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Permutations and Combinations Worksheet - Page 4

Permutations and Combinations Worksheet
  • Page 4
 31.  
What is the value of the expression 1805P3?
a.
4
b.
3
c.
5
d.
None of the above


Solution:

5P3 = 5×4×3 = 60
[Expand 5P3 .]

1805P3 = 180 / 60 = 3
[Substitute the value of 5P3 in the expression and simplify.]

The value of 1805P3 is 3.


Correct answer : (2)
 32.  
What is the value of 5!?
a.
5
b.
124
c.
120
d.
11
e.
13


Solution:

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

5! = 5 × 4 × 3 × 2 × 1
[Expand 5!.]

5! = 120
[Multiply.]

The value of 5! is 120.


Correct answer : (3)
 33.  
Find the number of ways of choosing 4 members from a team of 15 members.
a.
179
b.
1365
c.
1265
d.
169


Solution:

Number of ways of choosing 4 members from 15 members = 15C4

15C4 = 1 / 4! × 15P4
[Expand 15C4.]

15P4 = 15 × 14 × 13 × 12
[Expand 15P4 .]

4! = 4 × 3 × 2 × 1 = 24
[Expand 4!.]

15C4 = 1 / 24 × (15 × 14 × 13 × 12)
[Substitute the values of 4! and 15P4 .]

= 15 × 7 × 13 = 1365
[Simplify.]

Number of ways of choosing 4 members from a team of 15 members = 1365


Correct answer : (2)
 34.  
What is the value of 2! + 5!?
a.
127
b.
124
c.
122
d.
None of the above


Solution:

2! = 2 ×1 = 2
[Expand 2! and multiply.]

5! = 5 ×4 ×3 ×2 ×1 = 120
[Expand 5! and multiply.]

2! + 5! = 2 + 120 = 122
[Substitute and add.]

The value of (2! + 5!) is 122.


Correct answer : (3)
 35.  
In how many ways can 7 books be arranged in a shelf?
a.
8!
b.
7C7
c.
77
d.
7!


Solution:

Number of books = 7

Number of ways in which 7 books can be arranged = 7P7

7P7 = 7!

There are 7! ways to arrange 7 books in a shelf.


Correct answer : (4)
 36.  
Evaluate: 4! × (1! + 0!)
a.
53
b.
50
c.
48
d.
49


Solution:

4! = 4 ×3 ×2 ×1 = 24
[Expand 4! and multiply.]

1! + 0! = 1 + 1 = 2
[Substitute 1! = 1 and 0! = 1.]

4! x (1! + 0!) = 24 x 2 = 48
[Substitute the values of 4! and (1! + 0!).]


Correct answer : (3)
 37.  
What is the value of 9!(9-2)!?
a.
75
b.
72
c.
66
d.
77
e.
144


Solution:

9! = 9 × 8 × 7!
[Expand 9!.]

(9 - 2)! = 7!
[Simplify.]

9!(9-2)! = 9 × 8 × 7!7!
[Substitute the values of 9! and (9 - 2)!.]

= 9 × 8 = 72
[Simplify.]

The value of 9! / (9-2)! is 72.


Correct answer : (2)
 38.  
Marissa has 5 different toys and wants to display 3 of them. In how many ways can they be selected and arranged?
a.
30
b.
20
c.
55
d.
60


Solution:

Number of ways of selecting 3 out of 5 toys = 5C3

Number of ways in which 3 toys can be arranged = 3!

Number of ways in which 3 out of 5 toys can be selected and arranged = 5C3 × 3! = 5P3

= 5! / (5-3)! = 5! / 2! = 60
[Simplify.]

There are 60 different ways to select 3 of the 5 toys and arrange them for display.


Correct answer : (4)
 39.  
Find the number of arrangements of the letters of the word MATH.
a.
12
b.
24
c.
9
d.
10


Solution:

The word 'MATH' has 4 letters.

There are 4 letters to put in the first place, 3 letters to put in the second place, 2 letters to put in the third place and 1 letter to put in the fourth place.

By counting principle, number of arrangements = 4 × 3 × 2 × 1 = 24.

The letters of the word 'MATH' can be arranged in 24 different ways.


Correct answer : (2)
 40.  
A 13-member club is choosing its President, Vice president, Secretary and Treasurer. Find the number of ways in which they can be chosen.
a.
156
b.
1320
c.
1716
d.
17160


Solution:

President is to be chosen from 13 members, Vice president is to be chosen from 12 members, Secretary is to be chosen from 11 members and Treasurer is to be chosen from 10 members.

From the counting principle, the four people can be chosen in 13 x 12 x 11 x 10 ways.

= 17160
[Multiply.]

There are 17160 different ways in which they can be chosen.


Correct answer : (4)

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