Question

n!          12n!
__    =  ___
(n-4)!    (n-2)!

Answer 1

[tex] \frac{n!}{(n-4)!} = \frac{12n!}{(n-2)!} \\\\ \frac{(n-4)!(n-3)(n-2)(n-1)n}{(n-4)!} = \frac{12(n-2)!(n-1)n}{(n-2)!} \\\\ (n-3)(n-2)(n-1)n = 12n(n-1) \\ (n-2)(n-3) = 12 \\ n^2 - 5n + 6 - 12 = 0 \\ n^2 - 5n - 6 = 0 \\\\ \mbox{Se rezolva ecuatia de gradul 2 si obtinem solutiile 6 si -1.}\\ n! \: exista <=> n \in\:N => n = 6[/tex]