Question

Determinati n∈N,n≥2, stiind ca aranjamente n luate cate 2 este egal cu 110.
Va rog ma puteti ajuta cu rezolvarea pas cu pas

Answer 1

Răspuns:

Explicație pas cu pas:

$\frac{n!}{(n-2)!} =110\\\frac{1*2*3*...*(n-2)*(n-1)*n}{1*2*3*...*(n-2)} =110\\\\\\(n-1)*n=110\\n^2-n=110\\n^2-n-110=0\\delta=(-1)^2-4*1*(-110)=1+440=441\\n1=\frac{-(-1)-\sqrt{delta} }{2*1} =\frac{1-\sqrt{441} }{2}=\frac{1-21}{2} =\frac{-20}{2} =-10\\n2=\frac{-(-1)+\sqrt{delta} }{2*1} =\frac{1+\sqrt{441} }{2}=\frac{1+21}{2} =\frac{22}{2} =11$

dar n e numar natural rezulta doar solutia n2=11

Answer 2

$\it\ A^2_n=110 \Rightarrow \dfrac{n!}{(n-2)!}=110 \Rightarrow \dfrac{(n-2)!(n-1)\cdot n}{(n-2)!}=110 \Rightarrow (n-1)\cdot n =110\\ \\ (n-1)\cdot n= 10\cdot11 \Rightarrow n=11$