Question

An employer interviews 12 people for four openings at a company. Five of the 12 people are women. All 12 applicants are qualified. In how many ways can the employer fill the four positions when (a) the selection is random and (b) exactly two selections are women?

Answer 1

When the selection is random, the employer can fill the four positions in 12 choose 4 ways, which is 495 ways. When exactly two selections are women, there are 210 ways to fill the positions.

Step-by-step explanation:

(a) When the selection is random, the employer can fill the four positions in 12 choose 4 ways. To calculate this, we can use the formula for combinations: nCk = n! / (k!(n-k)!), where n is the total number of applicants and k is the number of positions to be filled. In this case, n = 12 and k = 4, so the number of ways to fill the positions is 12! / (4!(12-4)!) = 12! / (4!8!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495 ways.

(b) When exactly two selections are women, we need to choose 2 women from the 5 women applicants and 2 men from the 7 men applicants. This can be done in 5 choose 2 * 7 choose 2 ways. Using the combination formula, we have (5! / (2!(5-2)!)) * (7! / (2!(7-2)!)) = (5*4/2*1) * (7*6/2*1) = 10 * 21 = 210 ways.

Answer 2

a) The employer can fill the four positions in 495 ways.

b) The employer can fill the four positions in 210 ways.

Step-by-step explanation:

The order is not important.

For example:

Selecting John, Laura, Mary and Tre'Davious is the same as selecting Laura, John, Mary and Tre'Davious.

So we use the combinations formula to solve this problem.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

(a) the selection is random

There are 12 people, and 4 are selected. So

`C_(12,4) = (12!)/(4!8!) = 495`

(b) exactly two selections are women?

There are 7 men and 5 women. We want to select 2 men and 2 women. So

`C_(7,2)*C_(5,2) = (7!)/(2!5!)*(5!)/(2!3!) = 210`