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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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New post 26 Dec 2010, 09:47
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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)?

Answer (i) - 3840

Answer (ii) - 212160

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Re: Another tricky Circular Permutation Problem  [#permalink]

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New post 26 Dec 2010, 11:03
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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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A group of 8 friends sit together in a circle. Alice, Betty  [#permalink]

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New post 01 May 2016, 01:02
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!
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Re: A group of 8 friends sit together in a circle. Alice, Betty  [#permalink]

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New post 23 Jun 2017, 04:50
Bunuel wrote:
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


Bunuel

I could solve the problem, I have a question about the interpretation of language.

The condition is that Alice refuses to sit beside Betty unless Candy sits on the other side of Alice as well.

So this means that Alice will sit with Betty only if Candy is also there, but Alice can sit with Candy even if Betty is not there. Now this translation took me a while, can you please suggest somewhere to read about how to translate this logic language into something understandable? it will be great help. And if GMAT were to write this question..will it use the same language to express this situation?
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Re: A group of 8 friends sit together in a circle. Alice, Betty  [#permalink]

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New post 23 Jun 2017, 06:22
ShashankDave wrote:
Bunuel wrote:
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


Bunuel

I could solve the problem, I have a question about the interpretation of language.

The condition is that Alice refuses to sit beside Betty unless Candy sits on the other side of Alice as well.

So this means that Alice will sit with Betty only if Candy is also there, but Alice can sit with Candy even if Betty is not there. Now this translation took me a while, can you please suggest somewhere to read about how to translate this logic language into something understandable? it will be great help. And if GMAT were to write this question..will it use the same language to express this situation?


This is not a GMAT question, hence the tag Out of scope/Too hard. You can ignore it altogether and move on.
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Re: A group of 8 friends sit together in a circle. Alice, Betty   [#permalink] 23 Jun 2017, 06:22
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