Randomly take $51$ numbers from the set 1, 2, ..., 159. Find variance of their sum. We randomly take $51$ numbers from 159 natural numbers $1,...,159$ without replacement. Let $\alpha$ be a random variable equal to the sum of the selected numbers. Find the variance of $\alpha$.
Firstly I need to understand something about $\alpha$ destribution. There are totally $$C^{51}_{159} = \frac{159!}{51!108!}$$ kinds of sums. A lot of them are equal, because $$\sum_{i=1}^{51}i = 1326\leq\alpha\leq\sum_{i=109}^{159}i=6834$$
Consequenlty, I want to know how many subsets of $51$ numbers have the sum equal to $N$, where  $1362\leq N\leq6834$. I'm stuck here because I don't know how to do it.

Replace 51 and 159 with $n, M$ respectively. We have a vector $\mathbf{x}_{n\times 1}$ which follows a multivariate distribution, and $\alpha = \sum_{i=1}^n x_i$ where $x_i$ is the $i^{th}$ component of $\mathbf x$.
Then, by symmetry,
$E(\alpha)=E(\sum x_i)=\sum_i E(x_i) =nE(x_1)= \frac{n(M+1)}{2}$.
$$E(\alpha^2)=E\left(\sum_i x_i\right)^2 = E\left(\sum_i x_i^2\right)+E\left(\sum_{i\neq j} x_i x_j \right)$$
Again by symmetry
$$
E\left(\sum_i x_i^2\right)=nE(x_1^2)=\frac 16 n(M+1)(2M+1)
$$
$$
E\left(\sum_{i\neq j} x_i x_j \right)=(n^2-n)E(x_1 x_2)=\frac{n^2-n}{M^2-M}\sum_{i\ne j}ij = \frac{n^2-n}{M^2-M}\left(\left(\frac{M(M+1)}{2}\right)^2 - \frac{M(M+1)(2M+1)}{6}\right) \\= \frac{1}{12} (n^2-n)(M+1)(3M+2)
$$
Therefore $$\text{var } \alpha = E(\alpha^2) - (E(\alpha))^2 = \cdots = 73440$$
