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Senior Manager
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Today is Angelina birthday. She invited four friends to her [#permalink]
 01 Feb 2006, 14:56
Question Stats:
33% (03:02) correct
66% (00:31) wrong based on 0 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]
01 Feb 2006, 15:36
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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I also got 48.
4! , then the two switch seats, another 4!
4!+4!=48
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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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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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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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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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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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12.
AD Fixed , 3 variable, 3*2*1 =6
DA fixed 3 variable 3*2*1 =6
6+6 =12
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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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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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