Explanation
Official Answer: СStatement (1):

Given that

|x| = |y+1|If

x and

(y+1) both have same signs (either positive or negative), then

x = y+1 and

x>y, which makes

x and

y consecutive integers.

If

x and

(y+1) both do not have the same signs (one is negative and the other is positive), then

x = -(y+1) or

(x + y + 1) = 0. In that case, either one could be greater in absolute value. S1 is not sufficient by itself.

Statement (2):

If

x^y = x! + |y|,

x and

y both have to be positive because

x! + |y| is always a positive integer. If

y is negative,

x^y can never be an integer. So for

x^y to be an integer,

y has to be positive. Let's see what values satisfy the equation:

If

x = 1, and

y = 0,

x^y = x! + |y| = 1.

If

x = 2, and

y = 2,

x^y = x! + |y| = 4.

So

x and

y have at least two possibilities. S2 is not sufficient by itself either.

Statement (1)+Statement (2):

Because of Statement (2),

x and

y have to be positive, in which case, by Statement (1)

x > y. Sufficient. Therefore, the answer is C.

I don't get it..Statement 2 gives you 2 solutions 1,0 and 2,2 so x and y are non-negative .. even combining with statement 2 you get 2 solutions.. so answer is E..