Random Sum of random variables: Without replacement I have the numbers between 1 and 100 in an urn. I randomly draw $h$ many ($h$ fixed parameter) of them, without replacement. My random variable of interest is their sum. 
All approaches I could think of are iterative logic ("If drawn this, then draw that second time,...") and become unmanageable when $h$ is large. Is there a generic way to compute the pdf of this RV? 
I'm not familiar with characteristic functions, so I'd appreciate an approach more if it wouldn't require those. 
 A: Not an answer but on simulation with $h = 50$, the pmf plot is,

Note that the sample space is integer values not real values.
Edit 1: 
I am not sure whether a closed form solution for the pmf can be obtained. But here is an approach I tried. Since the order of samples doesn't matter when the sum is taken, therefore,
$$x_1 + x_2 + \ldots, x_h = x_{(1)} + x_{(2)} + \ldots, x_{(h)}$$
where $x_{(i)}$ is the $i$th order statistic.
Let us fix the sum to be $S$. We need to find the number of possible combinations $x{(i)}$'s such that,
$$x_{(1)} + x_{(2)} + \ldots, x_{(h)} = S$$
subject to
$$1 \leq x_{(1)} < x_{(2)} < \ldots < x_{(h)}$$
Introduce $a_1, a_2, \ldots a_{h-1}$ such that $a_i \geq 1$ and $\forall i\neq 1$,  $x_{(i)} = x_{(1)} + a_{1} + a_{2} + \ldots + a_{i-1}$. Then,
$$x_{(1)} + (x_{(1)} + a_{1}) + (x_{(1)} + a_{1} + a_{2}) + \ldots, (x_{(1)} + a_{1} + a_{2} + \ldots + a_{h-1}) = S$$
$$\implies hx_{(1)} + (h-1)a_{1} + \ldots + 2a_{h-2} + a_{h-1} = S$$
Take $y_1 = x_{(1)}-1$ and $\forall i\geq 2, y_i = a_{i-1}-1$. Then, our problem reduces to finding the combinations of $y_i$'s, such that,
$$hy_{1} + (h-1)y_{2} + \ldots + 2y_{h-1} + y_{h} = S - \frac{h(h+1)}{2}$$
subject to $y_i \geq 0$. The number of combinations will be equal to the coefficient of $z^{S - \frac{h(h+1)}{2}}$ in 
$$(1-z^{h})^{-1}(1-z^{h-1})^{-1}\ldots(1-z)^{-1}$$
Now, I am not sure how to proceed further i.e. whether a closed form solution of the above coefficient can be obtained. If the value of coefficient is $C_S$, then $P(x_1+x_2+\ldots+x_h=S) = \frac{C_S}{\binom{100}{h}}$.
Edit 2:
$C_S$ can be calculated computationally using the recurrence given in this wiki link but there is no general formula to compute it.
A: This is a partial answer representing maybe the most naive approach (no tricks).
Call $X_1,\dots, X_h$ your draws and $S_h = X_1 + \dots + X_h.$ Now 
\begin{align}
p(S_h=n) &= h!\sum_{x_1,\dots x_h} \mathbf{1}(1\leq x_1 < \dots < x_h \leq 100)\mathbf{1}( \textstyle\sum x_i =n)p(x_1,\dots,x_h) \\
&=h!\frac{(100-h)!}{100!} \sum_{x_1,\dots x_h} \mathbf{1}(1\leq x_1 < \dots < x_h \leq 100)\mathbf{1}( \textstyle\sum x_i =n) \\
&= {100 \choose h}^{-1}\left(\text{partitions of }n\text{ with }h\text{ unique parts of size at most 100}\right).
\end{align}
For $n\leq 100 + (\textstyle\sum_{k=1}^{h-1}k)=100+\frac{h\cdot(h-1)}{2}$, the summation term can be computed through a recurrence relation defined here: $p_k(n)$.
(note that the support is $[\sum_{k=1}^hk,\sum_{k=100-h+1}^{100}k]=[h\cdot(h+1)/2, 100k +(h - h^2)/2]$ so this is not much at all.)
