Derive the binomial distribution in the following algebraic way, which does not involve any explicit combinstorial analysis. One is again interested in finding the probability $W(n)$ of $n$ successes out of a total of $N$ independent trials. Let $w_{1}=p$ denote the probability of s success, $u_{2}=1-p=q$ the corresponding probability of a failure. Then $W(n)$ can be obtained by writing
$W(n)=\sum_{i=1}^{2} \sum_{j=1}^{2} \sum_{k=1}^{2} \ldots \sum_{m=1}^{2} u_{j} w_{j} v_{h} \cdots w_{m}$
Here each term contains $N$ factors and is the probability of a particular combination of successes and failures. The sum over all combinations is then to be taken only over those terms involving $w_{1}$ exactly $n$ times, i.e., only over those terms involving $\varphi_{1}{ }^{n}$.
By rearranging the sum (1), show that the unrestricted sum can be written in the form
$$
W(n)=\left(w_{1}+w_{2}\right)^{N}
$$
Expanding this by the binomial theorem, show that the sum of all terms in (1) involving $w_{1}^{2}$, i.e., the desired probability $W(n)$, is then simply given by the one binomial expension term which involves $w_{1}^{n}$.