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.

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.

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.

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!).]

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.

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.

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.

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.

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.

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.