Question

Buna ziua!
Stie cineva cum se rezolva ecuatia:  n!÷(n-4)!=12n!÷(n-2)!

Answer 1

$\frac{n!}{(n-4)!}= \frac{12n!}{(n-2)!}=> \frac{(n-4)!(n-3)(n-2)(n-1)n}{(n-4)!}= \frac{12(n-2)!(n-1)n}{(n-2)!} =>$
$(n-3)(n-2)=12=> n^{2}-5n+6=12=> n^{2}-5n-6=0$
cu solutiile $n_{1}= 3; n_{2}=2$