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...
 A: Here is a different way to break it down
$$
x_1\in\{1,2,3,4,5\}
$$
and given $x_1$ we then have $x_1+x_2<15$ and $x_2>6$ combined as
$$
6<x_2<15-x_1
$$
And whenever $x_1$ and $x_2$ are given, the value of $x_3$ follows from them.
For $x_1=5$ we then have $x_2\in\{7,8,9\}$ so three choices for $x_2$. Each time $x_1$ is decreased by $1$ we gain one option for $x_2$. Thus we have a total of
$$
3+4+5+6+7 = 25
$$
sets of integer solutions under the given constraints.

I ran the following code snippet in Python which confirmed the figure of 25:
n = 0
for x1 in range(1,16):
    for x2 in range(1,16):
        for x3 in range(1,16):
            if x1 < 6 and x2 > 6 and x1+x2+x3 == 15:
                n += 1
                print n, ":", x1, x2, x3

I understand that I did not answer the question using the method required, but I wonder why I find the number of solutions to be $25$ whereas the OP and the other answer find it to be $49$. Did I misunderstand the question in the first place?
A: (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 
$$\binom {8}2-\binom 32 =28-3=25\qquad\blacksquare$$
A: 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 
$$\binom{8}{2} - \binom{3}{2}$$
