Question 478404: Dear math teacher,
I am having difficulties with the following problem.
How many baseball nines can be chosen from 13 candidates if A, B, C, D are the only candidates for two positions and can play no other position?
I think I need to select 9 canditates from 13 but I also have 4 candidates (A, B, C, D) for 2 positions. If A, B, C, D can play no othe position but can fill only 2 positions,
13C9 times 4P2 = 8580 selections but my answer is supposed to be only 216. Where did I make a mistake?
Thank you for your time.
Ivanka
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! knowing what the answer should be makes it a lot easier.
you have a nice target to shoot for.
13C9 would be the answer if you had no restrictions.
given that you have restrictions, you have to modify the equations to fit those restrictions.
4 people can only play 2 positions.
the number of ways this can happen are 4C2 = 6
that leave 9 people for the remaining 7 positions.
the number of ways that can happen is 9C7 = 36
the total number of ways both can happen is 36*6 = 216
bring the numbers down low and you can see what is happening.
suppose 5 people for 2 positions.
2 people can only play 1 of the positions.
that means your equation becomes:
2C1 * 3C1 = 2*3 = 6
2C1 is the 2 people for the 1 position.
3C1 is the remaining 3 people for the other position.
now assume the people were abcde
a is person 1
b is person 2, etc.
suppose a and b can only play the first position.
your possible choices for the first position are:
a
b
your possible choices for the other position are:
c
d
e
you have 2 possible choices for position 1 and 3 possible choices for position 2.
the total number of possible ways you can outfit this team is 2*3 = 6
those ways are shown below:
position 1 position 2
a c
a d
a e
b c
b d
b e
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