Question

Rezolvati exercitiul 3 de la permutatii din poza atasata

Answer 1

$a)\:\frac{(n+2)!}{n!}=72\\\\\frac{n!(n+1)(n+2)}{n!}=72\\(n+1)(n+2)=72\\n^2+2n+n+2-72=0\\n^2+3n-70=0\\\Delta=9+280=289\rightarrow\sqrt{\Delta}=17\\n_{1,2}=\frac{-3\pm17}{2}\\n_1\ \textless \ 0\\n_2=7$

$b)\:\frac{n!}{(n-4)!}=\frac{12n!}{(n-2)!}\\\\\frac{(n-4)!(n-3)(n-2)(n-1)n}{(n-4)!}=\frac{(n-2)!(n-1)n}{(n-2)!}\\n(n-1)(n-2)(n-3)=12n(n-1)\\n(n-1)(n^2-3n-2n+6)=12n^2-12n\\n(n-1)(n^2-5n+6)=12n^2-12n\\n(n^3-5n^2+6n-n^2+5n-6)=12n^2-12n\\n(n^3-6n^2+11n-6)=12n^2-12n\\n^4-6n^3+11n^2-6n=12n^2-12n\\n^4-6n^3+11n^2-6n-12n^2+12n=0\\n^4-6n^3-n^2+6n=0\\n(n^3-6n^2-n+6)=0\\n(n^2(n-6)-1\cdot(n-6))=0\\n(n-6)(n^2-1)=0\\n(n-6)(n-1)(n+1)=0\rightarrow(n=0\text{ sau }n=6\text{ sau }n=1\text{ sau }n=-1)\\n\in\{0,\:1,\:6\}$\\
$c)\:\frac{1}{n!}-\frac{1}{(n+1)!}=\frac{n^3}{(n+2)!}\\\\\frac{(n+1)(n+2)}{n!(n+1)(n+2)}-\frac{n+2}{(n+1)!(n+2)}=\frac{n^3}{(n+2)!}\\\\n^2+2n+n+2-n-2=n^3\\n^2+2n=n^3\\n^3-n^2-2n=0\\n(n^2-n-2)=0\\n(n-1)(n-2)=0\rightarrow(n=0\text{ sau } n=1\text{ sau }n=2)\\n\in\{0,\:1,\:2\}$