Combinatorics: Distributing tasks in class Suppose you sit in a class room with $n$ students. There are $k$ question and every student can pick $5$ of them. How many different ways are there such that no question is left out?
In total we have $5\cdot n$ decisions. This is a really hard question, I tried simulating some cases but I don't see a common pattern.
 A: There are $\binom{k}5$ different sets of $5$ questions, so there are $\binom{k}5^n$ different ways for each of $n$ students to choose a set of $5$ questions. From this figure we need to subtract the number of ways that leave out at least one question. For any question $Q$ there are $\binom{k-1}5$ sets of $5$ questions that do not include $Q$, so $\binom{k-1}5^n$ of the $\binom{k}5^n$ ways leave out $Q$. Since there are $k$ questions, this might suggest that the answer is $$\binom{k}5^n-k\binom{k-1}5^n\,,\tag{1}$$ but that fails to take into account the fact that every combination of choices that leaves out two questions has been subtracted twice, once for each of the omitted questions, and is therefore counted $-1$ times in $(1)$. There are $\binom{k}2$ pairs of questions, and there are $\binom{k-2}5^n$ ways for the class to leave out a specific pair of questions, so in $(1)$ we’ve subtracted $\binom{k}2\binom{k-2}5^n$ too much; restoring it gives us
$$\binom{k}5^n-k\binom{k-1}5^n+\binom{k}2\binom{k-2}5^n\,.\tag{2}$$
We’re not done, since any set of class choices that leaves out three questions is counted $1-3+3=1$ time in $(2)$, and it shouldn’t be counted at all. Can you complete the inclusion-exclusion calculation from here?
