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:
$$x\_1 + x\_2 + x\_3 = 6,\quad x\_4 + x\_5 = 9$$
**Hint:**

$x\_1+x\_2+x\_3=6$ and is a constant in the second equation.

You get $x\_4+x\_5 =9 \implies$Number of solutions is $9+2-1 \choose 9$

You can continue from here.