<s>[INST] How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow  x_{2} + x_{3} + x_{4} = 40 - x_{1} \quad (*)$
Consider:
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 0, x_{3} \geq 4, x_{4} \geq 0 \quad (**)$
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 5, x_{3} \geq 4, x_{4} \geq 6 \quad (***)$
Let $f$ is the function that compute the number of non-negative solutions of an equation.
$\implies f(*) = f(**) - f(***)$ 
Thus, the number of non-negative solutions of (*) is $\sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1  \choose 3 - 1} - {25-x_{1}+3 - 1 \choose 3 - 1}) = 3045$
I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!
 [/INST] Math answer
Note that the given constraints for $x_1, x_2$ and $x_4$ and $\sum\limits_{i = 1}^4 x_i = 40$ allows us to define
$$
\begin{aligned}
x_3 &= 40 - x_1 - x_2 - x_4 \\
&\ge 40 - 8 - 4 - 5 \\
&= 23.
\end{aligned}$$
This renders the constraint $x_3 \ge 4$ redundant.  As a result, the required answer is $(8-2+1) \times (4+1) \times (5+1) = 210$.

Julia Programming Script
x1 = 2:8
x2 = 0:4
x4 = 0:5
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
println(minimum(x3))  # returns 23
println(length(x3))   # returns 210

Test this script on Tutorial's Point's online compiler.
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