Arbitrary team in moba games. Imagine you have a moba-game where are N characters and team consists from M players. You have found a team with M players and for a tournament you need to be able to take any combination of M (out of N) different characters and distribute those M characters between your team-members in a way that everyone knows how to play his character.
The question is how many characters in total your team-members have to learn for this?   
For example, if M-1 team-members know all character, the last member need to learn only N-M+1 character and your team are good. But this would take M*N-M+1 characters learned in total and most probably you can decrease this number.
If general case is to hard, then what if M=3?
 A: As Ian pointed out, a lower bound is $M(N - M + 1)$, because if any player does not know $M$ or more characters, then choosing any subset of $M$ of those unknown characters renders that player useless.
This lower bound is in fact tight.  Suppose the characters are imaginatively named with numbers, so the set of characters is $C = \{1, 2, ..., N \}$.
First we handle the case when $N \ge 2(M-1)$.  Consider the subsets $S_i = \{ i, i+1, i+2, ..., i + M - 2 \} \subset C$ (hence each $S_i$ is a set of $M-1$ characters).  Suppose the $i$th player learns all characters except those in $S_i$.  A total of $M(N-M+1)$ characters are thus learnt.
We now show that a team can be formed for any choice of $M$ characters.  Note that this entails matching each character to a player in the team that knows the character.  We do this by appealing to Hall's Marriage Theorem.  Hall's theorem says that such a matching exists if for every subset $T$ of the $M$ characters, there are at least $|T|$ players who know at least one of the characters in $T$.
Equivalently, we need to ensure that any set of $k$ players has at most $M - k$ characters none of them know; that is, for any subset $I$ of the players, $| \cap_{i \in I} S_i | \le M - |I|$.  However, this is readily seen to be true.  For a set $I = \{ i_1 < i_2 < ... < i_k \}$ of $k$ players, the set of characters none of them knows is $\cap_{i \in I} S_i = \{i_k, i_k + 1, ..., i_1 + M - 2 \}$, which has size $M - 1 - (i_k - i_1) \le M - 1 - (k-1) = M - k$, as required.
Thus by Hall's theorem, we can match each of the chosen characters to a different player, and hence form a team.
If $N < 2(M-1)$, then each player learns fewer than half the characters, so it is more convenient to focus on the $N - M + 1$ characters he or she learns.  Let $L_i = \{i,i+1,...,i+N-M\} \subseteq C$ be the characters the $i$th player learns.  One can similarly verify that Hall's condition holds in this set-up as well.
