Let m n be positive integers Prove that the following number

Let m, n be positive integers. Prove that the following numbers (2m)!(2n)!/m!n!(m + n)!, (2n)!/(n!)^2, (2n)!/n!(n + 1)! are all integers, and that the second number is even.

Solution

Given m, n be positive integers

(2m)! (2n)! / [ m! n!(m+ n)!]

let m= 1 and n= 1

2! 2! / (2! 2! 4!) = 0.0416

let us consider any positive integer values of m and n , so let m = 2, n =3

(2* 2)! (2 *3)! / [ 2! 3! (2 + 3)!]

= 4! 6! / 2! 3! 5!

= 12

(2n)! / (n!)^2

let n=1 , 2! / ( 1!)^2 = 2

n=2 , 4! / (4!)^2

= 0.0416

(2n)! / n! (n +1) !

let n= 1,

2! / 1! 2! = 1

n= 2, 4! / 2! 3! = 2

For above terms, all are positive integer and the second number is even.