Question

A baseball team has 15 field players and 10 pitchers. Each field player can take any of the 8 non-pitching positions. What is the number of possible starting lineup (8 field player 1 pitcher)

Answer 1

Answer:

If order matters: 2,594,592,000 ways.

If order does not matter:  64,350 ways

Step-by-step explanation:

Assuming that the order matters when picking the non-pitching positions (since they are different positions), the number of possible different starting lineups is given by the permutation of picking 8 players out of 15, multiplied by 10 (pick one out of 10 pitchers):

$n =10*\frac{15!}{(15-8)!}\\ n=10*15*14*13*12*11*10*9*8\\n=2,594,592,000$

Now if the order of the field players is not important, the number of possible starting lineups is given by the combination of picking 8 players out 15, multiplied by 10:

$n = n =10*\frac{15!}{(15-8)!8!}\\ n=\frac{10*15*14*13*12*11*10*9*8}{8*7*6*5*4*3*2} \\n=64,350$

Therefore, the number of ways to pick the starting lineup is:

If order matters: 2,594,592,000 ways.

If order does not matter:  64,350 ways