A group of 8 friends sit together in a circle. Alice, Betty : GMAT Problem Solving (PS)
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# A group of 8 friends sit together in a circle. Alice, Betty

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A group of 8 friends sit together in a circle. Alice, Betty [#permalink]

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26 Dec 2010, 08:47
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A group of 8 friends sit together in a circle. Alice, Betty and Candy are three members of the group of friends.

(i) If Alice refuses to sit beside Betty unless Candy sits on the other side of Alice as well, how many possible seating arrangements can there be?

(ii) 2 latecomers then come to join the group, and they have to sit apart from each other. How many possible seating arrangements can there be, bearing in mind
the condition from (i)?

[Reveal] Spoiler:
Answer (i) - 3840

Answer (ii) - 212160

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Re: Another tricky Circular Permutation Problem [#permalink]

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26 Dec 2010, 10:03
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subhashghosh wrote:
Hi

I am a bit perplexed by this another tricky Circular Permutation problem :

A group of 8 friends sit together in a circle. Alice, Betty and Candy are three members of the group of friends.

(i) If Alice refuses to sit beside Betty unless Candy sits on the other side of Alice as well, how many possible seating arrangements can there be?

(ii) 2 latecomers then come to join the group, and they have to sit apart from each other. How many possible seating arrangements can there be, bearing in mind the condition from (i)?

Answer (i) - 3840

Answer (ii) - 212160

The phrase "sit apart from each other" is confusing me a lot

Regards,
Subhash

Note that this is not the GMAT type question and it's beyond the GMAT scope.

A group of 8 friends sit together in a circle. Alice, Betty and Candy are three members of the group of friends.

(i) If Alice refuses to sit beside Betty unless Candy sits on the other side of Alice as well, how many possible seating arrangements can there be?

Total ways to arrange 8 people around the table is (8-1)!=7!;
Arrangement when Betty and Candy do sit together is (7-1)!*2=6!*2 ({1}, {2}, {3}, {4}, {5}, {6}, {AB} these 7 units can be arranged around the table in (7-1)!=6! and A and B can be arranged within the unit in 2 ways). Now, these arrangements will also include arrangements when Candy sits on the other side of Alice as well, # of these arrangement is (6-1)!*2=5!*2 ({1}, {2}, {3}, {4}, {5}, {CAB} these 6 units can be arranged around the table in (6-1)!=5! and {CAB} can be arranged within the unit in 2 ways: {CAB} and {BAC});

So we'll have 7!-(6!*2-5!*2)=3,840.

(ii) 2 latecomers then come to join the group, and they have to sit apart from each other. How many possible seating arrangements can there be, bearing in mind the condition from (i)?

When A and B does not sit together: (7!-6!*2)*8*7;

When A sits between C and B:
one latecomer sits between A and B: (5!*2)*2*7;
latecomers don't sit between these 3: (5!*2)*6*5;

Total: (7!-6!*2)*8*7+(5!*2)*2*7+(5!*2)*6*5=211,160
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Re: A group of 8 friends sit together in a circle. Alice, Betty [#permalink]

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21 Sep 2013, 12:34
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A group of 8 friends sit together in a circle. Alice, Betty [#permalink]

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01 May 2016, 00:02
Hi Bunuel - Can you please explain (ii) in more detail:

When A and B does not sit together: (7!-6!*2)*8*7; (I didn't understand why you multiply by 8 and 7 here?)

When A sits between C and B:
one latecomer sits between A and B: (5!*2)*2*7;
latecomers don't sit between these 3: (5!*2)*6*5;

Total: (7!-6!*2)*8*7+(5!*2)*2*7+(5!*2)*6*5=211,160

Thanks for your help!
A group of 8 friends sit together in a circle. Alice, Betty   [#permalink] 01 May 2016, 00:02
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# A group of 8 friends sit together in a circle. Alice, Betty

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