Hi LUANDATO,

We're told to choose three competitors from a group of five men and three women. We're asked for the number of different trios that include AT LEAST one woman. The answer choices to this question are 'spaced out' enough that you can do a little bit of work to eliminate all of the wrong answers.

To start, if the group has just 1 woman, then there will be 2 men. The number of groups of 2 men is 5c2 = 5!/(2!)(3!) = 10 different pairs of men.

With 3 women to choose from, there would be 3(10) = 30 groups with just 1 woman. There would then be additional groups with 2 women or all 3 women, so the total number of groups MUST be greater than 30. There's only one answer that's possible...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorLU wrote:The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios content at least one woman?

A. 10
B. 15
C. 16
D. 30
E. 46

The OA is E.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

We can use the formula:

Total number of ways to select the group - number of ways with no women = number of ways with at least one woman.

Total number of ways to select 3 people from a group of 8 people:

8C3 = 8!/[3!(8-3)!] = 8!/3!5! = (8 x 7 x 6)/3! = 56

Number of ways with no women:

5C3 = 5!/[3!(5-3)!] = (5 x 4 x 3)/3! = (5 x 4 x 3)/(3 x 2 x 1) = 10

Thus, the number of ways with at least one woman is 56 - 10 = 46.

Answer: E