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Counting methods # 17

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Konstantin Lynov
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Counting methods # 17 [#permalink] New post   11 Aug 2003, 07:43
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In how many ways could be eight shooting marks distributed between four archers in equal pairs?



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Re: Counting methods # 17 [#permalink] New post   11 Aug 2003, 07:59
Well, a pair of marks can be made in 8C2 ways. To distribute them among the four shooters, you would multiply this by 4!.

So 8C2 * 4! = 28 * 24 = 672 ways.
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Re: Counting methods # 17 [#permalink] New post   12 Aug 2003, 07:44
I would go with 8C2 * 4 = 28*4 = 112.

Could the author post the answer?
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Re: Counting methods # 17 [#permalink] New post   12 Aug 2003, 08:20
My reasoning here. What of the question had said "In how many ways can four marks be distributed among four shooters?"

In this case the answer would have been 4 * 3 * 2 * 1. Do you agree?

Now we want to know how 8C2 (or 28) can be distributed.

Let's number the shots 12345678.

What if the groupings were 12 34 56 78
Then we would have 4! different distributions

What if they were 13 24 56 78
Then 4! more

How about 18 24 36 45
4! more

So in the end there are 8C2 (or 28) groupings * 4!

So I think 28 * 4!



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Re: Counting methods # 17 [#permalink]
12 Aug 2003, 08:20
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