# How many ways are there to choose 31 members, out of 40 women and 40 men, while having a majority of women?

Assuming we have 2 groups of people: 40 men, and 40 women.

How many ways are there to choose 31 members (out of the 80), such that the majority is women?

When I saw the final answer, I was pretty shocked how simple it looked, even though I did not really understand how they got to it.

The way I looked at this problem is this way:

If we want to keep the women as a majority, we'll have to choose at least 16 women. So basically, the group could have {16,17,18...31} women.

Then, we have the rest of the people (aka men), to choose from. So, if we chose 16 women, we'll choose 15 men etc.

Then, I though about taking those options and sum them:

$\sum_{i=16}^{31} \binom{40}{i} \cdot \binom{40}{31-i}$

But Obviously I'm not really able to extract a real number of that.

Is there any sense of what I did? is there a simpler way to look at it?

*The final answer was actually pretty simple, but I didn't understand how they got it.

As there are equal numbers of men and women, there is a symmetry. Specifically: absent a tie, there is an equal chance than men or women will hold the majority. Thus the probability that women hold the majority is $$\frac 12 \times (1-p_t)$$ where $p_t$ denotes the probability of a tie. As the number is odd, a tie is impossible...hence the probability is simply $\frac 12$.
It follows that exactly half of the possible combinations have the desired feature, so the final answer is $$\frac 12 \times \binom {80}{31}$$
• It's correct but, as you point out, computationally unkind. Of course you could use some binomial identities to compare this to the probability that men hold the majority, but that's just the symmetry argument again. To be sure, if you say "find the number of combinations in which women hold more than $60\%$ of the positions", or anything that breaks the symmetry, then your method may well be the best. – lulu Mar 27 '16 at 14:28