Question

Cine ma poate ajuta?

Answer 1

Formula pentru aranjamente:
$A_n^k=\frac{n!}{(n-k)!}$

Pentru factorial vom avea nevoie de aceasta proprietate:
$n!=1\cdot2\cdot3\cdot...\cdot(n-1)\cdot n=(n-1)!\cdot n$


Inlocuim la fiecare folosind formula:
a)
[tex]A_{x+2}^2=56(x+2)\\\\ \frac{(x+2)!}{((x+2)-2)!}=56(x+2)\\\\ \frac{(x+2)!}{x!}=56(x+2)\\\\ \frac{(x+2)\cdot(x+1)!}{x!}=56(x+2)\\\\ \frac{(x+2)(x+1)\cdot x!}{x!}=56(x+2)\\\\ (x+2)(x+1)=56(x+2)\\\\ x+1=56\rightarrow \boxed{x=55}[/tex]

b)
[tex]A_{x+1}^2=30\\\\ \frac{(x+1)!}{((x+1)-2)!}=30\\\\ \frac{(x+1)!}{(x-1)!}=30\\\\ \frac{(x-1)!\cdot(x)\cdot(x+1)}{(x-1)!}=30\\\\ x(x+1)=30\\\\ x\in N\rightarrow \boxed{x=5}[/tex]

c)
[tex]A_{n-1}^5=18\cdot A_{n-3}^4\\\\ \frac{(n-1)!}{((n-1)-5)!}=18\cdot\frac{(n-3)!}{((n-3)-4)!}\\\\ \frac{(n-1)!}{(n-6)!}=18\cdot\frac{(n-3)!}{(n-7)!}\\\\ \frac{(n-1)!}{(n-6)\cdot(n-7)!}=18\cdot\frac{(n-3)!}{(n-7)!}\\\\ \frac{(n-1)!}{n-6}=18\cdot(n-3)!\\\\ \frac{(n-1)(n-2)(n-3)!}{n-6}=18\cdot(n-3)!\\\\ \frac{(n-1)(n-2)}{n-6}=18\\\\ (n-1)(n-2)=18(n-6)\\\\ n^2-3n+2=18n-108\\\\ n^2-21n+110=0\\\\ n^2-10n-11n+110=0\\\\ n(n-10)-11(n-10)=0\\\\ (n-10)(n-11)=0\rightarrow \boxed{n\in\{10,11\}}[/tex]