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Permutations and Combinations Worksheet - Page 9

Permutations and Combinations Worksheet
  • Page 9
 81.  
Which of the following is true for any positive number?
a.
(n-1)! = n!- 1!
b.
n! = n
c.
n! = n x (n-1) x (n-2) x (n - 3) x . . . x 1
d.
None of the above


Solution:

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

n! = n x (n - 1) x (n - 2) x (n - 3) x ....x 1

The value of n! is n x (n - 1) x (n - 2) x (n - 3) x ... x 1.


Correct answer : (3)
 82.  
Find the value of r, if 3P2 + r x 3C2 = 0.
a.
-2
b.
-3
c.
2
d.
1


Solution:

3P2 + r x 3C2 = 0
[Original equation.]

3P2 + r x 1 / 2! x 3P2 = 0
[Expand 3C2.]

6 + r x 1 / 2 x 6 = 0
[Substitute 3P2 = 3×2 = 6 and 2! = 2 ×1 = 2.]

6 + r x 3 = 0
[Cancel 6 with 2.]

r x 3 = -6
[Subtract -6 from each side.]

r = - 2
[Divide each side by 3.]

The value of r is - 2.


Correct answer : (1)
 83.  
What is the value of the expression 60/(5P3)?
a.
2
b.
3
c.
1
d.
None of the above


Solution:

5P3 = 5×4×3 = 60
[Expand 5P3 .]

60/(5P3) = 60 / 60 = 1
[Substitute the value of 5P3 in the expression and simplify.]

The value of 60/(5P3) is 1.


Correct answer : (3)
 84.  
Evaluate: 2! x (1! + 0!)
a.
9
b.
4
c.
6
d.
5


Solution:

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

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

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


Correct answer : (2)
 85.  
In how many ways can 7 people sit in 7 chairs?
a.
5041
b.
720
c.
5040
d.
40320


Solution:

Number of ways in which 7 people can sit in 7 chairs = 7P7

= 7! = 7 ×6 ×5 ×4 ×3 ×2 ×1 = 5040
[Simplify.]

7 people can sit in 7 chairs in 5040 ways.


Correct answer : (3)
 86.  
In how many ways can 9 questions be selected out of 14 questions such that the selection always contains the questions numbered 1, 4 and 9?
a.
472
b.
452
c.
467
d.
462


Solution:

Number of questions given in the test paper = 14

Number of questions remaining after selecting questions numbered 1, 4 and 9 = 14 - 3 = 11

The remaining 6 questions have to be selected from these remaining 11 questions.

Number of ways in which 6 questions can be selected out of 11 questions = 11C6

11C6 = 1 / 6! x 11P6

6! = 6 ×5 ×4 ×3 ×2 ×1 = 720
[Expand 6! and multiply.]

11P6 = 11×10×9×8×7 = 332640
[Expand 11P6.]

11C6 = 332640 / 720 = 462
[Substitute the values of 6! and 11P6.]

The number of ways in which 9 questions can be selected out of 14 questions, always selecting questions numbered 1, 4 and 9 is 462.


Correct answer : (4)
 87.  
A teacher wants 4 of her 10 students to write a skill test. In how many ways can she select 4 students?
a.
210
b.
220
c.
215
d.
205


Solution:

Number of ways of selecting 4 students out of 10 students = 10C4

10C4 = 14! x 10P4
[Expand 10C4]

10P4 = 10×9×8×7
[Expand 10P4]

4! = 4 ×3 ×2 ×1 = 24
[Expand 4!]

10C4 = 124 x 10×9×8×7
[Substitute the values of 4! and 10P4.]

10C4 = 210
[Simplify the above product.]

The teacher can select 4 students out of 10 students in 210 ways.


Correct answer : (1)
 88.  
What is the value of 4P6?
a.
4
b.
6
c.
Not defined
d.
24


Solution:

The expression nPr can be defined as, the number of all permutations of n different things taken r at a time.

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

In 4P6, the value of r = 6 is greater than the value of n = 4.

The value of 4P6 is not defined.


Correct answer : (3)
 89.  
What is the value of (-2)!?
a.
1
b.
2
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.

(- 2)! is not defined.


Correct answer : (4)
 90.  
What is the value of (5P2)2?
a.
410
b.
800
c.
420
d.
400


Solution:

5P2 = 5×4 = 20
[Expand and multiply.]

(5P2)2 = (20)2 = 400
[Substitute the value of 5P2.]

The value of (5P2)2 is 400.


Correct answer : (4)

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