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
 A: 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 $z < 0.$
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 $x = 0.$
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.
Finally add $1$ for the case $x = y = z = 0,$ which was not covered by any of the other cases.
The total will be $377$ if you do all these calculations correctly.
A: This is $f(6)$ where $f(n)$ for integers $n\ge0$ is the number of lattice
points in the polyhedron $nP$. This is the $n$-fold dilate of $P$, the polyhedron with vertices $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$. By Ehrhart's theorem, $f$ is a degree $3$ polynomial in $n$. Moreover, its leading coefficient is the volume of $P$, namely $\frac43$. We also have $f(0)=1$
and $f(1)=7$. But by Macdonald's reciprocity law $f(-n)$ is the negative
of the number of lattice points in the interior of $nP$, so that $f(-1)=-1$.
The only polynomial satisfying all these conditions is
$$f(n)=\frac43n^3+2n^2+\frac83n+1.$$
Then $f(6)=377$.
A: Try it first for the 2-dimensional $(x, y)$-plane and for smaller upper bounds: $|x| + |y| \leq 1$, then  $|x| + |y| \leq 2$, then $|x| + |y| \leq 3$.  In each case, sketch a picture.
Now try it in the $(x, y, z)$-space, with smaller upper bounds: $|x| + |y| + |z| \leq 1$, then $|x| + |y| + |z| \leq 2$, etc..  Sketch a picture in every case.
By the time you are up to $|x| + |y| + |z| \leq 3$, you will be seeing enough of a pattern to solve your problem--maybe even much earlier.
A: Here is a brute force solution. Letters $x,y,z$ stand for nonzero integers.
Case 1. Permutations of $(x,x,x)$.
$(1,1,1)$
$(2,2,2)$
$2$ combinations, one permutation of each, $2^3=8$ ways to assign $\pm$ signs, so $2\cdot1\cdot8=\boxed{16}$ solutions.
Case 2. Permutations of $(x,x,y)$, $x\ne y$.
$(1,1,2)$
$(1,1,3)$
$(1,1,4)$
$(2,2,1)$
$4$ combos, $3$ perms each, $2^3=8$ sign choices, so $4\cdot3\cdot8=\boxed{96}$ solutions.
Case 3. Perms of $(x,y,z)$ with $x,y,z$ all different.
$(1,2,3)$
Just $1$ combo, $6$ perms, $8$ sign choices, so $1\cdot6\cdot8=\boxed{48}$ solutions.
Case 4. Perms of $(x,x,0)$.
$(1,1,0)$
$(2,2,0)$
$(3,3,0)$
$3$ combos, $3$ perms each, $2^2=4$ sign choices, so $3\cdot3\cdot4=\boxed{36}$ solutions.
Case 5. Perms of $(x,y,0)$, $x\ne y$.
$(1,2,0)$
$(1,3,0)$
$(1,4,0)$
$(1,5,0)$
$(2,3,0)$
$(2,4,0)$
$6$ combos, $6$ perms each, $4$ sign choices, so $6\cdot6\cdot4=\boxed{144}$ solutions.
Case 6. Perms of $(x,0,0)$.
$(1,0,0)$
$(2,0,0)$
$(3,0,0)$
$(4,0,0)$
$(5,0,0)$
$(6,0,0)$
$6$ combos, $3$ perms each, $2$ sign choices, so $6\cdot3\cdot2=\boxed{36}$ solutions.
Case 7. Perms of $(0,0,0)$.
$(0,0,0$.
Just $1\cdot1\cdot1=\boxed{1}$ solution
The final answer is
$$16+96+48+36+144+36+1=\boxed{377}$$
provided $0$'s are allowed. However, if you really mean "negatives and positives" (no $0$'s) then you only get
$$16+96+48=\boxed{160}$$.
A: 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
$$
\sum_{k=0}^n\binom{n}{k}\binom{m+k}{n}=\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}
$$
