Cheating in multiple choice tests. What is the probability that in a multiple choice test exam session, where $k$ people took the test (that contains $n$ questions with 2 possible answers each and where exactly one answer to each question is the correct one) cheating has occurred, i.e. there exists at least two tests that are identical ?
My solution is $$1-\frac{\binom{2^{n}}{k}}{\binom{2^{n}+k-1}{k}},$$ since there are $2^n$ possible tests and the set of all multisets with $k$ elements from $2^n$ is the size of all possible "exam sessions". For the numerator I counted only those exam sessions where cheating didn't occur, i.e. all tests are different from eachother. Since what I'm looking for is the complement of this set I get the above.

It is not reasonable to consider two identical tests as evidence of cheating. And people do not choose answers at random. So let us reword the problem as follows.
We have $k$ people who each toss a fair coin $n$ times. What is the probability that at least two of them will get identical sequences of heads and tails? 
The required probability  $1$ minus the probability that the sequences are all different. We go after that probability.
There are $2^n$ possible sequences of length $n$ made up of H and/or T. In order to make typing easier, and also not to get confused, let's call this number $N$. So each of our $k$ people independently produces one of these $N$ sequences. 
Write down the sequences chosen by the various people, listed in order of student ID. There are $N^k$ equally likely possibilities.
Now we count the number of ways to obtain a list of $k$ distinct sequences. This is $N(N-1)(N-2)\cdots (N-k+1)$. So the probability the sequences are all different is 
$$\frac{N(N-1)(N-2) \cdots (N-k+1)}{N^k}.\tag{A}$$
So the answer to the original problem is $1$ minus the expression in (A).
The numerator can be written in various other ways, for example as $k!\dbinom{N}{k}$.  
Remark: I prefer to think of this in elementary Birthday Problem terms. The probability that the sequence obtained by Student $2$ is different from the one obtained by Student $1$ is $\frac{N-1}{N}$. Given that fact, the probability that the sequence obtained by Student $3$ is different from both of the first two sequences is $\frac{N-2}{N}$. And so on. So the probability in (A) can be thought of as
$$\frac{N-1}{N}\cdot\frac{N-2}{N}\cdots \frac{N-k+1}{N}.$$ 
