Question

aranjamente de n luate cate 8 plus aranjamente de n luate cate 7 egal cu 9 aranjamente de n luate cate 6

Answer 1

[tex] A^8_{n} + A^7_{n}=9 A^6_{n} \\ \frac{n!}{(n-8)!}+ \frac{n!}{(n-7)!} =9 \frac{n!}{(n-6)!} \\ n \geq 8 \\ \frac{n!(n-7)(n-6)+n!(n-6)-9n!}{(n-6)!}=0 \\ n^{2}-13n+42+n-6-9=0 \\ n^{2}-12n+27=0 \\ n=3;n=9 n=3-nu .poate.fi.solutie. [/tex]
S={9}