combination problems?

1) If A and B are selected for the committee, there are 4 slots still to be filled and 8 people available to fill them. For the first slot, we can select from the 8. Once that position is filled, we have 7 people left for the next, etc. So, having pre-selected A and B, we have 8 * 7 * 6 * 5 ways to pick the rest of the group. BUT, some of those are the same. We could have selected C first and D next or D first and C next. Either way, we would have the same people. So we have to divide by 4*3*2*1.
Use the same method and decide how people can be selected if A is not included but B is....Show more

1.
i. Choose remaining 4 committee members from remaining 8 people: C(8,4)=8!/4!4! choices.
ii. If B is in the committee then choose remaining 5 committee members from the remaining 8 people which exclude A: C(8,5)=8!/3!8! choices.
[understood as: "choices such that B is included and A is excluded". If instead this means "choices such that A and B do not sit on the committee together" then add on the C(8,5) ways in which the committee contains A but not B and the C(8,6) ways in which the committee contains neither A nor B.]

2.
Take the product of the choices of "3 men from 6" with "4 women from 7": C(6,3)×C(7,4)=(6!/3!3!)×(7!/3!4!) choices.
If one of the 6 men and two of the 7 women are already on the committee then there are 2 male and female places left and 5 men and 5 women left. Take the product of the choices: "2 men from remaining 5" and "2 women from remaining 5". C(5,2)²=(5!/3!2!)² choices.

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C(n,r)=n!/(n-r)!r!
n!=n×(n-1)×(n-2)×...×3×2×1...Show more

2)
3 men out of 6 men in 6C3 = 20 ways
4 women out of 7 women in 7C4 = 35 ways
20 x 35 = 700 ways

If 1 man and 2 women were always included, then the other 2 men from 5 men in 5C2 = 10 ways
and the other 2 women in 5C2 = 10 ways
10 x 10 = 100 ways
Proportion = 100/700 = 1/7...Show more