Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how? 
Edit:
Drawing from David K's answer:

One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
  That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
  which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
  Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$

$x,y,z >0 => x + y + z <= 3, n = Cr(3+4-1,3) * (8) = 160$

Now let $x = 0,$ $y > 0,$ and $z > 0.$
  Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$

$x=0, x,y>0 => y+z <=6, n = Cr(4+3-1, 4) * 3 * 14 = 180$

Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.

$x,y = 0, z>0 => z<=6, n = 6 * 2 * 3 = 36$
$x,y,z=0, n = 1$
Sum of all of them is : 160 + 180 + 36 + 1 = 377

Let $a_{n,m}$ be the number of $n$-tuples of integers so that
$$
\sum_{k=1}^n|x_k|\le m\tag1
$$
Pretty simply, we have
$$
a_{1,m}=2\binom{m}{1}+\binom{m}{0}\tag2
$$
Furthermore, we have the recurrence
$$
\begin{align}
a_{n+1,m}
&=\overbrace{\quad a_{n,m}\quad\vphantom{\sum_1^m}}^\text{$0$ in position $n+1$}+\overbrace{2\sum_{k=1}^ma_{n,m-k}}^\text{$\pm k$ in position $n+1$}\\
&=a_{n,m}+2\sum_{k=0}^{m-1}a_{n,k}\tag3
\end{align}
$$
Thus,
$$
a_{2,m}=4\binom{m}{2}+4\binom{m}{1}+\binom{m}{0}\tag4
$$
and
$$
a_{3,m}=8\binom{m}{3}+12\binom{m}{2}+6\binom{m}{1}+\binom{m}{0}\tag5
$$
In general, we get by induction, using $(2)$ and $(3)$ and $\sum\limits_{k=0}^{n-1}\binom{k}{j}=\binom{n}{j+1}$,
$$
a_{n,m}=\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}\tag6
$$
Plugging in $n=3$ and $m=6$ gives
$$
\bbox[5px,border:2px solid #C0A000]{a_{3,6}=377}\tag7
$$

I just came across this answer which proves
$$
\sum_{k=0}^n\binom{n}{k}\binom{m+k}{n}=\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}
$$
