SOLUTION: A baseball manager chooses his lineup 1-9. How many ways can he organize his lineup if the pitcher doe not hit in the first four? I did 9*8*7*6*6*5*4*3*2 is this correct?

Algebra ->  Permutations -> SOLUTION: A baseball manager chooses his lineup 1-9. How many ways can he organize his lineup if the pitcher doe not hit in the first four? I did 9*8*7*6*6*5*4*3*2 is this correct?      Log On


   

Question 397991: A baseball manager chooses his lineup 1-9. How many ways can he organize his lineup if the pitcher doe not hit in the first four?
I did 9*8*7*6*6*5*4*3*2
is this correct?
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the top of the lineup is considered first, there are only
8 choices for 1st batter, then 7 choices, 6, and 5
So the number of choices for the top 4 positions are
8%2A7%2A6%2A5 then the pitcher becomes available plus the
4 others left. 4+%2B+1+=+5, next here are 4 choices, then 3,
2, and 1
8%2A7%2A6%2A5%2A5%2A4%2A3%2A2%2A1+=+201600+ is the sum of all the choices
Hope I got it right

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A baseball manager chooses his lineup 1-9. How many ways can he organize his lineup if the pitcher does not hit in the first four?
I did 9*8*7*6*6*5*4*3*2
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Since the pitcher is not in the 1st four,
8 ways to choose 1st
7 ways to choose 2nd
6 ways to choose 3rd
5 ways to choose 4th
5 ways to choose 3rd
4 ways
3 ways
2 ways
1 ways
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Cheers,
Stan H.