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.
Let $S_{a, b, c, d, e} $ be the number of solutions with $x_1\ge a$, $x_2 \ge b$, $x_3\ge c$, $x_4\ge d$ and $x_5\ge e$.
The equation can be written as
$$(x_1-a)+(x_2-b)+(x_3-c)+(x_4-d)+(x_5-e)=21-a-b-c-d-e$$
So we have
$$S_{a, b, c, d, e} =\binom{21-a-b-c-d-e+4}{4}$$
if $a+b+c+d +e\le21$ and is $0$ otherwise.
The answer to this question is
$$S_{0, 1,15,0,0}-S_{4,1,15,0,0}-S_{0,4,15,0,0}+S_{4,4,15,0,0}=\binom{9}{4}-\binom{5}{4}-\binom{6}{4}+0=106$$
Best Answer
A particular solution to the equation in the non-negative integers corresponds to the placement of four addition signs in a row of $28$ ones. For instance, $$1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 + + 1 1 1 1$$ corresponds to the solution $x_1 = 9$, $x_2 = 8$, $x_3 = 7$, $x_4 = 0$, and $x_5 = 4$. Thus, the number of solutions of the equation in the non-negative integers is the number of ways four addition signs can be placed in a row of $28$ ones, which is $$\binom{28 + 4}{4} = \binom{32}{4}$$ since we must choose which four of the $32$ symbols (four addition signs and $28$ ones) will be addition signs.
In general, the number of solutions of the equation $$x_1 + x_2 + \cdots + x_k = n$$ in the non-negative integers is equal to the number of ways $k - 1$ addition signs can be inserted into a row of $n$ ones, which is $$\binom{n + k - 1}{k - 1} = \binom{n + k - 1}{n}$$ since we must choose which $k - 1$ of the $n + k - 1$ symbols ($n$ ones and $k - 1$ addition signs) will be addition signs or, alternatively, which $n$ of the symbols will be ones.
A particular solution of the equation in the positive integers corresponds to the placement of four addition signs in the $27$ spaces between successive ones in a row of $28$ ones. For instance, $$1 1 1 1 1 1 + 1 1 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1$$ corresponds to the solution $x_1 = 6$, $x_2 = 7$, $x_3 = 6$, $x_4 = 5$, and $x_5 = 4$. Thus, the number of solutions of the equation in the positive integers is the number of ways four addition signs can be placed in the $27$ gaps between successive ones in a row of $28$ ones, which is $$\binom{27}{4}$$ In general, the number of solutions of the equation $$x_1 + x_2 + \cdots + x_k = n$$ in the positive integers is the number of ways $k - 1$ addition signs can be placed in the $n - 1$ gaps between successive ones in a row of $n$ ones, which is $$\binom{n - 1}{k - 1}$$
Hint: Let $y_k = x_k - k$, $1 \leq k \leq 5$. Substitute $y_k + k$ for $x_k$, $1 \leq k \leq 5$, then solve the resulting equation in the non-negative integers.