Avem relatia de factorial : n!=1*2*3*4*...*(n-2)(n-1)n, adica produsul a ''n'' factori nincepand de la 1, ex. 3!=1*2*3=6, 4!=1*2*3*4=3!*4=24, 6!=1*2*3*4*5*6=720 mai putem scrie si: 6!=5!*6=4!*5*6...
(n-2)!=1*2*3*...*(n-5)(n-4)(n-3)(n-2)=(n-4)!(n-3)(n-2).
Deci: [tex] \frac{(n-2)!}{(n-4)!}=6,sau, \frac{(n-4)!(n-3)(n-2)}{(n-4)!}=6,simplificand, (n-3)(n-2)=6 [/tex].
Inmultim si obtinem ecuatia n²-5n=0, cu radacinile 0 si 5, dar conditia era n≥4, deci solutia x=5,
