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6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in

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6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in  [#permalink]

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New post 19 Jul 2018, 08:56
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Question Stats:

33% (02:47) correct 67% (02:19) wrong based on 115 sessions

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6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in a row of six chairs numbered from 1 to 6. Adam does not want to be seated in chair numbered 1 , Bob does not want to be seated in chair numbered 2, and Charlie does not want to be seated in chair numbered 5. How many ways can they be seated so that the all the considerations of Adam, Bob, and Charlie have been taken into consideration.
(a) 720
(b) 714
(c) 426
(d) 120
(e) None of the above

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Re: 6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in  [#permalink]

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New post 20 Jul 2018, 11:00
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We have to consider all the ways/arrangements in which Adam, Bob & Charlie do not have to compromise to seat on the chairs they don't like.
Total Number of arrangements without any restrictions = 6!= 720

We have to exclude the case when at-least one of Adam, Bob or Charlie has to seat on the chair he doesn't like.
This is infact the union of set containing the set A (when Adam has to sit on Chair 1), set B (when Bob has to sit on Chair 1),and set C (when charlie has to sit on Chair 3).
n(AUBUC)= n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

Now , n(A and B and C) = 3! = 6 ( as these 3 are fixed on their chair, it is simply the arrangement of remaining 3)
Similarly n(A and B) = 4!= 24, n(B and C) = 4!= 24, n(C and A) = 4!= 24
n(A) = 5! =120 ( only position of A fixed, remaining 5 can be arranged in 5!)
n(B) = 5! =120 ( only position of B fixed, remaining 5 can be arranged in 5!)
n(C) = 5! =120 ( only position of C fixed, remaining 5 can be arranged in 5!)

So, n(AUBUC) = 120+120+120-24-24-24+6 = 294
Hence the number of arrangements in which atleast one of A, B or C has to sit on chair that he doesnt like = 294

So required arrangements = total ways (without any restrictions) - number of arrangements in which atleast one of A, B or C has to sit on chair that he doesn't like.
= 6!-294 = 720-294 =426.
Answer = C
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Re: 6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in  [#permalink]

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New post 14 Sep 2018, 23:48
isn't the total number of combinations possible in circular arrangement=5!
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Re: 6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in  [#permalink]

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New post 14 Sep 2018, 23:50
also, in my approach, Im using slots method, filling up the three restrictions first which gives 5*4*3 and then filling up unrestricted seats which gives 3!. So I'm getting 360 as my answer. Can someone please point out the flaw in my approach which is getting me the wrong answer?
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Re: 6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in  [#permalink]

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New post 19 Sep 2018, 21:53
Total number of cases is 6! = 720

Because A cannot choose 1 spot, so we have less cases by 1/6 of the total number cases. Same applies to B and C.

So the total cases approximation is: 720 x (5/6)^3 = 416 closest to 426. Answer C.
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Re: 6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in   [#permalink] 19 Sep 2018, 21:53
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6 people, Adam, Bob, Charlie, Don, Eva, and Fiona have to be seated in

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