Bunuel wrote:
If x, y, and z are positive integers, is (x!)(y!) > z! ?
(1) x^y > z
(2) x + y = z
Hi,
lets see the choices and substitue values to see answer
(1) \(x^y > z\)
x = 3, y=2 and z=6... 3!2!>6!-- NO
x=3, y=2 and z= 3...3!2!>3!-- YES
Different answers -- Insuff
(2) x + y = zOR x!y!>(x+y)!
It will always be NO, since
x! = 1*2*3...(x-1)*x
y!= 1*2*3...*(y-1)*y
(x+y)!= 1*2*3*...X..y.....(x+y)
x!*y! gives us square of the lower integers, whereas (x+y)! gives us product of higher values..Examples
x=1 y=2 and z=3..
1!2!>3!----NO
x=3,y=3 and z=6
3!3!>6!---NO
Suff
B