An algebra problem by Megha Malyala

Algebra Level 4

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

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

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1 solution

Marco Brezzi
Jul 31, 2017

Expanding ( 2 n ) ! (2n)! we get

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

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

If we factor out a 2 2 from each of them

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

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

Therefore the answer is

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

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