Sejam $\,n, p\;\in\;\mathbb{N}^*\,$ e $\,x\;\in\;\mathbb{Z}\,$. Resolver a equação em $\,x\,$.
$\phantom{X}x^2\,-\,{\large \binom{n}{p}}x\,+\,{\large \binom{n\,-\,1}{p\,-\,1}}\centerdot{\large \binom{n\,-\,1}{p}}\;=\;0\phantom{X}$
resposta: $\,\left\{ {\large \binom{n\,-\,1}{p\,-\,1}}\; ; \; {\large \binom{n\,-\,1}{p}} \right\}\,$
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