Question

How do you simplify `2^n/[(2n)!]`?

Answer 1

`1/((2n-1)!! * n!)`

Explanation:

`2^n/((2n)!)` = #(2*2*2*2* ....) / [2n* (2n-1) *(2n-2) ....4*3* 2*1]#

Notice that each of the n 2's in the numerator will cancel with each of the n even terms in the denominator.

`2^n/((2n)!)` = #cancel(2*2*2*2* ....) / [cancel(2)n *(2n-1) *cancel(2) (n-1) ....cancel(2)2*3* cancel(2)1*1]#

`= 1/(n*(2n-1)*(n-1) ...2*3*1*1)`

Rearranging terms
`= 1/{[(2n-1)*(2n-3) ....3*1] * [n*(n-1)*(n-2) .... 2*1]`

The 1st term in the denominator is the product of odd integers from (2n-1) to 1. This is the known as the double factorial `(2n-1)!!`. The 2nd term is simply `n!`

Hence:

`2^n/((2n)!)` = `1/((2n-1)!! * n!)`