$a , b, c$ are positive integers such that ${a}\geq{b+c}$.
Prove that $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\binom{a+b}{b+c}$$
$a , b, c$ are positive integers such that ${a}\geq{b+c}$.
Prove that $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\binom{a+b}{b+c}$$
$\binom{a}{i+c}$ is the coefficient of $x^{i+c}$ from $(x+1)^{a}$ while $\binom{b}{i}$ is the coefficient of $\frac{1}{x^{i}}$ from $(\frac{1}{x}+1)^{b}$.
$\sum_{i=0}^{b}\binom{a}{i+c}\binom{b}{i}$ is the coefficient of $x^{c}$ from $(x+1)^{a}(\frac{1}{x}+1)^b=\frac{(x+1)^{a+b}}{x^{b}}$ which is $\binom{a+b}{b+c}$.
Write the first term on the left side as $\binom{a}{ a - c - i}$ and the term in the right side as $\binom{a + b}{ a - c}$. (Hypergeometric distribution?)
By Vandermonde, $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\sum_{i=0}^{b}{\binom{a}{a-c-i}\binom{b}{i}}=\binom{a+b}{a-c}=\binom{a+b}{b+c}$$