# Given 9 starting batters on a baseball team, how many batting orders are possible if the shortstop must bat first and the right fielder must bat fourth?

Since two of the positions are filled (position 1 and position 4), that leaves the remaining 7 positions open for permutation ...

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To calculate the number of possible batting orders, we first consider the positions that must be filled in a specific order: the shortstop (first position) and the right fielder (fourth position). Once those positions are filled, the remaining players can be arranged in any order.

- Choose the shortstop: There is only 1 shortstop to choose from.
- Choose the players for the second and third positions: There are 8 remaining players after selecting the shortstop.
- Choose the player for the fourth position (right fielder): There is only 1 right fielder to choose from.
- Choose the players for the fifth to ninth positions: After selecting the shortstop and right fielder, there are 7 remaining players.

Now, we multiply the number of choices for each step:

[ 1 \times 8 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 ]

[ = 1 \times 8 \times 1 \times 7! ]

[ = 8 \times 7! ]

[ = 8 \times 5040 ]

[ = 40,320 ]

Therefore, there are 40,320 possible batting orders if the shortstop must bat first and the right fielder must bat fourth.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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