2 corrected exams are being returned to each of n students. How many ways can the teacher give those 2 exams back to each student such that everyone receives at least 1 exam that is not his.
I know how to do this if 1 exam is being returned to each student, but this question says everyone receives at least 1 bad exam out of 2 exams, can anyone please enlighten me on how to achieve this?
What I have tried: Let each 2 bad exams be a and b. I used the property |a ∪ b| = |a| + |b| + |a ∩ b|, and applied the derangement formula for each of the term. Am I on the right track?
Thanks a lot!
Instead, find the probability of the opposite event, that there is at least one student who got both their tests back. Letting $A_i$ be the event that the $i$th student got both their tests back, this means finding $P(\bigcup_1^n A_i)$, which can be done using inclusion exclusion.
We have $$ P(A_i)=\frac{2}{(2n)(2n-1)} $$ since there is a $\frac{2}{2n}$ chance the first exam student $i$ receives is one of hers, and a $\frac1{2n-1}$ chance the other one is as well. Similarly, $$ P(A_i\cap A_j) = \frac{2}{(2n)(2n-1)}\cdot\frac{2}{(2n-2)(2n-3)} $$ and in general, $$ P(A_{i_1}\cap\dots\cap A_{i_k})=\frac{2^k(2n-2k)!}{(2n)!} $$ Thus, the inclusion exclusion principle gives $$ P\left(\bigcup_1^n A_i\right) = \sum_{k=1}^n(-1)^{k+1}\binom{n}{k}\cdot\frac{2^k(2n-2k)!}{(2n)!} $$