TITLE: Non-probabilistic proofs of a binomial coefficient identity from a probability question
QUESTION [8 upvotes]: Combining the answers given by me and Ralth to the probability question at Probability of getting three zeros when sampling a set, we get the following identity:
$$
\sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose m} p_j^m (1 - p_j )^{k - m} } = {n \choose m} (p p_j)^m (1 - p p_j)^{n-m},
$$
where $m \geq 0$ and $n \geq m$ are arbitrary integers, and $p$ and $p_j$ are arbitrary probabilities (in Ralth's notation $m$ is $k$, $p$ is $p_s$, and $p_j$ is $p_h$). Can you prove this identity directly (i.e., in a non-probabilistic setting)? Whether you'll find this identity interesting or not, at least Ralth's answer may now gain its due recognition.

REPLY [7 votes]: We can prove this using the Multinomial Theorem.
We have, using the Multinomial Theorem, that
$$\displaystyle (a+b+c)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} a^m b^{k-m} c^{n-k}$$
Set $\displaystyle a = pp_j x$, $b = p(1-p_j)$, $c = 1-p$.
We get
$$\displaystyle (pp_jx + p(1-p_j) + 1-p )^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} (pp_jx)^m (p(1-p_j))^{k-m} (1-p)^{n-k}$$
i.e.
$$\displaystyle (pp_jx + 1 - pp_j)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} p_j^m(1-p_j)^{k-m} p^{k}(1-p)^{n-k} x^m$$
Expanding the left hand side using binomial theorem and comparing the coefficients of $\displaystyle x^m$ gives the result.