$\displaystyle n\in<1,1004>$
Find how many n exist when k = 7
$\displaystyle 2n! \over n!n!2^k$
For :
$\displaystyle 2n! \over n!n!2^{k+1}$
Result isn't integer value.
I not clear what your question is. Do you mean that n must be an integer such that $\displaystyle 1\le n\le 1004$ such that $\displaystyle \frac{2n!}{n!n!2^7}$ is an integer but $\displaystyle \frac{2n!}{n!n!2^{8}$ is not? And is that really "2n!" and not "(2n)!"?
If so it is asking how many "n"s are such that [tex]\frac{2n!}{n!n!} has 7 factors of 2 but not 8.