Permutations and Combinations Worksheet

**Page 5**

41.

Using counting principle, find the number of three-letter permutations of the six letters S, P, R, I, T and E.

a. | 120 | ||

b. | 60 | ||

c. | 90 | ||

d. | 100 |

First letter can be selected in 6 ways, second letter can be selected in 5 ways and third letter can be selected in 4 ways.

By counting principle,

1

6 x 5 x 4 = 120

There are 120 three-letter permutations for the letters S, P, R, I, T and E.

Correct answer : (1)

42.

What is the value of $\frac{\mathrm{5!+4!}}{\mathrm{5!-4!}}$ ?

a. | $\frac{3}{2}$ | ||

b. | $1\frac{1}{4}$ | ||

c. | 1 | ||

d. | $1\frac{1}{3}$ |

[Expand 5! and multiply.]

4! = 4 ×3 ×2 ×1 = 24

[Expand 4! and multiply.]

5! + 4! = 120 + 24 = 144

[Substitute the values of 5! and 4!.]

5! - 4! = 120 - 24 = 96

[Substitute the values of 5! and 4!.]

[Substitute the values of (5! + 4!) and (5! - 4!).]

[Simplify the fraction.]

The value of

Correct answer : (1)

43.

In how many ways can 2 players be chosen from a group of 7 players to form a team?

a. | 25 | ||

b. | 23 | ||

c. | 19 | ||

d. | 21 |

Number of ways to select 2 players from 7 players =

[Expand

2! = 2 ×1 = 2

[Expand 2!.]

[Expand

[Substitute the value of

[Simplify the product.]

There are 21 ways to select 2 players from a group of 7 players to form a team.

Correct answer : (4)

44.

Find the number of combinations of choosing 2 fruits from 4 fruits.

a. | 6 | ||

b. | 3 | ||

c. | 4 | ||

d. | 2 |

List all the posible arrangements of choosing 2 fruits from 4 fruits:

(1, 2), (1, 3), (1, 4)

(2, 1), (2, 3), (2, 4)

(3, 1),(3, 2), (3, 4)

(4, 1), (4, 2), (4, 3)

As choosing the fruits (1, 2) is same as (2, 1), cancel the duplicate arrangements as shown below.

List all the combinations of 2 fruits without duplicate arrangements:

(1,2)(1,3)(1,4)

(2,3)(2,4)

(3,4)

So, the number of combinations of choosing 2 fruits from 4 fruits is 6.

Correct answer : (1)

45.

In how many ways can you assign the letters A, B, C, D, P, Q, R, and S to the corners of a square?

a. | 1660 | ||

b. | 1200 | ||

c. | 1500 | ||

d. | 1680 |

A square has 4 corners.

First corner can be assigned any of the 8 letters; second corner can be assigned any of the 7 letters and so on.

By counting principle,

1

8 × 7 × 6 × 5 = 1680

There are 1680 different ways in which the letters can be assigned to the four corners of the square.

Correct answer : (4)

46.

Find the number of combinations of choosing 2 marbles from 3 marbles.

a. | 4 | ||

b. | 6 | ||

c. | 3 | ||

d. | 2 |

List all the possible arrangements of choosing 2 marbles from 3 marbles:

(r, b), (r, g)

(b, r), (b, g)

(g, r), (g, b)

As choosing marbles (r, b) is same as choosing marbles (b, r), cancel the duplicate arrangements as shown below.

List all the combinations of 2 marbles without duplicate arrangements:

(r, b), (r, g)

(b, g)

The number of combinations of choosing 2 marbles from 3 marbles = 3

Correct answer : (3)

47.

There are 5 teams participating in a soccer tournament. Each team should play a match with all other teams in the preliminary rounds. How many matches should be played in the preliminary round?

a. | 15 | ||

b. | 5 | ||

c. | 20 | ||

d. | 10 |

List all the possible arrangements of choosing two teams from 5 teams:

(team 1, team 2), (team 1, team 3), (team 1, team 4), (team 1, team 5)

(team 2, team 1), (team 2, team 3), (team 2, team 4), (team 2, team 5)

(team 3, team 1), (team 3, team 2), (team 3, team 4), (team 3, team 5)

(team 4, team 1), (team 4, team 2), (team 4, team 3), (team 4, team 5)

(team 5, team 1), (team 5, team 2), (team 5, team 3), (team 5, team 4)

As the match between (team 1, team 2) is same as the match between (team 2, team 1), cancel the duplicate arrangements of matches as shown below.

List all the combinations of matches without duplicate arrangements:

(team 1, team 2), (team 1, team 3), (team 1, team 4), (team 1, team 5)

(team 2, team 3), (team 2, team 4), (team 2, team 5)

(team 3, team 4), (team 3, team 5)

(team 4, team 5)

So, the number of matches in the preliminary round = 10

Correct answer : (4)

48.

In how many different ways, can four students, Jim, Paul, Justin, and Bill, be seated in four chairs for a group discussion?

a. | 22 | ||

b. | 24 | ||

c. | 16 | ||

d. | 20 |

Total number of chairs available = 4

First chair can be filled up in 4 ways, second chair can be filled up in 3 ways, third chair can be filled up in 2 ways and fourth chair can be filled up in 1 way.

By counting principle,

1

4 × 3 × 2 × 1 = 24

There are 24 different ways in which Jim, Paul, Justin and Bill can be seated for the group discussion.

Correct answer : (2)

49.

What is the value of $\frac{5!}{(5-3)!\times 3!}$?

a. | 12 | ||

b. | 10 | ||

c. | 20 | ||

d. | 24 |

[Expand 3!.]

(5 - 3)! = 2!

[Simplify.]

2! = 2 × 1

[Expand 2!.]

3! = 3 × 2 × 1

[Expand 3!.]

[Substitute the values of 5!, 2! and 3!.]

= 10

[Simplify the fraction.]

The value of

Correct answer : (2)

50.

In how many different ways can the four alphabets A, B, C, and D be arranged?

a. | 16 | ||

b. | 12 | ||

c. | 36 | ||

d. | 24 |

A can take one of the 4 places, B can take one of the remaining 3 places, C can take one of the remaining 2 places and D can take the remaining one place.

By counting principle,

1

4 × 3 × 2 × 1 = 24

There are 24 arrangements of the four letters A, B, C, and D.

Correct answer : (4)