Number of increasing integer sequences with sum $n$ (stars and bars variant) The classic stars and bars example says the number of distinct ordered $k$-tuples of non-negative integers with sum $n$, is $n+k-1$ choose $k-1$. I have thought of a slightly more difficult variant:
How many ordered $k$-tuples of non-negative integers $(x_1,..,x_k)$ have sum $n$ and satisfy $x_1<x_2<...<x_k$?
 A: You are essentially asking for $Q(n, k)$, the number of partitions of $n$ into $k$ distinct parts. Given a partition into distinct parts, we can order it into the form you want. We have to be a little careful because you said "non-negative", meaning that the first number can be zero. This gives:
$$ Q(n, k) + Q(n, k-1) $$
It is well known that:
$$ Q(n, k) = P\left(n - \binom{k}{2}, k\right) $$
where $P(n,k)$ is the number of ways to partition $n$ into $k$ parts. If you need to calculate these numbers, the following recurrence relationship may help:
$$ P(n, k) = P(n-1, k-1) + P(n-k, k) $$

Partitions are much harder to deal with than compositions, so I would be surprised if there was an easier way to express this sum.
A: Let us compute over positive integers, any "good" k-tuple can be converted to a "good" (k+1)-tuple over non-negative integers by simply prefixing it with a zero.
And to start with, let us see how it works out for a triple
As an example, suppose $a + b + c = 50, \;1\leq a <b <c$
All three numbers obviously can't be the same.
Of the $\binom{49}2 = 1176$ solutions given by stars and bars,
there will be $24\;\;$ with $\;\;2-1\;of\; a\; kind:\; 1-1-48\;\; to\;\; 24-24-0$
each with $3$ permutations
So distinct triples with $6$ permutations each $= 1176 - 3*24 = 1104$
and final answer $= \frac{1104}6 =\boxed{184}$

This obviously won't do very well as the k-tuples become larger, then you can use the generating function approach.
For the problem we just solved using stars and bars,
we can write $a+(a+b)+(a+b+c) = 50$ over the positive integers
or $3a+2b+c = 50$
Generating functions of $3a,\; 2b,\; c\;$ are respectively $\frac{x^3}{1-x^3}\,,\;\; \frac{x^2}{1-x^2}\,,\;\; \frac{x}{1-x}$
Finally, find the coefficient of $x^{50}\;$ in the expansion of
$\frac{x^3}{1-x^3}\cdot\frac{x^2}{1-x^2}\cdot \frac{x}{1-x}= \boxed{184}$

