A baseball manager decides to put his strongest batter in either position.... REST BELOW please help I'm desperate

Question

A baseball manager decides to put his strongest batter in either position 1 or 4 and his pitcher in positions 6 or 9. If the other players may be put into any remaining positions, how many possible batting orders are there?

Answer 1

 

strongest in 1, pitcher in 6

strongest in 4, pitcher in 6

strongest in 1, pitcher in 9

strongest in 4, pitcher in 9

you have 7 players left to be put in 7 possible positions

player 3 goes in 1 of 7 positions

player 4 goes in 1 of 6 positions

player 5 goes in 1 of 5 positions

player 6 goes in 1 of 4 positions

player 7 goes in 1 of 3 positions

player 8 goes in 1 of 2 positions

player 9 goes in 1 of 1 position left

7*6*5*4*3*2*1=5040 batting orders for each of the 4 scenarios above

4*5040=20,160 possible batting orders

the simplest thing to do is the following...

(strongest in 1 of 2)*(pitcher in 1 of 2)*7*6*5*4*3*2*1

2*2*7*6*5*4*3*2*1=20,160 batting orders

if it did not matter who went where the answer would be...

9*8*7*6*5*4*3*2*1=362,880 batting orders but you put conditions on the pitcher and the strongest batter