1. ## Permutations

I encountered this question in some past papers & was wondering if anyone could guide me with it:

A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

Any help would be greatly appreciated.

2. Originally Posted by callie
I encountered this question in some past papers & was wondering if anyone could guide me with it:

A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

Any help would be greatly appreciated.
1) 4 if i were to answer you question.
but i think you mean how many different ways of chosing the committees.

so it would be 12C4 = 495.

2) same again 4 if i were to answer you question.

but the no. of different ways is 8C2 x 6C2 = 420.

3. Originally Posted by callie
I encountered this question in some past papers & was wondering if anyone could guide me with it:

A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

Any help would be greatly appreciated.
You can choose any one of the women to be on the committee. There are, of course, 8 ways to do that. Now you can choose any one of the remaining 7 women to be the second woman on the commitee. There are 7 ways to do that and so 8*7 ways to make those choices. But that counts "Betty first then Sarah second" as different from ""Sarah first, then Betty" which you don't want to do. So divide 2 since for each pair of women there are two orders.

Do the same thing for the two men.