A group consists of six men and seven women. Four people are selected to attend a conference.

Question

a. In how many ways can four people be selected from this group of thirteen​?

b. In how many ways can four women be selected from the seven women?

c. Find the probability that the selected group will consist of all women.

Answer 1

Parts (a) and (b) are combination questions. In a combination, the order of the items does not matter. Read carefully so that you can see that.

The formula is: _nC_r = n!/(r!(n-r)!), where n is the size of the total group and r is the size of the subgroup.

So for part (a) we have ₁₃C₄ = 13!/(4!(13 - 4)!) = 13!/(4! • 9!) = (13 • 12 • 11 • 10 • 9!)/(4 • 3 • 2 • 1 • 9!) = 715

~ 715 ways to pick 4 from 13 if order doesn't matter.

For part (b) we have ₇C₄ = 7!/(4!(7 - 4)!) = 7!/(4! • 3!) = (7 • 6 • 5 • 4!)/(4! • 3 • 2 • 1) = 35

~ 35 ways to pick 4 from 7 if order doesn't matter.

Part (c) is a probability question.

Probability is a ratio: (desired outcomes)/(possible outcomes).

In this case, we can use the previous answers:

(b)/(a) = 35/715 = .04895

~ The probability is .04895, or there is a 4.895% chance of choosing all women from this group.