<s>[INST] Prove that $\frac {n!}{(k! 2^k)(n-2k)!} \neq {n \choose 2}$ for $\forall k \neq 1$ where $n \neq 6$. Prove that $$\frac {n!}{(k! 2^k)(n-2k)!} \neq {n \choose 2}$$ for $\forall k > 1$ where $n \neq 6$.
I have expanded it it seems to be not equal but couldn't find the proper logic.
 [/INST] $$\dfrac {n!}{(k! 2^k)(n-2k)!} = \dfrac{(2k)!}{(k! 2^k)}\dfrac {n!}{(2k)!(n-2k)!} = \dfrac{(2k)!}{(k! 2^k)}{n\choose 2k}$$
$\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)$ is a positive integer strictly greater than $1$ when $k>1$
${n\choose 2k}\geq{n\choose 2}$ when $ k >1 $ and $2k\neq n-1,n$
So $\dfrac{(2k)!}{(k! 2^k)}{n\choose 2k} > {n\choose 2}$ when $ k >1 $ and $2k\neq n-1,n$
If $2k = n-1$(so $n = 2k+1 >3$), we have ${n\choose 2k} = {n\choose n-1} = n$, $\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)\geq (n-2)$, it's easy to verify $n(n-2) > {n\choose 2}$ when $n > 3$
If $2k = n$(so $n\geq 4$), we have ${n\choose 2k} = {n\choose n} = 1$, $\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)\geq (n-3)(n-1)$, it's easy to verify $(n-3)(n-1) > {n \choose 2}$ when $n >6$
Now verify the result for $n=4, k=2$ to conclude 
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