$$x_1 + x_2 + x_3 = 6$$
$$\dbinom{3+6-1}{6} = \dbinom{8}{6} = 28 \text{ possible integer solutions} $$
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15$$
$\dbinom{5+15-1}{15} = \dbinom{19}{15} = 3876 \text{ possible integer solutions}$
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?
Best Answer
Hint: $$x_1 + x_2 + x_3 = 6,\quad x_4 + x_5 = 9$$