﻿ Permutations and Combinations Worksheet - Page 8 | Problems & Solutions

# Permutations and Combinations Worksheet - Page 8

Permutations and Combinations Worksheet
• Page 8
71.
What is the value of (-3)!?
 a. 1 b. 6 c. Not defined

#### Solution:

The factorial of a number is the product of all the whole numbers from 1 to that number.

The range of whole numbers is {0, 1, 2, 3, 4....}.

Negative numbers do not come in the range of whole numbers.

The factorial of a negative number is not defined.

(- 3)! is not defined.

72.
Marissa has 8 coats, 6 pair of shoes, 5 jewelry sets, and 3 watches. State whether the way she can dress up is a permutation or a combination.
 a. Combination b. Permutation

#### Solution:

A permutation is an arrangement in which order is imporatnt and a combination is an arrangement in which order does not matter.

Marissa selects one from 8 coats, one from 6 pair of shoes, one from 5 jewelry sets, and one from 3 watches to dress up.

As order of selection does not matters, selection is always a combination.

So, the way Marissa can dress up is a combination.

73.
A detective wants to break a 3-lettered code, which has no repetition of alphabets. State whether the arrangement of alphabets in the code is a permutation or a combination.
 a. Combination b. Permutation

#### Solution:

A permutation is an arrangement in which order is important and a combination is an arrangement in which order does not matter.

As different orders of same letters form different words, ordering of alphabets is important in finding the code.

So, the arrangement of alphabets in the code is a permutation.

74.
In how many ways can you distribute a king, a queen and a jack card to three persons?

 a. 2 b. 3 c. 1 d. 6

#### Solution:

Number of cards to distribute = 3

The list of all possible distributions:

1st Person  2nd Person  3rd Person

King          Queen          Jack

Jack          King          Queen

King          Jack          Queen

Jack          Queen          King

Queen          Jack          King

Queen          King          Jack

Total number of possible distributions = 6

A king, queen and a jack card can be distributed to 3 persons in 6 different ways.

75.
What is the value of $\frac{5!}{3!×2!}$ ?
 a. 10 b. 6 c. 2 d. 30

#### Solution:

5! = 5 ×4 ×3 ×2 ×1
[Expand 5!.]

5! = 5×4×3 × 2!
[Simplify 5!.]

3! = 3 ×2 ×1
[Expand 3!.]

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

= 5×4×33 ×2 ×1
[Divide the numerator and the denominator by 2!.]

= 10
[Simplify the fraction.]

The value of 5! / 3!×2! is 10.

76.
What is the value of $\frac{n!}{n}$?
 a. $n$ b. 1! c. ($n$ - 1)! d. $n$!

#### Solution:

n! = n x (n - 1) x (n - 2) x .... x 2 x 1
[Expand n!.]

n!n = (n x (n - 1) x (n - 2) x .... x 2 x 1)/n
[Substitute the value of n!.]

n!n = (n - 1) x (n - 2) x .... x 2 x 1
[Simplify.]

(n - 1) x (n - 2) x .... x 2 x 1 = (n - 1)!
[Expansion of (n - 1)!.]

The value of n! / n is (n - 1)!.

77.
What is the value of 2C5?
 a. 5 b. 2 c. Not defined d. 7

#### Solution:

The expression nCr can be defined as, arranging n things taking r at a time in which order does not matter.

The value of r must be less than or equal to the value of n.

In 2C5, the value of r is 5, which is greater than the value of n given as 2.

The value of 2C5 is not defined.

78.
How many meaningful 4-letter words are there in the list of all possible arrangements of the letters in the word EARN, if each letter is used only once?
 a. 5 b. 1 c. 3 d. 2

#### Solution:

The possible arrangements of the letters in the word EARN are

The meaningful words in the above list are EARN and NEAR.

There are 2 meaningful words in the list of all possible arrangements of the letters in the word EARN.

79.
What is the value of 9P2 + 9C2?
 a. 108 b. 36 c. 118 d. 72

#### Solution:

9P2 = 9×8 = 72
[Expand and multiply.]

9C2 = 12! x 9P2

9C2 = 12 x 72
[Substitute the values of 2! and 9P2.]

9C2 = 36
[Simplify the product.]

9P2 + 9C2 = 72 + 36 = 108
[Substitute the values of 9P2 and 9C2.]

80.
What is the value of (4P2)2?
 a. 4 b. 144 c. 12 d. 156

#### Solution:

4P2 = 4×3 = 12
[Expand and multiply.]

(4P2)2 = (12)2 = 144
[Substitute the value of 4P2.]

The value of (4P2)2 is 144.