Rozwiązujemy równanie:

$\left(\begin{array}{c}n+2\\ 4\end{array}\right)=5\cdot \left(\begin{array}{c}n\\ 3\end{array}\right),n\ge 3,n\in \mathbb{N}$

$\frac{\left(n+2\right)!}{4!\cdot \left(n+2-4\right)!}=5\cdot \frac{n!}{3!\cdot \left(n-3\right)!}$

$\frac{\left(n+2\right)!}{4!\cdot \left(n-2\right)!}=5\cdot \frac{n!}{3!\left(n-3\right)!}$    

$\frac{\left(n+2\right)\left(n+1\right)\cdot n\cdot \left(n-1\right)\left(n-2\right)!}{4!\cdot \left(n-2\right)!}=5\cdot \frac{n\cdot \left(n-1\right)\left(n-2\right)\left(n-3\right)!}{3!\cdot \left(n-3\right)!}$  

$\frac{\left(n+2\right)\left(n+1\right)\cdot n\cdot \left(n-1\right)}{4\cdot 3\cdot 2}=5\cdot \frac{n\cdot \left(n-1\right)\left(n-2\right)}{3\cdot 2}\mid \cdot 6$  

$\frac{\left(n+2\right)\left(n+1\right)\cdot n\cdot \left(n-1\right)}{4}=5\cdot n\cdot \left(n-1\right)\left(n-2\right)\mid \cdot 4$  

$\left(n+2\right)\left(n+1\right)\cdot n\cdot \left(n-1\right)=20\cdot n\cdot \left(n-1\right)\left(n-2\right)\mid :\left(n-1\right)\ne 0$  

$\left(n+2\right)\left(n+1\right)\cdot n=20n\cdot \left(n-2\right)$  

$\left(n^2+3n+2\right)\cdot n=20n^2-40n$

$n^3+3n^2+2n=20n^2-40n\mid -20n^2+40n$

$n^3-17n^2+42n=0$  

$n\left(n^2-17n+42\right)=0$    

$n=0<3\text{lub}{n}^2-17n+42=0$

$\Delta ={\left(-17\right)}^2-4\cdot 1\cdot 42=289-168=121,\sqrt{\Delta }=11$

$n=\frac{17+11}{2}=\frac{28}{2}=14\text{lub}n=\frac{17-11}{2}=\frac{6}{2}=3$    

Wnioskujemy, że rozwiązaniem równania jest n=14 lub n=3.