How many batting lineups of the nine players can be
made for a baseball team if the catcher bats first, the
shortstop second, and the pitcher last?
9-3= 6
We have then only 6 positions left to play with:
6 * 5 * 4 * 3 * 2 * 1 = 6! = 720
To determine the number of batting lineups for a baseball team with nine players, where the catcher bats first, the shortstop bats second, and the pitcher bats last, we need to calculate the number of possible arrangements for the remaining 6 players in the lineup.
Here's how we can do it:
Step 1: Determine the number of possibilities for the remaining 6 positions in the lineup.
Since the first and second positions are fixed with the catcher and the shortstop, we have 6 remaining positions to fill with the remaining 7 players (including the pitcher). The number of possibilities for each position decreases by one as each position is filled.
So, for the first remaining position, there are 7 possibilities.
For the second remaining position, there are 6 possibilities.
For the third remaining position, there are 5 possibilities.
For the fourth remaining position, there are 4 possibilities.
For the fifth remaining position, there are 3 possibilities.
For the sixth remaining position, there are 2 possibilities.
Step 2: Calculate the total number of lineups.
To calculate the total number of lineups, we need to multiply the number of possibilities for each remaining position together.
Total lineups = 7 x 6 x 5 x 4 x 3 x 2 = 5040
Therefore, there are 5040 possible batting lineups for the baseball team with the given conditions.