# Math Help - function

1. ## function

$n\in<1,1004>$
Find how many n exist when k = 7

$2n! \over n!n!2^k$

For :

$2n! \over n!n!2^{k+1}$
Result isn't integer value.

2. Originally Posted by Day
$n\in<1,1004>$
Find how many n exist when k = 7

$2n! \over n!n!2^k$

For :

$2n! \over n!n!2^{k+1}$
Result isn't integer value.
I not clear what your question is. Do you mean that n must be an integer such that $1\le n\le 1004$ such that $\frac{2n!}{n!n!2^7}$ is an integer but $\frac{2n!}{n!n!2^{8}$ 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.

3. n is integer.
how many results have $2n! \over n!n!2^k$. How many is n in this function? This is question.

All is up.

$

2n! \over n!n!2^k$
= integer

$2n! \over n!n!2^{k+1}$ isn't integer