Solution: 2015 Fall Midterm - 9

Author: Michiel Smid

Question

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)

${59 \choose 3}$

(b)

${58 \choose 3}$

(c)

${58 \choose 4}$

(d)

${59 \choose 4}$

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.