Chapter 9.6, Problem 66E

(a)

To determine

To Find: How many ways can the employer fill the four positions when the selection is random.

Answer to Problem 66E

  $495$ Ways

Explanation of Solution

Given:An employer interviews $12$ people for four openings at a company. Five of the $12$ people are woman. All the $12$ applicants are qualified.

Given:

Total applicants= $12$

Number of women= $5$

Number of men= $7$

Total positions= $4$

If the selection is random

Then

  $\begin{array}{l}C{}_4\\ =\frac{12!}{(12-4)!4!}\\ =\frac{12\times 11\times 10\times 9\times 8!}{8!\times 4\times 3\times 2\times 1}\\ =495\end{array}$

Hence, $495$ ways

(b)

To determine

To Find: How many ways can the employer fill the four positions when exactly two selections are women.

Answer to Problem 66E

  $210$ Ways

Explanation of Solution

Given: An employer interviews $12$ people for four openings at a company. Five of the $12$ people are woman. All the $12$ applicants are qualified.

Given:

Total applicants= $12$

Number of women= $5$

Number of men= $7$

Total positions= $4$

If the selection haveexactly two selections are women.

Then

  $\begin{array}{l}C{}_2\times C{}_2\\ =\frac{5!}{(5-2)!2!}\times \frac{7!}{(7-2)!2!}\\ =\frac{5\times 4\times 3!}{3!\times 2\times 1}\times \frac{7\times 6\times 5!}{5!\times 2\times 1}\\ =10\times 21\\ =210\end{array}$

Hence, $210$ ways