University clubs - Counting two ways 
Consider a university with 2000 male and 2000 female students. Suppose that none of the 4000 students signed up for 100 or more clubs (Each student signed up for at most 99 clubs). You also know that each pairing of male,female student shares a club, that they're both signed up for.
Decide if it's possible for every club to have either at most 10 male students or at most 10 female students, or if there must exist at least one club, for which at least 11 male and at least 11 female students signed up.

The general idea I guess would be to count the number of male/female pairs $(2000^2?)$ and then try to get the number of clubs utilizing counting two ways.
I believe that once I get the number of clubs I should be able to verify if it's possible to reach the desired number of pairings with the limitation for male/female students (at most 10 of one or the other). But I'm not sure how to proceed to actually get the number of clubs.
 A: To minimize the number of members in clubs, assume that for any man-woman pairing, there exists only for one club that has both of them as members.
Define the matrix with elements $C_{mw}$ being the club id that has man $m$ and women $w$ both as members. All elements of the matrix need to be filled, and the question asks whether the number of different ids in any row or column can be less than 100 when the number of men or women in all clubs is 10 or less.
Without loss of generality, we can populate the upper 10 rows of the matrix with club id = 1. There are 10 men and 2000 women in club #1. There is no benefit in breaking that block into 2 or more clubs.
If we continue and populate the next 10 rows with club id = 2, and repeat, then the women will all belong to 200 clubs, violating the requirement. Instead, the leftmost 10 columns could be filled with club id=2, corresponding to 10 women and 2000 men. 
With both, the upper left block of 10x10 has man-woman pairing for two clubs. We can eliminate one of the clubs. The most sparse solution (reducing number of clubs everyone belongs to) leaves a checker-board pattern for the matrix.
$$
\begin{bmatrix}
1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 4 &  ... \\
2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 &  ... \\
1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 4 &  ... \\
2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 &  ... \\
1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 4 &  ... \\
2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 &  ... \\
1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 4 &  ... \\
2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 &  ... \\
1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 4 &  ... \\
2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 &  ... \\
3 & 2 & 3 & 2 & 3 & 2 & 3 & 2 & 3 & 2 & 3 & 4 &  ... \\
2 & 3 & 2 & 3 & 2 & 3 & 2 & 3 & 2 & 3 & 2 & 3 &  ... \\
\end{bmatrix}
$$
This is a solution in which every club has either 10 men and 1000 women or 10 women and 1000 men and everyone belongs to 100 clubs. It does not appear that there would be a solution where everyone belongs to at most 99 clubs.
