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5 Noble Knights are to be seated at a round table. In how

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5 Noble Knights are to be seated at a round table. In how [#permalink] New post 12 Jul 2007, 15:27
5 Noble Knights are to be seated at a round table. In how many ways can they be seated?

A)120
B)96
C)60
D)35
E)24
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Re: Challenge 1 Question - Noble Knights [#permalink] New post 12 Jul 2007, 17:03
leeye84 wrote:
5 Noble Knights are to be seated at a round table. In how many ways can they be seated?

A)120
B)96
C)60
D)35
E)24


(5-1)!= 4! = 24
Circular permutation formula is (n-1)!
This is solved by taking one knight in a fixed position and you have 4! ways to arrange the remaining 4.
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 [#permalink] New post 12 Jul 2007, 17:11
good variation of this problem:
how many ways can you seat 5 girls and 5 boys around a circular table if the girls and boys should alternate?
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 [#permalink] New post 12 Jul 2007, 17:32
Is n't it 4! * 2! = 48 ways
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 [#permalink] New post 12 Jul 2007, 18:27
WOW! This is no joke!

I thought it was just a tricky problem. I mean, ROUND TABLE, huh?

But now I see the other side of the coin.

Thank You Oops.

And Oops, can you explain the reasoning behind this circular permutation formula?
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 [#permalink] New post 12 Jul 2007, 18:42
I thought you had to do 5*4*3*2*1=120 or A
What is the OA, and what is the rule?
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 [#permalink] New post 12 Jul 2007, 23:58
oops wrote:
good variation of this problem:
how many ways can you seat 5 girls and 5 boys around a circular table if the girls and boys should alternate?


Oops how do you solve this one? Thanks.
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 [#permalink] New post 13 Jul 2007, 07:11
oops wrote:
good variation of this problem:
how many ways can you seat 5 girls and 5 boys around a circular table if the girls and boys should alternate?


I'm getting 4!*5! = 2880

What's the A ?
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 [#permalink] New post 13 Jul 2007, 10:50
oops wrote:
good variation of this problem:
how many ways can you seat 5 girls and 5 boys around a circular table if the girls and boys should alternate?


Is the answer 576?

The ways to seat 5 girls around the table = 24
The ways to seat 5 boys around the table = 24

24 x 24 = 576

Is that right?
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 [#permalink] New post 13 Jul 2007, 11:12
leeye84 wrote:
oops wrote:
good variation of this problem:
how many ways can you seat 5 girls and 5 boys around a circular table if the girls and boys should alternate?


Is the answer 576?

The ways to seat 5 girls around the table = 24
The ways to seat 5 boys around the table = 24

24 x 24 = 576

Is that right?



The way I see it is as follows:

Say we pick one of the girls (everything else will revolve around her)
Number of ways to arange 5 girls is 24. Boys go in between so they can be seated in 5! ways. I'm sticking with my answer, but would like to see what other people think.
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 [#permalink] New post 13 Jul 2007, 11:42
I think 24*24 is right. The rules of circular permutation still applies. We have basically two circles: one circle of boys and one circle of girls. Thoughts?
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 [#permalink] New post 13 Jul 2007, 15:33
Here it is:

G B G B G B G B G B

Now imagine this string as a circle. We take the first G as a starting point. The other Gs and all of the Bs are a regular factorial.

Just like 5 girls in a circle only. We take the first one as a starting point and the rest are 4!

I'd like to see some more opinions.
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 [#permalink] New post 13 Jul 2007, 15:51
leeye84, hope the other explanations have made it clear why (n-1)! is used for the roundtable pizza party :)

hayabusa, good thinking!
your answer is correct for the second problem - only need to fix one of them from one of the groups - the second group can be arranged in the regular way - hence n! * (n-1)! is the OA
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 [#permalink] New post 13 Jul 2007, 16:26
oops wrote:
leeye84, hope the other explanations have made it clear why (n-1)! is used for the roundtable pizza party :)

hayabusa, good thinking!
your answer is correct for the second problem - only need to fix one of them from one of the groups - the second group can be arranged in the regular way - hence n! * (n-1)! is the OA



That's exactly what I said! The second group (say the boys is 5!) The first is (5-1)! - cause we rotate around one of the chicks. Ans is 4!*5!

Enough BS for one day, its Friday!

Here is something to make your day:

http://video.google.com/videoplay?docid ... &plindex=6
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 [#permalink] New post 15 Oct 2007, 12:57
leeye84 wrote:
And Oops, can you explain the reasoning behind this circular permutation formula?


The number is instead of the usual factorial since all cyclic permutations of objects are equivalent because the circle can be rotated.

http://mathworld.wolfram.com/CircularPermutation.html
  [#permalink] 15 Oct 2007, 12:57
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