Formula for series of Binomial Coefficients Farmer John's hobby of conducting high-energy physics experiments on weekends has backfired, causing N wormholes (2 <= N <= 12, N even) to materialize on his farm, each located at a distinct point on the 2D map of his farm (the x,y coordinates are both integers).
According to his calculations, Farmer John knows that his wormholes will form N/2 connected pairs. For example, if wormholes A and B are connected as a pair, then any object entering wormhole A will exit wormhole B moving in the same direction, and any object entering wormhole B will similarly exit from wormhole A moving in the same direction. This can have rather unpleasant consequences.
For example, suppose there are two paired wormholes A at (1,1) and B at (3,1), and that Bessie the cow starts from position (2,1) moving in the +x direction. Bessie will enter wormhole B [at (3,1)], exit from A [at (1,1)], then enter B again, and so on, getting trapped in an infinite cycle!
*So no: of possible pairs would be this C(n,2)*C(n-2,2)*C(n-4,2)*C(n-6,2)C(n-8,2).......*
Sorry being so trivial,but i think there might be some kind of formula for this series
C(n,2)*C(n-2,2)*C(n-4,2)*C(n-6,2)C(n-8,2)....... like we have for C(n,2)+C(n-2,2)+C(n-4,2)+C(n-6,2)....=2^n-1
But i'am unable to arrive it,
if possible even suggest some kind of source for formulae for series of binomial coefficients 
So what does this expression amount to

C(n,2)*C(n-2,2)*C(n-4,2)*C(n-6,2)*C(n-8,2).......=E(n)

 A: Here we use the representation $\binom{n}{k}$ to denote the binomial coefficient $C(n,k)$. It is convenient to split the calculation in even and odd $n$.

Case: $2n$
We obtain for $n\geq 1$
  \begin{align*}
\binom{2n}{2}&\binom{2n-2}{2}\cdots\binom{2}{2}=\prod_{k=1}^n\binom{2k}{2}=\prod_{k=1}^n\frac{(2k)!}{(2k-2)!2!}\tag{1}\\
&=\frac{1}{2^n}\prod_{k=1}^n(2k)!\prod_{k=1}^n\frac{1}{(2k-2)!}\tag{2}\\
&=\frac{1}{2^n}\prod_{k=1}^n(2k)!\prod_{k=0}^{n-1}\frac{1}{(2k)!}\tag{3}\\
&=\frac{(2n)!}{2^n}
\end{align*}

Comment:


*

*In (1) we use the product sign, multiply thereby the binomial coefficients from right $\binom{2}{2}$ to left $\binom{2n}{2}$ and use the definition
\begin{align*}
\binom{n}{k}=\frac{n!}{k!(n-k)!}
\end{align*}

*In (2) we split the factors into single products and use $\prod_{k=1}^n\frac{1}{2!}=\prod_{k=1}^n\frac{1}{2}=\frac{1}{2^n}$.

*In (3) we shift the index of the second product by one and observe that by telescoping all factors with $1\leq k \leq n-1$ cancel.
In the same way we can do the calculation for odd values.

Case: $2n+1$
We obtain for $n\geq 1$
  \begin{align*}
\binom{2n+1}{2}&\binom{2n-1}{2}\cdots\binom{3}{2}=\prod_{k=1}^n\binom{2k+1}{2}=\prod_{k=1}^n\frac{(2k+1)!}{(2k-1)!2!}\\
&=\frac{1}{2^n}\prod_{k=1}^n(2k+1)!\prod_{k=1}^n\frac{1}{(2k-1)!}\\
&=\frac{1}{2^n}\prod_{k=1}^n(2k+1)!\prod_{k=0}^{n-1}\frac{1}{(2k+1)!}\\
&=\frac{(2n+1)!}{2^n}
\end{align*}

