Solution: 2018 Fall Midterm - 5

Author: Michiel Smid

Question

Consider the equation $$ x_1 + x_2 + x_3 + x_4 + x_5 = 17. $$ How many solutions $(x_1, x_2, x_3, x_4, x_5)$ does this equation have, where $x_1 \geq 0$, $x_2 \geq 0$, $x_3 \geq 0$, $x_4 \geq 0$, and $x_5 \geq 0$ are all integers?

(a)

${21 choose 5}$

(b)

${22 choose 4}$

(c)

${21 choose 4}$

(d)

${22 choose 5}$

Solution

We can use the dividers method to solve this problem.

We have 17 blocks (representing the sum of the 5 variables) and 4 dividers (representing the 4 partitions between the 5 variables).

We can place the 4 dividers into any of the 21 positions.

Everything before the first divider is $ x_1 $, everything between the first and second dividers is $ x_2 $, and so on.

Thus, there are $ \binom{21}{4} $ solutions to the equation.