Question
Determine if the situation describes a permutation or a combination. Justify your reasoning. Then solve the problem.
A Little League manager has 15 players on her team. In how many different ways can she form a 9-player batting order?
A Little League manager has 15 players on her team. In how many different ways can she form a 9-player batting order?
Answers
GPT-4o mini
The situation describes a permutation because the order in which the players are arranged matters. In a batting order, the sequence of the players is significant (e.g., player in position 1 is different from player in position 2).
To calculate the number of different ways to form a 9-player batting order from 15 players, we use the formula for permutations:
\[
P(n, r) = \frac{n!}{(n-r)!}
\]
where \( n \) is the total number of items to choose from (in this case, 15 players) and \( r \) is the number of items to arrange (in this case, 9 players).
Substituting the values into the formula:
\[
P(15, 9) = \frac{15!}{(15-9)!} = \frac{15!}{6!}
\]
Now, we can compute this step by step. Since calculating \( 15! \) directly can be cumbersome, we'll compute it in terms of \( 6! \):
\[
P(15, 9) = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7
\]
Now we carry out the multiplication:
\[
15 \times 14 = 210
\]
\[
210 \times 13 = 2730
\]
\[
2730 \times 12 = 32760
\]
\[
32760 \times 11 = 360360
\]
\[
360360 \times 10 = 3603600
\]
\[
3603600 \times 9 = 32432400
\]
\[
32432400 \times 8 = 259459200
\]
\[
259459200 \times 7 = 1816234400
\]
Thus, the total number of different ways the manager can form a 9-player batting order is:
\[
\boxed{1816234400}
\]
To calculate the number of different ways to form a 9-player batting order from 15 players, we use the formula for permutations:
\[
P(n, r) = \frac{n!}{(n-r)!}
\]
where \( n \) is the total number of items to choose from (in this case, 15 players) and \( r \) is the number of items to arrange (in this case, 9 players).
Substituting the values into the formula:
\[
P(15, 9) = \frac{15!}{(15-9)!} = \frac{15!}{6!}
\]
Now, we can compute this step by step. Since calculating \( 15! \) directly can be cumbersome, we'll compute it in terms of \( 6! \):
\[
P(15, 9) = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7
\]
Now we carry out the multiplication:
\[
15 \times 14 = 210
\]
\[
210 \times 13 = 2730
\]
\[
2730 \times 12 = 32760
\]
\[
32760 \times 11 = 360360
\]
\[
360360 \times 10 = 3603600
\]
\[
3603600 \times 9 = 32432400
\]
\[
32432400 \times 8 = 259459200
\]
\[
259459200 \times 7 = 1816234400
\]
Thus, the total number of different ways the manager can form a 9-player batting order is:
\[
\boxed{1816234400}
\]