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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)

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  • $\begingroup$ $\binom{n}{2}=\frac{n!}{2!(n-2)!}$ $\binom{n-2}{2}=\frac{(n-2)!}{2!(n-4)!}$... Can you see them cancel out? Try even and odd cases for $n$. $\endgroup$ – gambler101 Sep 2 '16 at 18:01
  • $\begingroup$ Take a look at Mathworld's page on multinomial coefficients, and specifically consider the case of all $n_k$'s equal to 2 (say, by taking equation 3 and applying it over-and-over again). $\endgroup$ – Semiclassical Sep 2 '16 at 18:08
  • $\begingroup$ @JonathanRichardLombardy i never thought of expanding expression in that way thanks $\endgroup$ – viru Sep 2 '16 at 18:09
  • $\begingroup$ @Semiclassical thanks i'll will look into that $\endgroup$ – viru Sep 2 '16 at 18:11
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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*}

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