# How many solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 21$ given the following restrictions …

How many solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 21$ where $x_i$, $i = 1, 2, 3, 4, 5$ is a nonnegative integer such that $0 \le x_1 \le 3$, $1 \le x_2 \lt 4$ and $x_3 \ge 15$,

My Approach

My idea is to find the total number of solutions without restrictions and from that subtract the solutions with restrictions to arrive at the answer.

Number of solutions without restrictions: $C(5-1 + 21, 21) = 12650$

Restriction 1: $x_3 \ge 15$ Number of solutions: $C(5-1+6,6)$ = 210

Restriction 2: $0 \le x_1 \le 3$ We can change the restriction to $x_1 \ge 4$ and subtract the number of solutions with this restriction from the total.

$C(5-1 + 21, 21) - C(5-1 + 17, 17) = 6665$

Restriction 3: $1 \le x_2 \lt 4$ We can break this restriction down into two parts: when $x \ge 1$ and when $x \ge 5$. Then subtract case $2$ from case $1$.

Case 1: $x \ge 1$. $C(5-1 + 20, 20) = 10626$

Case 2: $x \ge 5$. $C(5-1 + 21, 21) - C(5-1 + 16, 16) = 7805$.

Case $1$ - Case $2$ is: $2851$.

Now we can sum the restrictions and remove them from the total.

$$12650 - (2851 + 210 + 6665) = 2924$$

However, the answer in the textbook is: $106$.

Where is my reasoning incorrect?