Răspuns:
[tex]\frac{5!}{n!(5-n)!} \leq \frac{5!}{(n-2)!(5-n+2)!} \\
\frac{1}{n(n-1)} \leq \frac{1}{(6-n)(7-n)} \\
(6-n)(7-n)\leq n(n-1)\\
42-13n+n^{2} \leq n^{2} -n\\
42\leq 12n \\
\frac{42}{12} \leq n\\
3,5\leq n\\
n poate fi doar natural n=4, 5, 6.......[/tex]
Explicație pas cu pas:
