Variance of X which is a sum of random m draws of different numbers among n numbers (1,2,3,....n)

Question

You have a set of consecutive n natural numbers {1,2,3...,n}. m different numbers are drawn from these n numbers. Calculate the variance of the sum of these m numbers?.
My try:

Since each draw of m numbers are different and as likely as any other draw, would it not be true that it is a uniform distribution with range of X from the minimum of (1,2,3,...,m) with X being $\frac{m(m+1)}{2}$ to a maximum of (n,n-1,n-2,..n-(m-1)) with X being $\frac{m(2n-m+1)}{2}$ This mamimum and maximum form the range of the uniform distribution and that I can find the Expected value and Variance with these formulas $\frac{(a+b)}{2}$ and $\frac{(b-a)^2}{12}$ where a is the minimum and b is the maximum thus found.

Let me know if this is the right approach

Wasn't this asked at https://math.stackexchange.com/q/2813390/321264 ? — StubbornAtom, Oct 13 '18 at 09:00
There is only one way to get the minimum and maximum that you here described, but consider {2,3,...,m+1} and {1,3,4,5,...,m,m+2}. Both gives the same sum. — Dominik Kutek, Oct 13 '18 at 09:01

Satish Ramanathan · Accepted Answer · 2018-10-14T14:30:28.300

I figured a solution for this after some research which is quite complicated and would love for someone to simplify in the lines that I have presented.

$$Var(X) = \sum_{i=0}^{m(n-m)+1}\dfrac{ a_i q^{i}}{{n\choose m}}$$

$a_iq^i = {n\choose m}_q$ the gaussian binomial coefficient and $a_i$ is the coefficient representing the frequency of the sum starting from $\frac{m(m+1)}{2}$ to $\frac{m(2n-m+1)}{2}$

We define $q^i = \dfrac{\left(\frac{m(m-n)}{2} + i\right)^2}{4}$

Variance of X which is a sum of random m draws of different numbers among n numbers (1,2,3,....n)

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