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...

You solved the problem in the nonnegative integers rather than the positive integers.
You wish to determine the number of solutions of the equation 
$$x_1 + x_2 + x_3 = 15 \tag{1}$$
in the positive integers subject to the constraints $x_1 < 6$ and $x_2 > 6$.  
Let's deal with the constraint $x_2 > 6$ first.  Let $y_2 = x_2 - 6$.  Then $y_2$ is a positive integer.  Substituting $y_2 + 6$ for $x_2$ in equation 1 yields 
\begin{align*}
x_1 + y_2 + 6 + x_3 & = 15\\
x_1 + y_2 + x_3 & = 9 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers.  A particular solution of equation 2 in the positive integers corresponds to a placement of addition signs in two of the eight spaces between successive ones in a row of nine ones.  For instance, 
$$1 1 + 1 1 1 1 1 + 11$$
corresponds to the solution $x_1 = 2$, $y_2 = 5$, and $x_3 = 2$ (or $x_1 = 2$, $x_2 = 11$ and $x_3 = 2$ in equation 1), while 
$$1 1 1 1 + 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 4$, $y_2 = 2$, and $x_3 = 3$ (or $x_1 = 4$, $x_2 = 8$, and $x_3 = 3$ in equation 1).  Therefore, the number of  solutions of equation 2 is 
$$\binom{8}{2}$$
From these, we must exclude those solutions in which $x_1 \geq 6$.  Assume $x_2 \geq 6$.  Let $y_1 = x_1 - 5$.  Then $y_1$ is a positive integer.  Substituting $y_1 + 5$ for $x_1$ in equation 2 yields
\begin{align*}
y_1 + 5 + y_2 + x_3 & = 9\\
y_1 + y_2 + y_3 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers.  The number of solutions of equation 3 is the number of ways we can place two addition signs in the three spaces between successive ones in a row of four ones, which is 
$$\binom{3}{2}$$
Hence, the number of solutions of equation 1 subject to the constraints $x_1 < 6$ and $x_2 > 6$ is 
$$\binom{8}{2} - \binom{3}{2}$$
