Question

A baseball manager has 11 players of the same ability. How many 9 player starting lineups can he

Answer 1

Answer:

55

Step-by-step explanation:

Combination will be used here as teams are a group.

So using combination:

No. of combination of 'n' things taken 'r' at a time

$\limits^n {C} \,_r$= $\frac{n!}{r! (n-r)!}$

here n = 11 ,

r = 9

11C9 =  $\frac{11!}{9! (11-9)!}$ = $\frac{11!}{9! (2!)}$

11C9 = 55

Which are the possible 9 player starting lineups