SOLUTION: a committee of 5 people must be selected from 5 men and 8 women. 13 people in ALL. How many ways can a selection be done if there are at least 3 women on the committee?

Question 1019009: a committee of 5 people must be selected from 5 men and 8 women. 13 people in ALL.
How many ways can a selection be done if there are at least 3 women on the committee?
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
a committee of 5 people must be selected from 5 men and 8 women. 13 people in ALL.
How many ways can a selection be done if there are at least 3 women on the committee?

Solution:
There are three cases:
1. 3 women and two men
C(5,2)*C(8,3)=560 ways
2. 4 women and 1 man:
C(5,1)*C(8,4)=350 ways
3. 5 women and no men:
C(5,0)*C(8,5)=56 ways
Total=560+350+56=966 ways.

Following is the incorrect way to solve the problem:
Choose 3 women and two more from the rest:
C(8,3)*C(10,2)=2520 >>966
This way, we're overcounting, since some of the women chosen out of the remaining 10 could have been counted already, for example, ABC were chosen first, and G was chosen next. But then it would be the same if ABG was chosen first, and C was chosen later.

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