﻿ Permutations and Combinations Worksheet - Page 5 | Problems & Solutions

# Permutations and Combinations Worksheet - Page 5

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

#### Solution:

Total number of letters = 6

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,
1st letter  2nd letter  3rd letter
6      x      5      x      4     =    120

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

42.
What is the value of $\frac{5!+4!}{5!-4!}$ ?
 a. $\frac{3}{2}$ b. $1\frac{1}{4}$ c. 1 d. $1\frac{1}{3}$

#### Solution:

5! = 5 ×4 ×3 ×2 ×1 = 120
[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!.]

5!+4!5!-4! = 14496
[Substitute the values of (5! + 4!) and (5! - 4!).]

5!+4!5!-4! = 32
[Simplify the fraction.]

The value of 5!+4! / 5!-4! is 3 / 2.

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

#### Solution:

Total number of players = 7

Number of ways to select 2 players from 7 players = 7C2

7C2 = 1 / 2! × 7P2
[Expand 7C2.]

2! = 2 ×1 = 2
[Expand 2!.]

7P2 = 7×6 = 42
[Expand 7P2.]

7C2 = 1 / 2 × 42
[Substitute the value of 7P2 in 7C2.]

7C2 = 21
[Simplify the product.]

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

44.
Find the number of combinations of choosing 2 fruits from 4 fruits.
 a. 6 b. 3 c. 4 d. 2

#### Solution:

Label the four fruits as 1, 2, 3 and 4.

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.

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

#### Solution:

Total number of letters = 8

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,
1st corner   2nd corner   3rd corner   4th corner
8       ×        7        ×        6       ×        5       =    1680

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

46.
Find the number of combinations of choosing 2 marbles from 3 marbles.
 a. 4 b. 6 c. 3 d. 2

#### Solution:

Label the three marbles as r, b and g.

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

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

#### Solution:

Let the 5 teams be team 1, team 2, team 3, team 4 and team 5.

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

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

#### Solution:

Total number of students participating in the group discussion = 4

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,
1st chair  2nd chair  3rd chair  4th chair
4      ×      3     ×      2     ×     1    =    24

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

49.
What is the value of $\frac{5!}{\left(5-3\right)!×3!}$?
 a. 12 b. 10 c. 20 d. 24

#### Solution:

5! = 5 × 4 × 3 × 2 × 1
[Expand 3!.]

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

2! = 2 × 1
[Expand 2!.]

3! = 3 × 2 × 1
[Expand 3!.]

5!(5-3)!×3! = 5 × 4×3×2 × 1(2×1)(3×2 × 1)
[Substitute the values of 5!, 2! and 3!.]

= 10
[Simplify the fraction.]

The value of 5!(5-3)!×3! is 10.

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

#### Solution:

Total number of letters = 4.

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,
1st place 2nd place 3rd place 4th place
4    ×    3    ×    2    ×    1    =    24

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