Permutation and Combination

[vinay.pathak]

I was trying to solve this but not sure what I am doing wrong.

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women.

8C2 * 5C3 * 5C1 = 28 * 10 * 5 = 1400.

Other way

Case 1 4W 2M == 5C4 * 8C2 = 5 * 28 = 140
Case 2 3W 3M == 5C3 * 8C3 = 10 * 56 = 560
Total = Case 1 + Case 2 = 700

So the question is what am i doing wrong in 1st way?

[tim]

You're overcounting. Every situation you account for in your first example has been duplicated. When you choose one person from the five that remain after you've chosen two men and three women, you could swap that person with one of the people of that gender who had been chosen before in the gender-specific selections. The thing to keep in mind here is that if there are separate cases to consider, you are almost always safer to consider the cases separately and add them..

[vinay.pathak]

Thanks a lot for correctly pointing the issue.

[tim]

my pleasure..