How to simplify $\frac{(2K)!}{(K!)^2}$

Question

How to simplify $\frac{(2K)!}{(K!)^2}$?

Im learning from class that $$\frac{(2K)!}{(K!)^2}=\frac{(K+1)(K+2)\cdot ... \cdot2K}{(K!)}$$

But I dont understand how did can I factor $(2K)!$ ?

Answer 1

Note that$$(2K)!=(2K)\times(2K-1)\times\cdots\times(K+1)\times\overbrace{K\times(K-1)\times\cdots\times2\times1}^{\phantom{K!}=K!}.$$Therefore,$$\frac{(2K)!}{K!}=(2K)\times(2K-1)\times\cdots\times(K+1)$$and so$$\frac{(2K)!}{(K!)^2}=\frac{(2K)\times(2K-1)\times\cdots\times(K+1)}{K!}.$$

Answer 2

This is the $N=2K$ special case of $$\binom{N}{K}=\frac{1}{K!}\frac{N!}{(N-K)!}=\frac{\prod_{j=N-K+1}^{j=N}j}{K!}.$$

Answer 3

$$\frac{8!}{4!^2}=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1\cdot2\cdot3\cdot4\cdot1\cdot2\cdot3\cdot4}=\frac{5\cdot6\cdot7\cdot8}{1\cdot2\cdot3\cdot4}.$$

See the pattern ?

Answer 4

You can use falling and rising factorials: $$\small\frac{(2K)!}{(K!)^2}={2K\choose K}=\frac{(2K)^{\underline{K}}}{1^{\overline{K}}}=\frac{(2K)(2K-1)(2K-2)\cdots(K+2)(K+1)}{1\cdot 2\cdot 3\cdots(K-1)K}.$$ See more here.