Is there missing information to answer this question! I have two quantities $A_1$ and $A_2$, and I would like to compare them in order to know which one is bigger.
Knowing that $x$, $y$ are two constants such that $y \leq \frac{x}{2}$ 
$$A_1 = \sum_{z=1}^y (x-y+z)!(y-z+1)! (z+1)$$
$$A_2 = \sum_{z=1}^y (x-y+z+2)! (y-z)! (z+3)$$
Simply I want to specify if $A_1 > A_2$ or $A_1 < A_2$ ?
I guess that we can say $A_1 > A_2$ $n$ increases because the factorial increases faster, but which constraints I should put here? 

We have $x-y\ge y.$ So, for each given $z,$ the ratio of the "$z$-term" in $A_2$ to the "$z$-term" in $A_1$ is $$\frac {((x-y)+z+2)((x-y)+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}\ge \frac {(y+z+2)(y+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}=$$ $$=(y+z+2)\cdot  \frac {(y+1)+z}{(y+1)-z}\cdot \frac {z+3}{z+1}>$$ $$>(y+z+2)\cdot 1 \cdot 1\ge 4.$$ So $A_2>4A_1>0.$
