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?

You are asking for the number of partitions of the number $50$ into parts of size at most $5$. It is equal to the number of partitions of the number $50$ into at most $5$ parts, and is the coefficient of $X^{50}$ in the power series $\prod_{i=1}^5\frac1{1-X^i}$, but there is no easy formula for such coefficients. 
There is a lot of literature about counting partitions though.
