Another Ordered Quadruples Question How many ordered quadruples $(x_1,x_2,x_3,x_4)$ that are elements of the $\Bbb Z^4$ are there such that $|x_1|+|x_2|+|x_3|+|x_4|= 50$?
 A: If all of the $x_i$ are nonzero, then the count is $2^4$ times the number of solutions in positive integers to $x_1+x_2+x_3+x_4=50$. The latter number is $C(49,3)=18424$, and the $2^4$ factor is from choosing signs on the four $x_i$. This gives $2^4*C(49,3)=294784$ for this case.
Now if three of the $x_i$ are to be nonzero, there are 4 ways to choose which are nonzero, and then multiply by $2^3 C(49,2)$. This gives $4 \cdot 2^3 C(49,2)=37632$ for this case.
And if two of the $x_i$ are to be nonzero, there are 6 ways to choose which are nonzero, and then multiply by $2^2*C(49,1)$. This gives $6 \cdot 2^2 C(49,1)=1176$ for this case.
Finally if only one of the $x_i$ is nonzero, there are 4 ways to choose which is nonzero, and two ways to pick the sign (here $|x_i|=50$ of course). So 8 for this case.
Then we add $294784+37632+1176+8=333600$ for the total number asked in the question.
EDIT: the number of solutions to
$$|x_1|+...+|x_k|=n$$
may be found using the above method, and gives
$$\sum_{j=1}^k 2^j C(k,j) C(n-1,j-1),$$
where $C(m,t)$ denotes the binomial $m$ choose $t$. Maple could not find this sum in closed form.
A: This is the coefficient of $t^{50}$ in the expansion of $$\biggl(\sum_{-\infty}^{\infty}t^{|n|}\biggr)^4=\bigl(1+2\sum_1^{\infty}t^n\bigr)^4$$ Now $$1+2\sum_1^{\infty}t^n=1+{2t\over1-t}={1+t\over1-t}$$ so $$\bigl(1+2\sum_1^{\infty}t^n\bigr)^4=(1+t)^4(1-t)^{-4}=(1+4t+6t^2+4t^3+t^4)\left({3\choose3}+{4\choose3}t+{5\choose3}t^2+\cdots\right)$$ and the coefficient of $t^{50}$ is $${53\choose3}+4{52\choose3}+6{51\choose3}+4{50\choose3}+{49\choose3}=333600$$
