Deligne-Simpson problem in the symmetric group Question.
Let $C_1,\dots,C_k$ be conjugacy classes in the symmetric group $S_n$. (More explicitly,
each $C_i$ is given by a partition of $n$; $C_i$ consists of permutations whose cycles
have the length prescribed by the partition.) Give a necessary and sufficient condition on
$C_i$ that would ensure that there are permutations $\sigma_i\in C_i$ with
$$\prod\sigma_i=1.$$
Variant.
Same question, but now $\sigma_i$'s are required to be irreducible in the sense that
they have no common invariant proper subsets $S\subset\lbrace 1,\dots,n\rbrace$.
I am not certain how hard this question is, and I would appreciate any comments or observations. (I was unable to find references, but perhaps I wasn't looking for the right things.)
This question is inspired by Jonah Sinick's question via the simple
Geometric interpretation.
Consider the Riemann sphere with $k$-punctures $X=\mathbb{CP}^1-\lbrace x_1,\dots,x_k\rbrace$. 
Its fundamental group $\pi_1(X)$ is generated by loops $\gamma_i$ ($i=1,\dots,k$) subject to the relation
$$\prod\gamma_i=1.$$
Thus, homomorphisms $\pi_1(X)\to S_n$ describe degree $n$ covers of $X$, and the problem
can be stated as follows: Determine whether there exists a cover of $X$ with prescribed
ramification over each $x_i$. The variant requires in addition the cover to be irreducible.
Background.
The Deligne-Simpson Problem refers to the following question: 
Fix conjugacy classes $C_1,\dots,C_k\in\mathrm{GL}(n,\mathbb{C})$ (given explicitly by $k$ Jordan forms). What is the necessary and sufficient condition for existence of matrices
$A_i\in C_i$ with $$\prod A_i=1$$
(variant: require that $A_i$'s have no common proper invariant subspaces)?
There are quite a few papers on the subject; my favorite is Simpson's paper, which has references to other relevant papers. The problem has a very non-trivial solution (even stating the answer is not easy): first there is a certain descent procedure (Katz's middle convolution algorithm) and then the answer is constructed directly (as far as I understand, there are two answers: Crawley-Boevey's argument with parabolic bundles, and Simpson's construction using non-abelian Hodge theory).
The same geometric interpretation shows that the usual Deligne-Simpson problem is
about finding local systems (variant: irreducible local systems) on $X$ with prescribed local monodromy. 
So: any remarks on what happens if we go from $\mathrm{GL}(n)$ to $S_n$?
 A: This question is a more than 100 years old problem and it is called in the litearture "Hurwitz exitence problem". This is an open problem. Though many partial cases are solved. For example you can check the following article
On the existence of branched coverings between surfaces with prescribed branch data, I
Ekaterina Pervova, Carlo Petronio
http://arxiv.org/abs/math/0508434
Of course, you can give an obvious formal answer, (in the first case that is not irreducible) that the cover exists if and only if the product of the elements in the group algebra of $S_n$ corresponding to the permutation that you chose contains $1$ in their decomposition. But this is just a reformulation of the problem.
Here is a different example of a recent article on Hurwitz existence problem, it contains in partcular a lot of references on the research in this topic.
Solution of the Hurwitz problem for Laurent polynomials
http://arxiv.org/abs/math/0611776
Also notice that there is a whole branch of math nowadays where people try to compute the actual number of ramified covers, and not only to answer the quesiton wheather a cover exists or not.  Here is a typical example 
Gromov-Witten theory, Hurwitz theory, and completed cycles
Authors: Andrei Okounkov, Rahul Pandharipande
http://arxiv.org/abs/math/0204305
A: (wanted to make this a comment but couldn't figure how)
As Dmitri says, you can express the number of ramified covers using the group algebra of S_n, not just whether it's nonzero; once you do this, an inclusion/exclusion argument based on splitting up the conjugacy classes into smaller partitions lets you figure out the number of these that are irreducible.  this is an in-principle answer that's of course totally useless in practice
