Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$

How can we prove that

$$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$

(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.

Solutions Collecting From Web of "Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$"

Suppose you have to select n balls from a collection of $R$ black balls and $M$ white balls.

Then we must select $k$ black balls and $n-k$ white balls in whatever way we do.(for $0\le k\le n$)

For a fixed $k\in N,0\le k\le n$ we can do this in $\binom{R}{k}\binom{M}{(n-k)}$ ways.

so to get the total no. of ways we must add the above for all $k:0\le k\le n$

So we have the total no. of ways $=\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}$.

But if we think about it in a different way we can say that we have to select $n$ balls from a collection of $R+M$ balls and this can be done in $\displaystyle \binom{R+M}{n}$ ways.

So ,

$$\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}=\displaystyle \binom{R+M}{n}$$

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\begin{align}
\pars{1 + x}^{R + M}&=\sum_{k = 0}^{R + M}{R + M \choose k}x^{k}
\\
\pars{1 + x}^{R}\pars{1 + x}^{M}
&=\sum_{\ell = 0}^{R}{R \choose \ell}x^{\ell}\sum_{\ell’ = 0}^{M}{M \choose \ell’}x^{\ell’}\sum_{k = 0}^{R + M}\delta_{k,\ell + \ell’}
\\[3mm]&=
\sum_{k = 0}^{R + M}x^{k}\sum_{\ell = 0}^{R}{R \choose \ell}
\sum_{\ell’ = 0}^{M}{M \choose \ell’}\delta_{\ell’,k – \ell}
=
\left.\sum_{k = 0}^{R + M}x^{k}\sum_{\ell = 0}^{R}{R \choose \ell}
{M \choose k – \ell}\right\vert_{0 \leq k – \ell \leq M}
\end{align}

$$\color{#0000ff}{\large%
{R + M \choose k}
=
\sum_{\ell = 0 \atop {\vphantom{\LARGE A^{a}}k – M \leq\ \ell\ \leq k}}
^{R}{R \choose \ell}{M \choose k – \ell}}\,,
\qquad 0 \leq k \leq R + M
$$

By using the Chu-Vandermonde identity:

$$\sum_{k=0}^n \binom{R}k\binom{M}{n-k}=\sum_{k=0}^n \binom{R}k\binom{(R+M)-R}{n-k}=\binom{R+M}n$$

(Reproduced from there.)

Since ${R\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^R$ and ${M\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^M$, the sum $S(R,M,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^R\cdot(1+x)^M=(1+x)^{R+M}$.

This proves that $S(R,M,n)={R+M\choose n}$.