1.]

[Substitute the values of 4! and 0!.]

The value of

Correct answer : (3)

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 |

[Expand

[Expand

5! = 5 ×4 ×3 ×2 ×1 = 120

[Expand 5!]

[Substitute the values of 5! and

[Simplify the above product.]

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

Correct answer : (4)

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 |

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,

1

9 × 10 = 90

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

Correct answer : (2)

57.

What is the value of $\frac{(6+2)!}{(6-2)!}$ ?

a. | 840 | ||

b. | 3360 | ||

c. | 560 | ||

d. | 1680 |

[Add.]

8! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1

[Expand 8!.]

(6 - 2)! = 4!

[Subtract.]

4! = 4 ×3 ×2 ×1

[Expand 4!.]

[Simplify the above fraction.]

The value of

Correct answer : (4)

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 |

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

Correct answer : (2)

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 |

[Expand

[Expand

3! = 3 × 2 × 1

[Expand 3!.]

[Substitute the values of

[Simplify the above product.]

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

Correct answer : (3)

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 |

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.

Correct answer : (3)