hypedaj wrote:How many 5 person committees chosen at random from a group consisting of 3 men, 5 women, and 2 children contain at least 1 woman?

A) 250
B) 251
C) 252
D) 275
E) 300

Could someone explain to me how to solve this problem using Brett's FCP method? Thanks!

Shorter way to answer for at least 1 woman is to take 'no woman' number of ways out from all possible ways of choosing 5 out of a total of 10 people.

Total First

Make 5 blanks

_ _ _ _ _

Now, since first person can be chosen in 10 number of ways, second person can be chosen in 9 number of ways, and so on, hence place those options in order, over those blanks

(10)(9)(8)(7)(6)
_ _ _ _ _

Since order doesn't matter in selection process and we've 5 blanks up there, therefore we must divide the above product by (5)(4)(3)(2)(1) in order to get rid of repeat selections.

(10)(9)(8)(7)(6)
_ _ _ _ _ = 252 ways.
(5) (4)(3)(2)(1)

No Woman Next

'No woman' means either man or child and there are exactly 5 persons left if we don't consider women, hence there is exactly 1 way of forming a 5-person committee out of 5 persons.

Finally

Number of 5-person committee containing at least 1 woman = [spoiler]252 - 1 = 251

Pick B[/spoiler].