<s>[INST] Find the number of positive integer solutions to the equation: $ (x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77. $ 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.
 [/INST] 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:


*

*$A=1$, $B=77$

*$A=7$, $B=11$

*$A=11$, $B=7$

*$A=77$, $B=1$


Only two of these are possible with all $x_i$ and $y_j$ positive.
</s>