An algebra problem by Megha Malyala

Algebra Level 4

$\large \frac{(2n)!}{n!} = \ ?$

2n(2n-1)(2n-2)(2n-3)\cdots (n-1)

2^{n}

(1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1))2^n

2!

2n

1 solution

Marco Brezzi
Jul 31, 2017

Expanding $(2n)!$ we get

$(2n)!=(2n)\cdot (2n-1)\cdots 2\cdot 1$

Where there are $2n$ factors, $n$ of which are even

If we factor out a $2$ from each of them

$(2^n\cdot n\cdot (n-1) \cdots 2\cdot 1)((2n-1)\cdot (2n-3)\dotso 3\cdot 1)$

$2^n\cdot n!((2n-1)\cdot (2n-3)\cdots 3\cdot 1)$

Therefore the answer is

$\dfrac{(2n)!}{n!}=\boxed{2^n((2n-1)\cdot (2n-3)\cdots 3\cdot 1)}$