I am not sure why you have four variables in the statement of your question and three variables in your answer. I will assume you meant to work with four variables.
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers.
Method 1: We reduce the problem to one in the non-negative integers. Let $y_k = x_k - 1$ for $1 \leq k \leq 4$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 1$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to placing three addition signs in a row of eight ones. For instance,
$$1 1 1 1 1 + 1 + 1 1 1$$
corresponds to the solution $y_1 = 5$, $y_2 = 1$, and $y_3 = 3$, while
$$1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 6$. Thus, the number of solutions of equation 2 is the number of ways three addition signs can be inserted into a row of eight ones, which is
$$\binom{8 + 3}{3} = \binom{11}{3}$$
since we must choose which three of the eleven symbols (eight ones and three addition signs) will be addition signs.
Method 2: A particular solution of equation 1 in the positive integers corresponds to inserting three addition signs in the eleven spaces between successive ones in a row of $12$ ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance,
$$1 1 1 + 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 5$. Thus, the number of solutions of equation 1 in the positive integers is the number of ways three addition signs can be inserted into the eleven gaps between successive ones in a row of $12$ ones, which is
$$\binom{11}{3}$$
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
We reduce the problem to one in the non-negative integers. Since $x_1$ is an integer, $x_1 > 1 \implies x_1 \geq 2$. Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \implies x_2 \geq 2$ and $x_3 > 3 \implies x_3 \geq 4$. Let
\begin{align*}
y_1 & = x_1 - 2\\
y_2 & = x_2 - 2\\
y_3 & = x_3 - 4\\
y_4 & = x_4
\end{align*}
Then each $y_k$, $1 \leq k \leq 4$, is a non-negative integer. Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields
\begin{align*}
y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with
$$\binom{4 + 3}{3} = \binom{7}{3}$$
solutions.
This can be done with generating functions, as suggested by JMoravitz in a comment, and it's quite possible to do it with a pencil, though you'll probably want to use a calculator to do the arithmetic; I certainly did.
First, to understand the comment, note that $x_1$ and $x_2$ are each represented by the polynomial $$x^4+x^5+x^6+\dots+x^{29}.$$
This is because $4\leq x_1,x_2\leq29$.
Similarly, $x_3$ and $x_4$ are each represented by the polynomial $$x^{10}+x^{11}+\dots+x^{40}.$$ Finally, $x_5$ and $x_6$ are each represented by the polynomial $$1+x+x^2+\dots+x^{109}$$ Since the sum is to be $109$, neither $x_5$ nor $x_6$ can be more than $109$, and they must be $\geq0$.
To multiply these $6$ polynomials, we choose a term from each polynomial, multiply them together, and add up the products over all choices of terms. The coefficients in the polynomials are all $1$, so the coefficient of $x^n$ in the product, for some natural number $n$, is just the number of ways to pick one term from each polynomial such that the sum of their exponents is $n$. When $n=109$, this is just the solution to our problem. For example, the solution $x_=20,x_2=14,x_3=30,x_4=30,x_5=15,x_6=0$ corresponds to choosing the terms $x^{20},x^{14},x^{30},x^{30},x^{15},1$ from the polynomials, in order.
Now it's just a matter of figuring out the coefficient of $x^{109}$ without multiplying the polynomials.
I'll use the notation $[x^n]p(x)$ to mean the coefficient of $x^n$ in the formal power series $p(x)$. We want $$c=[x^{109}]\left(x^4+\cdots+x^{29}\right)^2
\left(x^{10}+\cdots+x^{40}\right)^2
\left(1+x+x^2+\cdots\right)^2
$$
Note that we don't need an upper bound on the $x_5$ and $x_6$. It doesn't matter if we include exponents $>109$ in the polynomial, because they won't contribute anything to the coefficient of $x^{109}$ in the product. As you'll see, this simplifies the calculation, because we have two fewer factor of the numerator this way.
Then, using the formula for a geometric series,$$
\begin{align}
c&=[x^{109}]\left(x^4-x^{30}\right)^2
\left(x^{10}-x^{41}\right)^2(1-x)^{-6}\\
&=[x^{81}]\left(1-x^{26}\right)^2
\left(1-x^{31}\right)^2
\sum_{n=0}^\infty\binom{-6}{n}(-x)^n\\
&=[x^{81}](1-2x^{26}+x^{52})(1-2x^{31}+x^{62})
\sum_{n=0}^{81}(-1)^n\binom{n+5}{5}(-x)^n\\
&=[x^{81}](1-2x^{26}-2x^{31}+x^{52}+4x^{57}+x^{62})
\sum_{n=0}^{81}\binom{n+5}{5}x^n\\
\end{align}$$
since we may ignore terms of degree $>81$.
We just need to pick out the terms in the product that result in a term of degree $81.$ We have $$
\binom{86}{5}-2\binom{60}5-2\binom{55}5+\binom{34}5+4\binom{29}5+\binom{24}5=17,741,536
$$
Best Answer
The argument boils down to the following:
Since $x_1+...+x_6$ is less that $10$, it is missing something from $10$. Denote this quantity by $x_7$. By doing this you transformed the inequality into an equation, which in this case (and actually often) is easier to solve than an inequality.
Now for the second part: since you need $x_1+...+x_6$ to be strictly less than $10$, it follows that $x_7 \geq 1$. But the technique which you learned (stars and bars probably) works for variables which are non-negative, it doesn't work with restrictions of this form . To fix this note that $x_7-1 \geq 0$, and denote this by a new variable.
It would had probably been more intuitive to observe that $x_1+..+x_6 <10$ in integers is equivalent to $x_1+..+x_6 \leq 9$. Now denote by $x_7$ (or $y_7$) the missing quantity from $9$, which could be $0$, and reduce the problem directly to the second equation.