[tex]\begin{aligned}&9P(n,5)=P(n,3)\cdot P(9,3)\\&9\cdot\dfrac{n!}{(n-5)!}=\dfrac{n!}{(n-3)!}\cdot\dfrac{9!}{6!}\\&\dfrac{n!}{(n-5)!}=\dfrac{n!}{(n-3)!}\cdot7\cdot8\\&\dfrac{1}{(n-5)!}=\dfrac{56}{(n-3)!}\\&(n-4)(n-3)=56\\&n^2-3n-4n+12-56=0\\&n^2-7n-44=0\\&n^2-11n+4n-44=0\\&n(n-11)+4(n-11)=0\\&(n+4)(n-11)=0\\&n=-4 \vee n=11\end[/tex]
[tex]n[/tex] can't be negative, therefore [tex]n=11[/tex]
