Question

Ma poate ajuta cineva la 4?

Answer 1

Cn(2)[combinari de n luate cate 2]=n!/2!(n-2)!=n(n-1)(n-2)!/2(n-2)!
Cn(2)=n(n-1)/2
n(n-1)/2=21
n²-n=42
n²-n-42=0
Δ=1+168=169
n1=(1-13)/2=-6∉N
n2=(1+13)/2=7 
Solutie ,n=7

Answer 2

[tex]\it C_n^2 =21 \Rightarrow \dfrac{n!}{2!(n-2)!} =21 \Rightarrow \dfrac{(n-2)!\cdot(n-1)\cdot n}{1\cdot2(n-2)!} =21 \Rightarrow \\\;\\ \Rightarrow \dfrac{(n-1)\cdot n}{2} =21 \Rightarrow (n-1)\cdot n=2\cdot21 \Rightarrow (n-1)\cdot n=42 \ \ \ \ (1) \\\;\\ Dar,\ 42 = 6\cdot7 \ \ \ \ (2) \\\;\\ (1),\ \ (2) \Rightarrow n = 7 [/tex]