Solution: 2015 Fall Midterm - 9
Author: Michiel SmidQuestion
How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 55$,
where $x_1 \geq 0$, $x_2 \geq 0$, $x_3 \geq 0$, and $x_4 \geq 0$ are integers?
(a)
${58 \choose 3}$
(b)
${59 \choose 4}$
(c)
${59 \choose 3}$
(d)
${58 \choose 4}$
Solution
We can use the dividers method.
We have 55 buckets and 3 extra spots for the dividers to go.
Thus, there are $\binom{55+3}{3}$ solutions.