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Today is Angelina birthday. She invited four friends to her

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Today is Angelina birthday. She invited four friends to her [#permalink] New post 01 Feb 2006, 13:56
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C
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Question Stats:

33% (03:02) correct 67% (00:31) wrong based on 4 sessions
Today is Angelina birthday. She invited four friends to her party: Barbara, Carol, Don, and Frank. Everyone will be seated on a round table. How many possible way can these people be seated if Don and Carol must always sit next to each other?

A) 120
B) 60
B) 48
C) 24
D) 12
E) 6
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Re: Prob: 5 people in birthday party [#permalink] New post 01 Feb 2006, 14:36
48

combine C and D into one.

4 people 4 seat = 4!

C and D can be sit in 2 ways.

so 4! * 2 = 48
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 [#permalink] New post 01 Feb 2006, 14:43
I also got 48.

4! , then the two switch seats, another 4!

4!+4!=48
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 [#permalink] New post 01 Feb 2006, 15:08
In circular permutations of n people we have (n-1)! permutations.

If I am not wrong then this should be (4-1)! * 2 = 12
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 [#permalink] New post 01 Feb 2006, 15:18
Same answer.....I considered two people as 1.

Then, 4 people can be arranged in 4*3*2*1 = 24 ways. But since the table is circular we need to divide 24 by the number of positions, 24/4 = 6.

4 people can be arranged on a round table in 6 ways...but wait...we considered 2 people as one set..they can be arranged among themselves in 2!.

Total number of ways = 6*2! = 12.



ps_dahiya wrote:
In circular permutations of n people we have (n-1)! permutations.

If I am not wrong then this should be (4-1)! * 2 = 12
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 [#permalink] New post 02 Feb 2006, 04:17
Barbara, Carol, Don and Frank --> 4 people

Since Don and Carol must always be next to each other, combine them both as 1 entity --> call it DonC

So # of arrangements involving Barbara, DonC and Frank = 3! = 6

But We can switch Don and Carol's position around.

So total # of arrangements = 12
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 [#permalink] New post 02 Feb 2006, 06:39
ps_dahiya wrote:
In circular permutations of n people we have (n-1)! permutations.

If I am not wrong then this should be (4-1)! * 2 = 12


dahiya,.....y u do *2 :?:

thanx
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 [#permalink] New post 02 Feb 2006, 07:44
i got 12 too (after some serious thought).
My initial thought was 4! x 2, but because the tabe is circular, you have to realize that ABCDF is the same as BCDFA, so only the combinations of the three other guests matter: 3!, then multipy by 2, for the cases in which CD are switched.

Is there a more mathematical explanation for this, or an analagous real-world situation (the way the electing 3 officers out of 9 people is analagous to creating 3-digit number out of 9 digits)?
ie- what are other expamples of circular permutations that aren't so blatanly worded as "x people sitting in a circle"?
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 [#permalink] New post 02 Feb 2006, 09:21
12.

AD Fixed , 3 variable, 3*2*1 =6
DA fixed 3 variable 3*2*1 =6

6+6 =12
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 [#permalink] New post 02 Feb 2006, 13:36
there are 5 people including b'day gal

since 2 people will be clubbed as one

5-1 = sitting arrangements for 4 * 2

sitting arrangements for 4 = (4-1)! (circular)

so (4-1)!*2 = 12
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 [#permalink] New post 02 Feb 2006, 16:37
jinesh wrote:
ps_dahiya wrote:
In circular permutations of n people we have (n-1)! permutations.

If I am not wrong then this should be (4-1)! * 2 = 12


dahiya,.....y u do *2 :?:

thanx


we take AB as one person and C,D and F as seperate. So there are (4-1)! arrangements. But now we take BA as one combination. Then there are (4-1)!

So total = (4-1)! + (4-1)! = (4-1)! * 2 = 12

Hope this helps.
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  [#permalink] 02 Feb 2006, 16:37
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Today is Angelina birthday. She invited four friends to her

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