Question: $$x\_1 + x\_2 + x\_3 = 6$$

$$\dbinom{3+6-1}{6} = \dbinom{8}{6} = 28 \text{ possible integer solutions} $$

$$x\_1 + x\_2 + x\_3 + x\_4 + x\_5 = 15$$

$\dbinom{5+15-1}{15} = \dbinom{19}{15} = 3876 \text{ possible integer solutions}$

I solved those individually, but the question asks for the number of solutions that solve BOTH. How would I solve for the union of the two?

Answer: 
Hint:
$$x\_1 + x\_2 + x\_3 = 6,\quad x\_4 + x\_5 = 9$$
This doesn't answer your question directly, but I can't help mentioning that [Generating functions](https://www.google.com/search?client=safari&rls=en&q=generating%20functions%20with%20counting&ie=UTF-8&oe=UTF-8#client=safari&rls=en&sclient=psy-ab&q=generating%20functions%20counting&oq=generating%20functions%20counting&gs_l=serp.3..0l2j0i30j0i7i30.5611.5611.0.5807.1.1.0.0.0.0.108.108.0j1.1.0...0.0...1c.1.12.psy-ab.rzbW_1mmNFc&pbx=1&bav=on.2,or.r_cp.r_qf.&bvm=bv.45960087,d.eWU&fp=cfd7e1693ca1d4fa&biw=1065&bih=635) are helpful with these kinds of problems. For the first problem, note that $x\_i$ can be either $0, 1, 2, 3, 4, 5$, or $6$.

Thus, you are after the coefficient of $x^6$ in the product

$$
(x^0 + x^1 + x^2 + x^3 + x^4+x^5+x^6)^3,
$$
[which is easily computed](http://www.wolframalpha.com/input/?i=%281%20%2b%20x%20%2b%20x%5E2%20%2b%20x%5E3%20%2b%20x%5E4%20%2b%20x%5E5%20%2b%20x%5E6%29%5E3) as $28$.

For the second problem, you would need to compute the coefficient of $x^{15}$ in the product

$$
\left( \sum\_{k=0}^{15} x^k \right)^5
$$

You can easily use generating functions to solve the problem about which you posted by using the other answerer's observation that you are really counting solutions to $x\_4 + x\_5 = 9$, where $0 \leq x\_i \leq 9$.