A committee of 3 men and 4 women , is to be formed from 5 men and 7 women . How many different committees can be formed if : A. a particular man must be on the seat
B .two particular must not be on the committee
C. There are no restrictions.

Topic : Permutation and combination

Part A

A particular man on the seat, then 2 men are out of 4:

4C2 = 4!/(4-2)!2! = 4*3*2/2*2 = 6

4 women out of 7:

7C4 = 7!/(7-4)!4! = 7*6*5*4!/3!4! = 35

Total combinations:

6*35 = 210Part B

Option 1. 2 men excluded

3 men out of 3 and 4 women out of 7

Total ways:

7C4 = 35 (as above)

Option 2. 2 women are excluded

3 men out of 5 and 4 women out of 5

5C3*5C4 = 5!/3!(5-3)! * 5!/4!(5-4)! = 5*4/2 * 5/1 = 10

Option 3. 1 man and 1 woman excluded

3 out of 4 men and 5 out of 6 women:

4C3*6C5 = 4!/3!(4-3)! * 6!/5!(6-5)! = 4*6 = 24 Part C

No restrictions, so the number of ways:

3C5 * 7C4 = 5!/3!(5-3)! * 35 = 4*5/2 * 35 = 10*35 = 350

I'm not sure i'm going to have to think about that but i don't have time right now sorry

Cbecause they are equal so that means it is the same on both sides? ! hope this