How many ways can the manager of a baseball team put together a batting order of his nine players, if the shortstop must bat 3rd?
If the shortstop must bat 3rd, we can fix the shortstop in the 3rd position. The rest of the players can be arranged in the remaining 8 positions in $(8-1)! = \boxed{7! = 5,\!040}$ ways.
To calculate the number of ways the manager can put together a batting order of nine players with the shortstop batting 3rd, we need to consider the arrangement of the other eight players.
Since the shortstop is fixed in the 3rd position, the batting order for the remaining eight players can be determined. There are 8 players left to choose from and 8 remaining positions to fill.
To calculate the number of ways we can arrange the remaining 8 players, we can use the concept of permutations. The formula for permutations is given by:
nPr = n! / (n - r)!
Where n is the total number of players to choose from and r is the number of remaining positions.
In this case, n = 8 (remaining players) and r = 8 (remaining positions). Plugging these values into the formula, we get:
8P8 = 8! / (8 - 8)! = 8! / 0! = 8!
The factorial (n!) of a number is the product of all positive integers less than or equal to n. In this case, 8! is equal to 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Calculating the value, we have:
8! = 40320
Therefore, there are 40,320 ways the manager can put together the batting order of the nine players, with the shortstop batting 3rd.