Conditional expectation of product of independent Bernoulli random variables Given $X_i$, $i=1,...,n$ are Bernoulli(p) independent and identically distributed.
I want to find $E[X_1X_2X_3X_4 |S=s]$ where $S=\sum_{i=1}^{n} X_i$.
But i'm stuck on how to find such expectation, should i find first the conjointy probability? And If so How can i?

Guide:
$X_1X_2X_3X_4$ only contains values in $\{0,1\}$ so that: $$\mathbb E[X_1X_2X_3X_4\mid S=s]=P(X_1X_2X_3X_4=1\mid S=s)=$$$$P(X_1=X_2=X_3=X_4=1\mid S=s)$$
So it comes to finding the probability that the first $4$ experiments are successes under the condition that in total $s$ of the $n$ experiments are successes.
Addendum:
Note that $\{1,\dots,n\}$ has $\binom{n}{s}$ subsets of cardinality $s$ and that by $s$ successes each of these sets has equal probability to be the "success"-set (i.e. the set $\{i\in\{1,\dots,n\}\mid X_i=1\}$). Further $\binom{n-4}{s-4}$ of these sets contains $\{1,2,3,4\}$ as a subset. From this we conclude that:$$E[X_1X_2X_3X_4\mid S=s]=P(X_1=X_2=X_3=X_4=1\mid S=s)=\frac{\binom{n-4}{s-4}}{\binom{n}{s}}$$
This under the convention that $\binom{n-4}{s-4}=0$ if $s<4$

Also you can go for: 
$$\begin{aligned}P\left(X_{1}=X_{2}=X_{3}=X_{4}=1\mid\sum_{i=1}^{n}X_{i}=s\right) & =\frac{P\left(X_{1}=X_{2}=X_{3}=X_{4}=1\wedge\sum_{i=1}^{n}X_{i}=s\right)}{P\left(\sum_{i=1}^{n}X_{i}=s\right)}\\
 & =\frac{P\left(X_{1}=X_{2}=X_{3}=X_{4}=1\wedge\sum_{i=5}^{n}X_{i}=s-4\right)}{P\left(\sum_{i=1}^{n}X_{i}=s\right)}\\
 & =\frac{p^{4}P\left(\sum_{i=5}^{n}X_{i}=s-4\right)}{P\left(\sum_{i=1}^{n}X_{i}=s\right)}\\
 & =\frac{p^{4}\binom{n-4}{s-4}p^{s-4}\left(1-p\right)^{n-s}}{\binom{n}{s}p^{s}\left(1-p\right)^{n-s}}\\
 & =\frac{\binom{n-4}{s-4}}{\binom{n}{s}}
\end{aligned}
$$
