In how many ways can we form a committee of positive size from $7$ women, $4$ men so that there are at least $2$ women in the committe?

Question

In how many ways can we form a committee of positive size from $7$ women, $4$ men so that there are at least $2$ women in the committee?

So the committee size must be $\geq 2$ and $\leq 11$ since it must contain at least 2 women.

for each size, we can have different combinations and it's really looking complex to me. Please help.

Answer 1

There are $2^{11}$ ways to form a committee.

There are $7 \cdot 2^4$ ways to form a committee with exactly one woman.

There are $2^4$ ways to form a committee with exactly zero women.

The answer is $2^{11} - 7 \cdot 2^4 - 2^4$.

Answer 2

On a case by case basis, if the no. of women in the committee is $k\leq7$, then there are $\binom7k$ ways to select which of them are in the committee. Corresponding to that, there are $2^4$ ways to assign men to that committee.

You sum the expression for $k=2,\ldots,7$.

Answer 3

First assume 2 women are in the committee, now there are 5 women and 4 men left. We can now do casework for the rest. 1 person:2 | 2 people:3 | 3 people:4 | 4 people:5 | 5 people:5 | 6 people:4 | 7 people:3 | 8 people:2 | 9 people:1. The answer is 29 combinations.

Note that each element is NOT distinct, only men and women are(if i'm reading this correctly).