Question

Sa se rezolvare ecuatia: C de n luate cate 8 = C de n luate cate 10

Answer 1

Conditiile de existenta:
n≥8 si n≥10=> n≥10, n numar natural.
[tex]C_n^8=C_n^{10}\\ \frac{n!}{8!(n-8)!} = \frac{n!}{10!(n-10)!}\\ \frac{n!}{8!(n-10)!(n-9)(n-8)} = \frac{n!}{8!\cdot9\cdot 10(n-10)!}\\ (n-9)(n-8)=9\cdot 10\\ n=18[/tex]

Answer 2

n≥10,n numar natural
[tex]C_n^k=\frac{n-k+1}{k}C_n^{k-1}\\ C_n^{10}=\frac{n-10+1}{10}C_n^9=\frac{n-9}{10}C_n^{9}=\frac{n-9}{10}\frac{n-9+1}{9}C_n^8=\frac{n-9}{10}\frac{n-8}{9}C_n^8=C_n^8\Leftrightarrow\\ \frac{n-9}{10}\frac{n-8}{9}=1\Leftrightarrow(n-9)(n-8)=90 [/tex]
Produsul a doua numere naturale consecutive este 90. Numerele sunt 9 si 10,
n-8=10 ⇒ n=18