If $n$ is even then find the value of $$S=\binom{n}{0}\binom{n+5}{n}+\binom{n}{2}\binom{n+5}{n-2}+\binom{n}{4}\binom{n+5}{n-4}+\cdots$$
If somehow I can find the value of $$A=\binom{n}{1}\binom{n+5}{n-1}+\binom{n}{3}\binom{n+5}{n-3}+\binom{n}{5}\binom{n+5}{n-5}+\cdots$$ I can found the value of original expression. Because, $$S=\binom{2n+5}{n}-A$$ Any help is greatly appreciated.
I considered coefficient of $x^n$ in the expansion of $(1+x)^n\cdot(1+x)^{n+5}$ but here we also counted some unwanted terms whose sum is $A$.
