$\left( {\begin{array}{*{20}{c}}
  n \\ 
  {r – 1} 
\end{array}} \right)\,:\,\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)\,\, = \,36\,:\,84\,$  અને $\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)\,:\,\left( {\begin{array}{*{20}{c}}
  n \\ 
  {r + 1} 
\end{array}} \right)\, = \,84\,:\,126$

$\frac{{n\,!}}{{(r – 1)(n – r + 1)\,!}}.\,\frac{{r\,!(n – r)\,!}}{{n\,!}} = \frac{3}{7}\,$ 

$\frac{{n\,!}}{{r\,!(n – r)!}}.\,\frac{{(r + 1)\,!\,(n – r – 1)\,!}}{{n\,!}} = \frac{2}{3}$ અને

$\frac{{n\,!}}{{r\,!(n – r)!}}.\,\frac{{(r + 1)\,!\,(n – r – 1)\,!}}{{n\,!}} = \frac{2}{3}$

$\frac{r}{{n – r + 1}} = \frac{3}{7}$  અને $\frac{{r + 1}}{{n – r}} = \frac{2}{3}$

$7r = 3n – 3r + 3$ અને $3r + 3 = 2n – 2r$

$3n – 10r = -3$ અને $2n – 5r = 3$ એટલે કે, $4n – 10r = 6$

બંને સમીકરણ ઉકેલતાં $n = 9, r = 3$