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:

  1. Each box can hold at most 1 object:$$\begin{align} C_r^n = \begin{pmatrix} n\\r \end{pmatrix} \end{align}$$

  2. Each box can hold any number of obects: $$\begin{align} H_r^n = \begin{pmatrix} {r+n-1}\\r \end{pmatrix} \end{align}$$

  3. 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.

  • 1
    $\begingroup$ Factor $77$, what does this imply about the values of $(x_1 + x_2 + x_3)$ and $(y_1 + y_2 + y_3 + y_4)$? From there on, your knowledge of partitioning numbers and combinatorics should come in handy $\endgroup$ – J. Marx-Kuo Sep 23 '16 at 4:12
  • $\begingroup$ Is it something like what @Nick Peterson suggested below? $\endgroup$ – Stoner Sep 23 '16 at 5:59
  • $\begingroup$ Yes, his last sentence is especially important $\endgroup$ – J. Marx-Kuo Sep 23 '16 at 6:52


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.

  • 1
    $\begingroup$ Oh @Nick Peterson do you mean $x_1+x_2+x_3=7$ and $y_1+y_2+y_3+y_4=11$ , for which in total there are 2 possibilities as stated above and so I have to consider all the aforementioned cases? (Since all $x_i, y_i > 0$) $\endgroup$ – Stoner Sep 23 '16 at 5:59
  • 1
    $\begingroup$ Alright I got it! Thank you for your suggestion! $\endgroup$ – Stoner Sep 23 '16 at 7:41

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