There are n men and n women. Each man chooses k women and each woman chooses k men. (k is given) Woman and man meet iff both of them chose each other

Solution 1:

I don't understand why you got $E[X]= \sum_{i=0}^n i {n\choose k}(\frac{k}{n})^i(1-\frac{k}{n})^{n-i}$.

I especially don't understand why there is a leading factor of $i$ in the sum.

Here is how I would approach it.

Choose any man. Call him $m_1$. He likes $k$ the women $w_1,\ldots, w_k$.

What is the probability $\boldsymbol w_1$ also likes him?

The expected number of men met by $m_1$ is $k$ times that probability, since it is the same probability that each of $w_1,\ldots, w_k$ like him.

The total expected number of meetings is $n$ times THAT probability, since the total number of expected meetings is the same for each man.