Permutation and counting problem

[cientifiquito]

Homework Statement

Let r be a positive integer. For any number x, let
(x)_r = x(x-1)(x-2)...(x-r+1)

Show that
(-1/2)_r = (-1)^rr!2^-2r(2r take r)

Homework Equations

by "2r take r" I mean what is usually denoted by (n / r) (written like a fraction but without the bar) and is calculated as: n!/(r!(n-r)!)

The Attempt at a Solution

If I start from the definition of (x)_r, plugging in -1/2, I get as far as:

(-1)^r(-1/2)^r(1)(1+2)(1+4)(1+6)...(1+2r-2)
i.e.,

(-1)^r(-1/2)^r(1)(1+2)(1+4)(1+6)...(2r - 1)

And if I start from what I'm supposed to be showing that (-1/2)_r is equal to, I can get to

(-1)^r(-1/2)^r[(2r)!/(r!(2r-r)!)]
i.e.,
(-1)^r(-1/2)^r[(r+1)(r+2)...(2r)]

but obviously I'm not seeing the connection between the two

 

[henry_m]

You've got a product of odd integers 1.3.5...(2r-1); the problem is how you write that compactly.
The trick here is that we can compactly write a product of even integers 2.4.6...(2r), and then you can write 1.3.5...(2r-1)=[1.2.3.4...(2r)]/[2.4.6...(2r)].

 

[cientifiquito]

I see now, thanks henry_m