﻿ Combinations Worksheet | Problems & Solutions

# Combinations Worksheet

Combinations Worksheet
• Page 1
1.
$\left(\begin{array}{c}9\\ 9\end{array}\right)$ = ?
 a. 9! b. 9 c. 1

#### Solution:

(99) = 9!(9 - 9)! · 9!
[(nr) = nCr = n!(n - r)!  r!.]

= 9!9! · 0!

= 1
[0! = 1.]

2.
A box contains 5 color pencils. How many different sets of 3 pencils can be chosen from it?
 a. 10 b. 4 c. 6 d. 20

#### Solution:

The number of ways of choosing 3 sets from 5 is 5C3.

= 5!(5 - 3)! · 3!
[nCr = n!(n - r)!  r!.]

= 10

10 sets of 3 pencils can be formed.

3.
Find the number of ways of selecting 2 vowels and 2 consonants from the letters of the set {I, N, T, E, R, V, A, L}.
 a. 21 b. 60 c. 13 d. 30

#### Solution:

The number of vowels in the set = 3
[The vowels in the set are I, E, A.]

The number of consonants = 5
[The consonants in the set are N, T, R, V, L.]

2 vowels can be selected from the 3 vowels in 3C2 ways.

2 consonants can be selected from the 5 consonents in 5C2 ways.

The number of ways of selecting 2 vowels and 2 consonants from the letters of the given set = 3C2 · 5C2
[Use the fundamental counting principle.]

= 3!(3 - 2)! · 2! · 5!(5 - 2)! · 2!
[Use nCr = n!(n - r)!  r!.]

= 3 · 2!1! · 2! · 5 · 4 · 3!3! · 2! = 30
[Simplify.]

The number of ways of selecting 2 vowels and 2 consonants from the letters of the given set = 30

4.
How many 4 card hands that contain exactly 3 aces and 1 queen can be chosen from a 52 card deck?
 a. 12 b. 4 c. 16 d. 8

#### Solution:

There are 4 aces and 4 queens in the 52-card deck. 3 aces can be chosen from 4 in 4C3 ways, and 1 queen can be chosen from 4 in 4C1 ways.

So, the number of 4 card hands that contain exactly 3 aces and 1 queen is (4C3)(4C1).

= 4!(4 - 3)! · 3! · 4!(4 - 1)! · 1!
[nCr = n!(n - r)!  r!.]

= 4 · 3!1! · 3! · 4 · 3!3! · 1! = 16
[1! = 1.]

5.
If ${}_{n}C{}_{2}$ = 28, then what is the value of $n$?
 a. 8 b. 7, - 8 c. - 7 d. 8, - 7

#### Solution:

nC2 = 28

n!(n - 2)!  2! = 28
[nCr = n!(n - r)!  r!.]

n(n - 1)(n Ã¢â‚¬â€œ 2)!(n - 2)!  2! = 28

n(n - 1) = 56

n2 Ã¢â‚¬â€œ n - 56 = 0

(n - 8)(n + 7) = 0
[Factor.]

n = 8 or n = - 7

n cannot be negative. So, n = 8.

6.
Find the number of permutations of the elements in the set {F, O, U, R}.
 a. 3 b. 4 c. 6 d. 24

#### Solution:

The number of elements in the set = 4

The number of permutations of 4 distinct elements of the set is 4P4

= 4! = 24
[nPn = n!.]

7.
Find the number of permutations of the elements in the set {17, 19, 21, 23, 25, 27}.
 a. 25 b. 720 c. 162 d. 150

#### Solution:

Number of elements in the set = 6

Number of permutations of 6 distinct elements of the set is 6P6.

= 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
[nPn = n!.]

8.
Find the number of 8 letter permutations from the letters of the word KEYBOARD.
 a. 40320 b. 1 c. 8 d. 64

#### Solution:

The number of letters of the word is 8.
[Write the number of letters of the word.]

The number of 8 letter permutations from the 8 distinct letters of the word is 8P8 = 8! = 40320
[nPn = n!.]

The number of 8 letter permutations from the 8 distinct letters of the word KEYBOARD is 40320.

9.
In how many ways can 2 different books be arranged on a shelf?
 a. 2! b. 3! c. 5! d. 4!

#### Solution:

The number of ways of arranging 2 different books on a shelf is 2P2

= 2!
[nPn = n!.]

10.
How many different 2 digit numbers can be formed using {2, 3, 4, 5, 6, 7, 8}, if repetition of a digit is not allowed?
 a. 120 b. 42 c. 72 d. 5040

#### Solution:

The number of permutations of 7 digits taken 2 at a time is: 7P2 = 7!(7 Ã¢â‚¬â€œ 2)!
[nPr = n!(n - r)!.]

= 7!5! = 42