Question

Hi, can someone check if this is right. Is (n!)/(2n)!=(1*2*3...(n-2)(n-1)(n))/(1*2*3...(2n-2)(2n-1)(2n))=((n-2)(n-1))/2((2n-2)(2n-1))

Answer 1

Vivian, not too many of us are going to be willing to wade through that long expression. Have you considered using Equation Editor, below, to present such problems for greater clarity?

\[ (n!)/(2n)!=(1*2*3...(n-2)(n-1)(n))/(1*2*3...(2n-2)(2n-1)(2n))=((n-2)(n-1))/2((2n-2)(2n-1))\]

Answer 2

Try typing each part of the equation separately, again, for easier reading:\[ (n!)/(2n)!=(1*2*3...(n-2)(n-1)(n))/(1*2*3...(2n-2)(2n-1)(2n))=\]

Answer 3

Note that n!=n(n-1)(n-2)...1, and
(2n)!=2n(2n-1)(2n-2).... 1.

Answer 4

You are typing essentially the same thing as I am, except that you're starting with 1, 2, 3, ... and working your way up to n, whereas I start with n and work my way down to 1. Doesn't matter...either is fine.

Answer 5

ok, so can you cancel the terms out like I did?