A variation of how many integer solutions are there for $a+b+...=n$ There was a combinatorics question last night that I thought I could answer, turns out I couldn't (so I discarded my draft) but I thought I could, I now want to find out if I got close.
The question was about splitting 50 things into any number of groups of 1,2,3,4 or 5 items. 
If I were asked how many integer solutions there were to $A+B+C+D+E=10$ say I'd be quite happy to conclude that you "encode" any such solution using... say X and -s, so "XXX-XXX--XXX-X" denotes 3As, 3Bs, 0Cs 3Ds and 1 E, any arrangement of 10Xs and 4 dashes is a solution. So it is simply $\frac{14!}{10!4!}$ using the axiom of choice and such, I am happy with this work.
You can also modify it to work out how many integer solutions there are when any number is greater than a certain value (by substituting something like $A=a+1$ then A>=0 means a>=1).
But I cannot use this to solve A+2B+3C+4D+5E=n, at least not directly, I had a play on paper but I could not find anywhere to go with this, was I on the right lines or is this a totally different problem?

$A+2B+3C+4D+5E=n$ is the same problem, and it says to divide n in groups of 
multiple of 1's, multiple of 2's ...upto multiple's of 5's
Which is given by coefficient of $x^n$ in the expansion of
$$(1+x+x^2+\cdots)\cdot(1+x^2+x^4+\cdots)\cdot(1+x^3+x^6+\cdots)\cdot(1+x^4+x^8+\cdots)\cdot(1+x^5+x^{10}+\cdots)$$
$$={(1-x)^{-1}}\cdot(1-x^2)^{-1}\cdot(1-x^3)^{-1}\cdot(1-x^4)^{-1}\cdot(1-x^5)^{-1}$$
