Number of integral solutions of the equation $x+y+z \le 25$ I came up with this equation
$x+y+z≤25$, such that $x≥-3$, $y≥-2$ and $z≥0$.

I introduced a dummy variable $w$ and the equation became $x+y+z+w=25$ where $w≤30$.

I used the Multinomial Theorom and I proceeded in this way,
Solutions are equivalent to finding coefficient of $x^{25}$
$(x^{-3}+x^{-2}+x^{-1}+1...x^{25})(x^{-2}+x^{-1}+1+...x^{25})(1+x+x^2+...x^{25})(1+x+x^2+x^3...x^{30})$

This can be converted to 
$\frac{(x^{-3})(1-x^{29})(x^{-2})(1-x^{28})(1-x^{26})(1-x^{31})}{(1-x)^4}$

Now When I found the coefficients of $x^{25}$ I found $^{23}C_3- ^4C_3 - ^7C_3 - ^5C_3$ Which is not the answer written in the textbook.

Can someone explain What did I do wrong?
Finding the number of integer solutions to $\begin{cases}x+y+z\leq 25\\x\geq -3\\y\geq -2\\ z\geq 0\end{cases}$ is equivalent to finding the number of integer solutions to

$$\begin{cases} x'+y'+z + w = 30\\x'\geq 0\\y'\geq 0\\z\geq 0\\w\geq 0\end{cases}$$

seen by a change of variables $x'=x+3, y'=y+2$ and introducing a dummy variable $w = 25-x-y-z$



Your attempt and the subtraction of binomial coefficients looks like you were trying to deal with upper bounds on one or more of the variables, like if it were $\begin{cases}x'+y'+z+w=30\\0\leq x'\leq 10\\0\leq y'\leq 15\\ 0\leq z\\0\leq w\end{cases}$ or similar...

Note in particular that the condition $-3\leq x$ is different than the condition $-3\leq x\color{red}{\leq 25}$.  If we were to be using a generating function here to describe $-3\leq x$, it would have involved $(x^{-3}+x^{-2}+x^{-1}+1+x+x^2+\dots+x^{25}+x^{26}+\dots+\dots)$ and would not have ended at the $x^{25}$ term.

Here, we have no upper bounds to worry about... and the answer is a straightforward textbook example of stars-and-bars

$$\binom{30+4-1}{4-1}$$
Nice question

$x+y+z ≤ 25$

Where $ x≥ -3$, $y ≥ -2$, $z ≥ 0$
Where $(x,y,z)€Z$

My intention here is to create a table of values, but since i can't type &amp;&amp;&amp;

Let's say the equation is 
$x+y+z ≤ w$
Where $w$, their sum is such that $w ≤ 25$
The range of values of $x$, $y$ and $z$ are $:$ 

$x → (-3,-2,-1,0,1,2,3,............)$

$y → (-2,-1,0,1,2,3,4,.......…....)$

$z → (0,1,2,3,4,5,6,7,............)$

Now the maximum value of $x$ will occur when both $y$ and $z$ are minimum, because they all have a fixed range $w$

$w → (-5,-4,-3,-2,-1,0,1,2...........,23,24,25)$

So also the maximum of $y$ will occur when $x$ and $z$ are minimum and vice versa

$x → (-3,-2,-1,0,1,2,..........25,26,27)$

$y → (-2,-1,0,1,2,.................25,26,,27,28)$

$z → (0,1,2,3,4,..............27,28,29,30)$

Since we are trying to find the number of integer solutions to the inequality, then number of integers in each set of its variables $x$, $y$ and $z$ will  hold the key

The number of elements in set $x$ is 31, same with $y$ and $z$
Therefore number of integer solutions $31×31×31$

$29,791$