Find number of integer solutions of linear equation We're given an equation.
$$
x_1 + x_2 + x_3 + x_4 + x_5 =21
$$
$$x_i \ge 0$$
aditionnal conditions are:
$$ 0\le x_1 \le 3$$
$$ 1 \le x_2 \le4$$
$$ 15 \le x_3$$
Task is to find all integer solutions to equation. This is a typical example of inclusion-exclusion principle. First number of solution, which satisfy
$$ 15 \le x_3$$ and $$ 1 \le x_2$$ are found. According to the formula, we calculate all selections with repetitions from a set of 5 elements $$x_1,x_2,x_3,x_4,x_5$$ and the length of selection is $$21 -15 -1 = 5$$ hence:  $$C(5 + 5 -1, 5 ) = 126$$
Assume, that given result is the universal set of solutions in this situation $U$
Now to use the inclusion-exclusion principle, opposite conditions are counted for $$x_1 \le 3$$ and $$x_2 \le4 $$ These conditions are respectively $$x_1\ge 4, x_2\ge 5$$ To find them calculate $$C_1(5+17-1,17) = 5985$$ and $$C_2(5+16-1,16) = 4845$$
Notice, that these results CAN NOT BE USED TO FIND THE ANSWER, as each of them includes the $x_3 < 15$ numbers and $C_1$ also includes $x_2 < 1$ numbers (this does not satisfy the initial restrictions). Hence, after including these two conditions: 
1) Number of solutions, which satisfy:
$$
4 \le x_1, 1 \le x_2, 15 \le x_3
$$
This is:
$$C_3(5 + (21-4-1-15) -1, (21-4-1-15)) = 5$$
2) Number of solutions, which satisfy:
$$
5 \le x_2, 15 \le x_3
$$
This is:
$$C_4(5 + (21-5-15) -1, (21-5-15)) = 5$$
Taking all of calculations into consideration, by principle of inclusion-exclusion we find that result is:
$$
C -C_3 - C_4= 126 -5 -5 = 116.
$$
 A: Your error is that you lost the condition $x_3\geq 15$. You still have to add that into the mix when you assume $x_1\geq 4$ and/or $x_2\geq 5$. In particular, there are no cases where $x_3\geq 15$ and $x_1\geq 4$ and $x_2\geq 4$.
Give a set $U$, and two subsets $V,W$, you get:
$$|U\setminus (V\cup W)| = |U|-|V| - |W| +|V\cap W|$$
But your $V,W$ are not subsets - they include cases where $x_3<15$, which your original count for $U$ was cases where $x_3\geq 15$.
I'd use generating functions to solve this problem.
As you indirectly noticed, you can think of these conditions as: $$y_1+y_2+y_3+y_4+y_5=5; y_i\geq 0\text{ and } y_1,y_2\leq 3.$$ 
You are trying to find the coefficient of $x^5$ in the expression:
$$(1+x+x^2+x^3)^2(1+x+x^2+\cdots)^3 = \frac{(1-x^4)^2}{(1-x)^5} = \frac{1-2x^4+x^8}{(1-x)^5}$$
And $$\frac{1}{(1-x)^5} = \sum_{k=0}^{\infty} \binom{k+4}{4}x^k$$
So the coefficient of $x^5$ in:
$$(1-2x^4+x^8)\sum_{k=0}^{\infty} \binom{k+4}{4}x^k$$
is $\binom{9}{4} - 2\binom{5}{4}=116$.
This is essentially the same result you'll get with inclusion-exclusion. We have $\binom{9}{4}=|U|$ and $\binom{5}{4}=|V|=|W|$, and $|V\cap W|=0$.
