# Probability with batting orders

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• Sep 27th 2012, 02:33 PM
jthomp18
Probability with batting orders
How many batting orders (the order in which baseball players face an opposing pitcher in a game) does a manager have available for the nine baseball players of a team, if the center fielder must bat 4th and the pitcher must bat 9th?

1. 5040

2. 362,880

3. 181,440

4. 40,320
• Sep 27th 2012, 04:53 PM
chiro
Re: Probability with batting orders
Hey jthomp18.

For this problem you are fixing two out of the nine which means you are left to re-order the rest how you please.

If there is no other constraints for this problem then think about how possibilities you get for each position.

The first position has 7 choices. You fix one of these choices and the second position has 6 (since you already used up 1 for the first). Then you have 5,4,3,2 and 1 choice(s) for the rest of them. These are all independent choices (once you choose the first choice, the second choice only depends on the number of people that are left) so you can multiply these choices.

What does this process give you as an answer?
• Sep 29th 2012, 09:05 AM
Tim28
Re: Probability with batting orders
This is permutations. 7! =5040
• Sep 29th 2012, 11:16 AM
HallsofIvy
Re: Probability with batting orders
Quote:

Originally Posted by Tim28
This is permutations. 7! =5040

chiro had explained HOW to get the answer. Do you think it is at all helpful to simply give the answer without any explanation?