Number of integral solutions for $|x | + | y | + | z | = 10$ How can I  find the number of integral solution to the
equation

$|x | + | y | + | z | = 10.$

I am using the formula,

Number of integral solutions for $|x| +|y| +|z| = p$ is $(4P^2) +2 $, So the answer is 402.

But, I want to know, How we can find it without using formula.

any suggestion !!!

Please help
The number $n_1(k)$ of solutions of 
$$|x|=k$$
is quite obviously given by
$$n_1(k)=\begin{cases}2&amp;\text{if }k&gt;0\\1&amp;\text{if }k=0\\0&amp;\text{if }k&lt;0.\end {cases}$$
Then the number of solutions $n_2(k)$ of 
$$|x|+|y|=k$$
can by obtained as 
$$n_2(k)=\sum_{i\in\mathbb Z}n_1(i)\cdot n_1(k-i)=\begin{cases}2+(k-1)\cdot 4+2=4k&amp; \text{if }k&gt;0\\1&amp;\text{if }k=0\\0&amp;\text{if }k&lt;0.\end{cases}$$
Finally, the number $n_3(k)$ of solutions of
$$|x|+|y|+|z|=k$$
is (using $\sum_{i=1}^n i=\frac{n(n+1)}2$)
$$n_3(k)=\sum_{i\in\mathbb Z}n_2(i)\cdot n_1(k-i)\\=\begin{cases}2+2\sum_{i=1}^{k-1}4i + 4k=2+4k(k-1)+4k=4k^2+2&amp; \text{if }k&gt;0\\1&amp;\text{if }k=0\\0&amp;\text{if }k&lt;0.\end{cases}$$



In general, the number of solutions of 
$$|x_1|+\cdots +|x_m|=k$$
is given by
$$n_m(k)=\begin{cases}P_m(k)&amp;\text{if }k&gt;0\\1&amp;\text{if }k=0\\0&amp;\text{if }k&lt;0,\end{cases}$$
where $P_m$ is some polynomial of degree $m-1$. The $P_m$ can be obtained recursively, e.g. via the relations
$$P_m(X+1)-P_m(X)= P_{m-1}(X)+P_{m-1}(X+1)$$
$$P_m(1)=2m$$
I think it must 8 times the number of intergral solution to equation:
$$x+y+z=10$$
So the result is :
$$8C_{12}^2=528$$