I could use some hlep on this one, anyone know?

Question

Assume that you have a committee of 11 members, made up of 5 men and 6 women.

In how many ways can the 3 tasks be assigned so that both men and women are given assignments?

Answer 1

There are 11P3 ways of assigning 3 tasks to 3 of the committee.

Of these, there are 5P3 ways using only men, and 6P3 using only women. All the rest of the ways must use a mixture of men and women, i.e. 11P3 - 5P3 - 6P3 which I think might be about 810.

Answer 2

I would say 135. There are two possible ways this can happen: women get two and men one or men get two and women one. In the first case, there are (5 comb 2) *(6 comb 1) ways this could happen; in the second case, (6 comb 2) *(5 comb 1). Add these both together to get 135.

Answer 3

Assumption: A task is assigned to 1 person

If so there are 11C3 ways of assigning this. We clearly dont want a situation where its all men or all women so we take off those situations (i.e., 6C3 and 5C3)

So the answer would be 11C3 - 6C3 - 5C3 = 135

Answer 4

it equals to:
number of way y may assign 1task to women(=>2 for men)+
number of way y may assign 2task to women(=>1 for men)+
=
6! / ((6-1) !* 1!) + 6! / ((6-2)! * 2!)=6+5*6/2=21

Answer 5

Do u want to subdivide your committe into three? In that case the formula is 11C3 or,
11 factorial devided by (three factorial into eight factorial into 30)

Not too sure about the 30 part.