# How many solutions does $x_1+x_2+x_3 = 11$ have if $0\le x_1 \le 3$, $0\le x_2 \le 4$, and $0\le x_3 \le 6$?

How many solutions does $$x_1+x_2+x_3 = 11$$ have if $$0\le x_1 \le 3$$, $$0\le x_2 \le 4$$, and $$0\le x_3 \le 6$$?

I tried to do it with method 2 but there is a problem;

\begin{align*} x_1+x_2+x_3 = 11\tag{1}\\ x_1\ge4, x_2\ge5, x_3\ge7 \end{align*}

then combining equation 1 and 2; \begin{align*} y_1 = x_1-4, y_2=x_2-5,y_3=x_3-7\tag{2}\\y_1+4+y_2+5+y_3+7 = 11\\y_1+y_2+y_3=-4\tag{3} \end{align*} how can I manage this problem in equation 3?

• We need to exclude those cases in which $x_1 \geq 4$ or $x_2 \geq 5$ or $x_3 \geq 7$. The reason you got a sum equal to a negative number is that you are considering cases in which all three of those conditions are violated simultaneously. Use the Inclusion-Exclusion Principle to eliminate those cases in which at least one of the restrictions is violated. Mar 30 at 22:24
• @N.F.Taussig yeah you are right in equation 3 it tries to take $N(C_1C_2C_3)$ Mar 30 at 22:28

I'll edit these equations tomorrow if needed. I need to study a bit more to exam lol..

$$S_0 = N = \binom{11+3-1}{3-1} = \binom{13}{2}=78\tag{0}$$

Equation (0) is Total Number of Solutions

$$N(C_1) = y_1+4+y_2+y_3 = 11 => y_1+y_2+y_3 = 7 => \binom{7 + 3 -1}{3-1} = \binom{9}{2} = 36 \tag{1}$$

$$N(C_2) = y_1+y_2+5+y_3 = 11 => y_1+y_2+y_3 = 6 => \binom{6 + 3 -1}{3-1} = \binom{8}{2} = 28 \tag{2}$$

$$N(C_3) = y_1+y_2+y_3+7 = 11 => y_1+y_2+y_3 = 4 => \binom{4 + 3 -1}{3-1} = \binom{6}{2} = 15 \tag{3}$$

$$S_1 = Equation(1)+Equation(2)+Equation(3) = 79$$

Equation(1) is Total Number of Solutions with restriction $$x_1\ge4$$

$$N(C_1C_2) = y_1+4+y_2+5+y_3 = 11 => y_1+y_2+y_3 = 2 => \binom{2 + 3 -1}{3-1} = \binom{4}{2} = 6 \tag{4}$$

$$N(C_1C_3) = y_1+4+y_2+y_3+7 = 11 => y_1+y_2+y_3 = 0 => \binom{0 + 3 -1}{3-1} = \binom{2}{2} = 1 \tag{5}$$

$$N(C_2C_3) = y_1+y_2+5+y_3+7 = 11 => y_1+y_2+y_3 = -1\tag{6}$$

$$S_2 = Equation(4)+Equation(5)+Equation(6) = 7$$

Equation(4) is Total Number of Solutions with restriction $$x_1\ge4 \cup x_2 \ge5$$

At Equation(6) $$y_1+y_2+y_3=-1$$ means there is no solution for this spesific case.

So finally; $$N(\overline C_1\overline C_2\overline C_3) = S_0-S_1+S_2 => 78-79+7 = 6$$ total solutions with these restrictions.

I believe that is correct.

• Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. Mar 30 at 22:29
• Please type your solution rather than posting a link to an image since links can get broken and images cannot be searched. Also, users who use screen readers may not be able to read images. Mar 30 at 22:31