Question

A=(1,2,3...10).Determinati nr.-ul submultimilor cu trei elemente ale lui A care au elemente in progresie geometrica.

Answer 1

$C^3_{10} = \frac{10!}{3!(10-3)!} = \frac{1*2*3...*10}{1*2*3 *7!}= \frac{4*5*6*7..*10}{1*2*3*..*7}= \\ =\frac{ 8*9*10}{1*2*3}= 4*3*10=120$ de submultimi