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?
To determine the number of ways the manager can put together a batting order with the shortstop batting 3rd, we can use the principle of counting.
Step 1: Identify the number of players who are not the shortstop.
In this case, there are 8 players who are not the shortstop.
Step 2: Determine the number of ways to arrange the non-shortstop players.
Since the shortstop will bat 3rd, we can treat the remaining 8 players as a group. The number of ways to arrange this group can be calculated as 8!.
Step 3: Determine the number of ways to arrange the batting order.
Since the shortstop is fixed in the 3rd position, we need to arrange the remaining 8 players and the shortstop in the remaining 8 positions. This can be calculated as 8! * 1! (as the shortstop is fixed).
Step 4: Calculate the number of ways.
To find the total number of ways, we multiply the number of ways to arrange the non-shortstop players by the number of ways to arrange the batting order, which gives us 8!.
Therefore, the manager of the baseball team can put together a batting order of his nine players, with the shortstop batting 3rd, in 8! ways, which is equal to 40,320 ways.
To find the number of ways the manager can put together a batting order of his nine players, with the shortstop batting 3rd, we can use the concept of permutations.
Let's break down the problem:
1. The shortstop must bat 3rd. This means that one spot in the batting order is fixed.
2. We have 8 remaining players to consider for the remaining 8 spots in the batting order. The order in which these players are placed matters.
To calculate the number of ways to order these 8 players, we use the concept of permutations.
The formula for permutations is:
nPr = n! / (n - r)!
Where n is the total number of players and r is the number of available spots in the batting order.
In this case, n = 8 (since the shortstop is already fixed at the 3rd spot) and r = 8 (since there are 8 remaining spots).
Plugging these values into the formula:
8P8 = 8! / (8 - 8)! = 8! / 0! = 8!
The factorial (!) represents the product of all positive integers less than or equal to the given number.
Calculating 8!:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
Therefore, there are 40,320 ways the manager can put together the batting order of his nine players, given that the shortstop must bat 3rd.