Question

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

A)120
B)792
C)95,040
D)3,991,680

Answer 1

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

Answer 2

Answer: B)792

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=(12!)/((12-5)!5!)............[^nC_r=(n!)/((n-r)!r!)}]\\=(12*11*10*9*8*7!)/(7!*5!)\\\\=(12*11*10*9*8)/(120)=792`

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