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?

Question

Audrey Andrews · Accepted Answer

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

distinct batting orders #=7"!"=5","040# hope that helped!

Madeline Adams · Answer

undefined 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.

1. Choose the shortstop: There is only 1 shortstop to choose from.
2. Choose the players for the second and third positions: There are 8 remaining players after selecting the shortstop.
3. Choose the player for the fourth position (right fielder): There is only 1 right fielder to choose from.
4. 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.