Given that (x+y)! = (x!)*(y!) is true for all x ∈ N and y ∈ N (integers = 0), find all x and y such that the equality holds:

Question

Given that (x+y)! >= (x!)*(y!) is true for all x ∈ N and y ∈ N  (integers >= 0), find all x and y such that the equality holds:

David C. · Accepted Answer

(x+y)! = (x!)*(y!) implies (x!)*(x+1)*...*(x+y) = (x!)*(y!), which,

in turn, implies (x+1)*(x+2)*...*(x+y) = y! = 1*2*...*y, so then we get

the following set of equations: x+1 = 1, x+2 = 2, ... , and x+y = y, so

x must = 0 for any, and therefore all, of these to be true. This works

for any y, so, also due to symmetry, our solutions are the following:

(x=0, y=0); (x=0, y>=1); & (x>=1, y=0).

Given that (x+y)! >= (x!)*(y!) is true for all x ∈ N and y ∈ N (integers >= 0), find all x and y such that the equality holds: