On the variance in the Bernoulli scheme I'm trying to solve the following problem. 
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?

With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] \prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] \prod_{q=1}^n (1+u)
\sum_{q=1}^n \frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u \frac{1}{2} n (n+1)
\\ = \frac{1}{2} n (n+1) {n-1\choose k-1}.$$
This gives for the expectation
$$\frac{1}{2} n (n+1) {n-1\choose k-1} {n\choose k}^{-1}
= \frac{1}{2} n (n+1) {n-1\choose k-1}
\frac{k}{n} {n-1\choose k-1}^{-1}$$
or
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[X] = \frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] \prod_{q=1}^n (1+ux^q)
\left(\sum_{q=1}^n \frac{q u x^{q-1}}{1+ux^q}\right)^2
\\ + [u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n 
\left(\sum_{q=1}^n \frac{q u}{1+u}\right)^2
\\ + [u^k] (1+u)^n
\sum_{q=1}^n \frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
\sum_{q=1}^n \frac{q u}{(1+u)^2} q u
\\ = {n-2\choose k-2} \frac{1}{4} n^2 (n+1)^2
\\ + {n-1\choose k-1} \frac{1}{3} (n-1) n (n+1)
- {n-2\choose k-2} \frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ \frac{k(k-1)}{n(n-1)} \frac{1}{4} n^2 (n+1)^2
\\ + \frac{k}{n} \frac{1}{3} (n-1) n (n+1)
- \frac{k(k-1)}{n(n-1)} \frac{1}{6} n (n+1) (2n+1)
\\ = \frac{k(k-1)}{n-1} \frac{1}{4} n (n+1)^2
+ k \frac{1}{3} (n-1) (n+1)
- \frac{k(k-1)}{n-1} \frac{1}{6} (n+1) (2n+1).$$
or
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[X(X-1)] = \frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that 
$$\mathrm{Var}[X] = 
\mathrm{E}[X(X-1)] + \mathrm{E}[X] - \mathrm{E}[X]^2$$
so that
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{Var}[X] = \frac{1}{12} k (n+1) (n-k).}$$
