Question

There are 12 people on a basketball team. How many different five-person starting lineups can be chosen

Answer 1

12!/5!7! = (12*11*10*9*8)/(5*4*3*2*1)=792

Answer 2

Answer: B)792

Step-by-step explanation:

Given: The number of people in basketball team = 12

To choose different five-person starting lineups , we will use combinations.

The number of different five-person starting lineups can be chosen  will be given by :-

$^{12}C_5=\frac{12!}{(12-5)!5!}............[^nC_r=\frac{n!}{(n-r)!r!}}]\\=\frac{12\times11\times10\times9\times8\times7!}{7!\times5!}\\\\=\frac{12\times11\times10\times9\times8}{120}=792$

Hence, 792 different five-person starting lineups can be chosen .