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
We obtain \begin{align*} \color{blue}{\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}} &=\sum_{k=0}^{18}\binom{18}{k}\binom{20}{k}\tag{1}\\ &=\sum_{k=0}^{18}\binom{18}{18-k}\binom{20}{k}\tag{2}\\ &\,\,\color{blue}{=\binom{38}{18}=33\,578\,000\,610}\tag{3} \end{align*}
Comment:
In (1) we omit the index $k=19$, since $\binom{p}{q}=0$ if $q>p$.
In (2) we apply the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
In (3) we apply Vandermonde's identity.
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 $(1+x)^{18} (1+x)^{20}.$
