Question

Combinari din n-2 elemente luate cate 2 + Combinari din n-3 elemente luate cate 2+ Combinari din n-4 elemente luate cate 2=19

Answer 1

   
[tex]\displaystyle\\ C_{n-2}^2+C_{n-3}^2+C_{n-4}^2=19\\\\ \frac{(n-2)(n-3)}{1\cdot2}+\frac{(n-3)(n-4)}{1\cdot 2}+\frac{(n-4)(n-5)}{1\cdot2}= 19\\\\ \frac{(n-2)(n-3)+(n-3)(n-4)+(n-4)(n-5)}{2}=19\\\\ (n-2)(n-3)+(n-3)(n-4)+(n-4)(n-5)=19\cdot 2\\\\ (n-2)(n-3)+(n-3)(n-4)+(n-4)(n-5)=38\\\\ (n^2-5n +6)+(n^2-7n+12)+(n^2-9n+20)=38\\\\ 3n^2-21n +38=38~~\Big|~-38\\\\ 3n^2-21n=0\\\\ n(3n-21)=0\\\\ n_1 = 0 ~\text{respins deoarece: } ~n;~n-2;~n-3;~n-4~~\ \textless \ 2\\\\ 3n-21=0\\\\ 3n=21\\\\ n_2=\frac{21}{3}=\boxed{7} [/tex]