Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$  twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom{18}{1k}\cdot \left ( \binom{19}{k}+\binom{19}{k-1} \right )=\sum_{k=0}^{19}\binom{18}{k}\cdot\left ( \binom{18}{k}+2\binom{18}{k-1}+\binom{18}{k-2} \right )$
Now have no idea what to do with that

Your sum is equal to $c_n = \sum_{k=0}^n \binom{18}{k} \binom{20}{n-k}$ for $n=20.$ Note that $c_n$ is the coefficient of $x^n$ in $(1+x)^{18} (1+x)^{20}.$
