Q: Find the number of positive integer solutions to the equation:
$$\begin{align} (x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77. \end{align}$$
The solution given is $\begin{pmatrix} 6\\2 \end{pmatrix}$ $\begin{pmatrix} 10\\3 \end{pmatrix}$$+$ $\begin{pmatrix} 10\\2 \end{pmatrix}$ $\begin{pmatrix} 6\\3 \end{pmatrix}$.
I am clueless as to how can I go about solving this, given that the method of solving this question should centre around the concepts of combinations, distribution problems and multisets, in particular the formulas for $r$ $\mathbb identical $ objects, $n$ $distinct$ boxes:
Each box can hold at most 1 object:$$\begin{align} C_r^n = \begin{pmatrix} n\\r \end{pmatrix} \end{align}$$
Each box can hold any number of obects: $$\begin{align} H_r^n = \begin{pmatrix} {r+n-1}\\r \end{pmatrix} \end{align}$$
Each box must hold at least one object - Put 1 object in each box first: r-n objects left. Apply (2), then number of ways: $$\begin{align} \begin{pmatrix} {r-n+n-1}\\{r-n} \end{pmatrix} = \begin{pmatrix} {r-1}\\{n-1} \end{pmatrix}\end{align}$$
Some help will be deeply appreciated.
Hint:
You probably have specific examples of how to find the number of integer solutions to either $x_1+x_2+x_3=k$ or $y_1+y_2+y_3+y_4=k$, for any constant $k$. (If you don't: think about trying to distribute those $k$ $1$'s into boxes, such that each box gets at least one.)
To reduce there: note that $77=7\cdot11$, which are both prime. So, if $A\cdot B=77$, you must have one of:
Only two of these are possible with all $x_i$ and $y_j$ positive.