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.
 A: 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
$$\frac{N-1}{N}\cdot\frac{N-2}{N}\cdots \frac{N-k+1}{N}.$$ 
