This would simplify to ^{2008}C_{1004} \sum{\left(^{1004}C_{k} \right)^{2}}
Which now can be evaluated using coefficient of xn in exp. of (1+x)2n , may be.
\hspace{-16}\mathbf{\displaystyle \sum_{k=0}^{1004} \frac{2008!}{k!\times k!\times (1004-k)!\times (1004-k)!}}
This would simplify to ^{2008}C_{1004} \sum{\left(^{1004}C_{k} \right)^{2}}
Which now can be evaluated using coefficient of xn in exp. of (1+x)2n , may be.