how many batting lineups of the nine players can be made for a baseball team if the cathcher bats first, the shortstop second and the pitcher last?
If we leave out the catcher, the shortstop and the pitcher, that leaves us with 6 positions to fill.
There will be a choice of 6 people for the first of the 6 position, and 5 for the next, until the last there is one.
How many lineups can you make?
720!
We have then only 6 positions to play with:
6 * 5 * 4 * 3 * 2 * 1 = 6! = 720
To find the number of batting lineups, we can use the concept of permutations. Since the catcher, shortstop, and pitcher have fixed positions in the lineup, we need to determine the arrangements for the remaining six players.
Step 1: Calculate the number of arrangements for the remaining six players.
There are six remaining positions in the lineup (excluding the catcher, shortstop, and pitcher). We want to determine the number of arrangements for these six players. This can be done using a permutation formula:
6P6 = 6!/(6-6)! = 6! = 6*5*4*3*2*1 = 720.
Step 2: Calculate the number of batting lineups.
Since the positions for the catcher, shortstop, and pitcher are fixed, the number of batting lineups is equal to the number of arrangements for the remaining six players.
Thus, the number of batting lineups is 720.
In conclusion, there are 720 batting lineups that can be made for a baseball team if the catcher bats first, the shortstop bats second, and the pitcher bats last.