If we are given to find the number of ways 10 positive integers can sum to 50, we simply find the coefficient of $x^{100}$ in $(x+x^2+...+x^{90})^{10}$, which turns out to be $\binom{99}{90}$.
But what do we do if we are asked to find the ways if the numbers are distinct? How do we proceed then? Can we use a generating functions approach? I have thought a lot, but didn't cup up with anything doable. Please help me out.
A generating function approach is possible, but first let's note that if $q(n,k)$ denotes the number of partitions of $n$ into exactly $k$ distinct parts and $p(n,k)$ denotes the partition of $n$ into exactly $k$ (not necessarily distinct) parts, then:
$$ q(n,k) = p\left(n - \binom{k}{2}, k\right) $$
For details see this Answer to the previously asked Partition an integer $n$ into exactly $k$ distinct parts, which however does not go into the generating function aspect.
We are therefore able to translate the generating function for $p(n,k)$ already noted in the Question to one for $q(n,k)$. In particular, since $q(100,10) = p(55,10)$, this should be the coefficient of $x^{55}$ in:
$$ \frac{x^{10}}{(1-x)(1-x^2)\ldots(1-x^{10})} $$
According to Henry Bottomley's Partition and composition calculator (may require browser permissions for Java applet to run), this number is $33401$.
Since the generating function for $p(n,k)$ is:
$$ P_k(x) = \sum_{n=k}^{\infty} p(n,k)x^n = \frac{x^k}{(1-x)(1-x^2)\ldots(1-x^k)} $$
the generating function for $q(n,k)$ is:
$$ Q_k(x) = \sum_{n=k(k+1)/2}^{\infty} p\left(n - \binom{k}{2},k\right)x^n $$
$$ = x^{\binom{k}{2}} \sum_{m=k}^{\infty} p\left(m,k\right)x^m $$
$$ = \frac{x^{k(k+1)/2}}{(1-x)(1-x^2)\ldots(1-x^k)} $$
The same conclusion is reached at Partitions Into Distinct Parts (MathPages).
