For integers $k$ and $n$ such that $1\le k<n$, let $C^n_k=\frac{n!}{k!(n-k)!}$. Then $\left(\frac{n-2k-1}{k+1}\right)C^n_k$ is an integer
$\textbf{(A) }\text{for all }k\text{ and }n\qquad \\ \textbf{(B) }\text{for all even values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\ \textbf{(C) }\text{for all odd values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\ \textbf{(D) }\text{if }k=1\text{ or }n-1,\text{ but not for all odd values }k\text{ and }n\qquad \\ \textbf{(E) }\text{if }n\text{ is divisible by }k,\text{ but not for all even values }k\text{ and }n$
We know $C^n_k = \binom{n}{k}$, so let's rewrite the expression as $\left(\frac{n-2k-1}{k+1}\right) \binom{n}{k}$. Notice that \[n-2k-1 = n-2(k+1)+1 = (n+1)-2(k+1).\]
This allows us to rewrite the expression as \[\left(\frac{(n+1)-2(k+1)}{k+1}\right) \binom{n}{k}.\]
From here, we just have to do some algebra to get
\[\left(\frac{(n+1)-2(k+1)}{k+1}\right) \binom{n}{k} = \left( \frac{n+1}{k+1}-2 \right) \frac{n!}{k!(n-k)!}\]
\[= \frac{(n+1)!}{(k+1)!(n-k)!} - 2 \cdot \frac{n!}{k!(n-k)!}\]
\[= \binom{n+1}{k+1}-2\binom{n}{k},\]
so $\left(\frac{n-2k-1}{k+1}\right)C^n_k$ is an integer $\boxed{\textbf{(A) }\text{for all }k\text{ and }n}$. ~[jiang147369](https://artofproblemsolving.com/wiki/index.php/User:Jiang147369)