Find the number of integer solutions for $x_1+x_2+x_3 = 15$ under some constraints by IEP. For equation
$$
x_1+x_2+x_3 = 15
$$
Find number of positive integer solutions on conditions:
$$
x_1<6, x_2 > 6
$$
Let: $y_1 = x_1, y_2 = x_2 - 6, y_3 = x_3$
than, to solve the problem, equation $y_1+y_2 +y_3 = 9$ where $y_1 < 6,0<y_2, 0<y_3 $ has to be solved. Is this correct?
To solve this equation by inclusion-exclusion, number of solution without restriction have to be found $C_1 (3+9-1,9)$ and this value should be subtracted by $C_2 (3+9-7-1,2)$ , (as the negation of $y_1 < 6$ is $y_1 \geq 7$).
Thus:
$$
55-6=49
$$
Is this the correct answer ? 
Problem must be solved using inclusion-exclusion...

(This corrects for the restriction that the integers are positive and not non-negative. Thanks for those who have commented to point this out.)
As the restrictions specify $x_1>0, x_2>6, x_3>0$ and $x\in\mathbb Z$, this is the same as $x_1\geq1, x_2\geq7, x_3\geq 1$.
First we "reserve" $1$ for $x_1$, $7$ for $x_2$ and $1$ for $x_3$ and apply stars-and-bars on the remaining. 
$$x_1+x_2+x_3=15\\
(\overbrace{y_1+1}^{x_1})+(\overbrace{y_2+7}^{x_2})+(\overbrace{y_3+1}^{x_3})=15\\
y_1+y_2+y_3=6
$$
$$\large\overbrace{*\;*}^{y_1}\;\bigl|\;\overbrace{*\;*}^{y_2}\;\bigl|\;\overbrace{*\;*}^{y_3}$$
Using stars-and-bars, without applying the upper constraint $x_1<6$, number of combinations is 
$$\binom {6+2}2$$
With the restriction $x_1<6$ which is the same as $y_1<5$ impossible combinations are for $y_1\geq 5$. From the $6$ elements, remove $5$, and use stars-and-bars. Number of impossible combinations is
$$\binom {1+2}2$$
Hence total number of combinations is 
$$\binom {8}2-\binom 32 =28-3=25\qquad\blacksquare$$
