Question 148291: How many ways can a baseball coach choose the first, second and third batters in the lineup of a team of 15 players.
Found 2 solutions by Alan3354, alnavy70:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! For the 1st, he can choose any of 15
For the 2nd, any of the remaining 14
For the 3rd, any of the remaining 13
So for the 1st, there are 15 choices, then 14, then 13.
# of combinations = 15*14*13 = 2,730.
Answer by alnavy70(1)  (Show Source):
You can put this solution on YOUR website! In this particular question, the different ways of possible players that a coach can choose from the 1st, 2nd, and 3rd batter. Therefore, using The Rule of Sum we can solve.
For the 1st player he has 15 different ways to choose from
For the 2nd player he has 14 different ways to choose from
For the 3rd player he has 13 different ways to choose from
Therefore,
For the first experiment or the second or the last is 15 + 14 + 13 = 42 possible ways the coach can choose the 1st, 2nd, and 3rd batters.
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