Combinations or Perumations? I have 27 arrows and a bow and 10 targets to shoot at.

joshgarzaa

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I have 27 arrows and a bow and 10 targets to shoot at:
(a) How many ways are there to distribute the arrows to the targets?
(b) How many ways are there to distribute the arrows to the targets ensuring at least one arrow per target (obviously a bullseye)?
(c) How many ways are there to shoot the targets such that every target has at least 2 arrows, and one target has only 1?

For A, would it be 271 ways via pigeonhole principle?
I don't really know where to start for B & C
 
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I have 27 arrows and a bow and 10 targets to shoot at:
(a) How many ways are there to distribute the arrows to the targets?
(b) How many ways are there to distribute the arrows to the targets ensuring at least one arrow per target (obviously a bullseye)?
(c) How many ways are there to shoot the targets such that every target has at least 2 arrows, and one target has only 1?
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

For part A, I'd assume we'd have to use the pigeonhole principle where the number of pigeons(K) is 27 and the number of pigeonholes is (n)10.
From the pigeonhole formula, we would get Kn+1. So, 27*10 + 1 is 271 ways?
 
I have 27 arrows and a bow and 10 targets to shoot at:
(a) How many ways are there to distribute the arrows to the targets?
(b) How many ways are there to distribute the arrows to the targets ensuring at least one arrow per target (obviously a bullseye)?
(c) How many ways are there to shoot the targets such that every target has at least 2 arrows, and one target has only 1?

For A, would it be 271 ways via pigeonhole principle?
I don't really know where to start for B & C
Hi, can you please state the pigeonhole principle in your own words?
 
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