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Circular Arrangement Question

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by viv_gmat » Mon Aug 08, 2011 11:57 am
Q. There are 6 people at a party sitting at a round table with 6 seats: A,B,C,D,E AND F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?

A. 720
B. 120
C. 108
D. 84
E. 48

C
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by Frankenstein » Mon Aug 08, 2011 12:33 pm
Hi,
If the question is: A cannot sit next to either D or F then the answer should be 36.
If the question is: A cannot sit next to both D and F(A cannot sit between D and F), then the answer should be 108. Please check the wording of this question because for this question answer should be 36, not 108.
Can you post the source.
Cheers!

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by viv_gmat » Mon Aug 08, 2011 12:39 pm
Frankenstein wrote:Hi,
If the question is: A cannot sit next to either D or F then the answer should be 36.
If the question is: A cannot sit next to both D and F(A cannot sit between D and F), then the answer would be 108. Please check the wording of this question because for this question answer should be 36, not 108.
Can you post the source.
Hey thanks for your reply...I too feel the same, but the wordings are as I posted.
Anyway, If the question is: A cannot sit next to either D or F then the answer should be 36.
Could you post your explanation.
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by Frankenstein » Mon Aug 08, 2011 12:45 pm
Hi,
Okay. Let's fix a position for A. It has two adjacent positions one on either side. D and F cannot take these two positions. So, D and F can be placed in the remaining 3 positions in 3P2 = 6ways.
The remaining three positions can be filled by B,C,E in 3! ways.
So, total number of ways = 6*6 = 36 ways.
Source plz?
Cheers!

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by viv_gmat » Mon Aug 08, 2011 12:47 pm
Frankenstein wrote:Hi,
Okay. Let's fix a position for A. It has two adjacent positions one on either side. D and F cannot take these two positions. So, D and F can be placed in the remaining 3 positions in 3P2 = 6ways.
The remaining three positions can be filled by B,C,E in 3! ways.
So, total number of ways = 6*6 = 36 ways.
Source plz?
Thanks!!
Manhattan
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by Frankenstein » Mon Aug 08, 2011 12:51 pm
Hi,
If it is from MGMAT, please send PM to MGMAT expert. He might give a better explanation, and he will also let us know whether our understanding of this question is correct.
Cheers!

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by viv_gmat » Mon Aug 08, 2011 8:55 pm
Frankenstein wrote:Hi,
Okay. Let's fix a position for A. It has two adjacent positions one on either side. D and F cannot take these two positions. So, D and F can be placed in the remaining 3 positions in 3P2 = 6ways.
The remaining three positions can be filled by B,C,E in 3! ways.
So, total number of ways = 6*6 = 36 ways.
Source plz?
Frankenstein, how is this explanation-
Since, it is a circular arrangement the solution for arranging adcdef = (6-1)! = 120. Then ad-(another 4people) = (5-1)! = 24 similarly AF(another 4 people) = (5-1)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF -(another 3peopl) = (4-1)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84.

I solved this problem like this initially...Your thoughts plz.
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by Frankenstein » Mon Aug 08, 2011 9:03 pm
viv_gmat wrote: Frankenstein, how is this explanation-
Since, it is a circular arrangement the solution for arranging adcdef = (6-1)! = 120. Then ad-(another 4people) = (5-1)! = 24 similarly AF(another 4 people) = (5-1)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF -(another 3peopl) = (4-1)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84.

I solved this problem like this initially...Your thoughts plz.
Hi,
You have subtracted the cases AD and AF but you have neglected the cases DA and FA. Each of which will be having 24 cases.
So, you need to subtract 2*24 from 84 which will give 84-48 = 36.
Cheers!

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by viv_gmat » Mon Aug 08, 2011 9:06 pm
Frankenstein wrote:
viv_gmat wrote: Frankenstein, how is this explanation-
Since, it is a circular arrangement the solution for arranging adcdef = (6-1)! = 120. Then ad-(another 4people) = (5-1)! = 24 similarly AF(another 4 people) = (5-1)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF -(another 3peopl) = (4-1)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84.

I solved this problem like this initially...Your thoughts plz.
Hi,
You have subtracted the cases AD and AF but you have neglected the cases DA and FA. Each of which will be having 24 cases.
So, you need to subtract 2*24 from 84 which will give 84-48 = 36.
O yess.....I missed those cases....it makes a perfect sense now..thanks again !!
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