How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$? 
How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$?

Based off how to solve these problems from my last question I believe I'm meant to take $x^\prime =x-1, y^\prime=y-2,z^\prime=z-3,w^\prime=w-4$
then substituting these values $x=x^\prime +1,y=y^\prime +2,z=z^\prime +3,w=w^\prime +4$
I get $x^\prime+y^\prime+z^\prime+w^\prime=35$
Which I use the formula, ${n+k-1\choose k-1}$ to solve with $n=35,k=4$
Gives me ${35+4-1\choose 4-1}={38\choose 3}=8436$
Is this correct? If it is why does this substitution not change the problem?
 A: Your answer is incorrect.  As a sanity check, notice that the number of integer solutions of the equation 
$$x + y + z + w = 25$$ 
satisfying $x \geq 1$, $y \geq 2$, $z \geq 3$, $w \geq 4$ should be smaller than the number of solutions of the equation in the nonnegative integers since we cannot substitute $0$ for $x$, $0$ or $1$ for $y$, $0$, $1$, or $2$ for $z$, or $0$, $1$, $2$, or $3$ for $w$.  
Your strategy is correct, but you made an error when you did your calculations.
\begin{align*}
x + y + z + w & = 25\\
x' + 1 + y' + 2 + z' + 3 + w' + 4 & = 25\\
x' + y' + z' + w' & = 15
\end{align*}
which is an equation in the nonnegative integers with
$$\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3}$$
solutions, which is smaller than the $\binom{25 + 4 - 1}{4 - 1} = \binom{28}{3}$ solutions in the nonnegative integers, as we would expect.  
Notice that by making these substitutions, we are first putting one object in the first box, two objects in the second box, three objects in the third box, and four objects in the fourth box, then distributing the remaining $25 - (1 + 2 + 3 + 4) = 25 - 10 = 15$ objects to the four boxes without restriction.
