Combinatorics circular sitting problem [on hold]
There are $3$ americans, $2$ britishers, $1$ portugese, $1$ chinese and they are allowed to sit around a circular table so that no two people of same nationality sit side by side. Answer is $3148$. I ve tried it many a times but...answer given in book is 3148
combinatorics permutations
asked 2 days ago
Harshit Raj
92
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put on hold as off-topic by amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo yesterday
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There are $3$ americans, $2$ britishers, $1$ portugese, $1$ chinese and they are allowed to sit around a circular table so that no two people of same nationality sit side by side. Answer is $3148$. I ve tried it many a times but...answer given in book is 3148
combinatorics permutations
asked 2 days ago
Harshit Raj
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago
add a comment |
There are $3$ americans, $2$ britishers, $1$ portugese, $1$ chinese and they are allowed to sit around a circular table so that no two people of same nationality sit side by side. Answer is $3148$. I ve tried it many a times but...answer given in book is 3148
combinatorics permutations
asked 2 days ago
Harshit Raj
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There are $3$ americans, $2$ britishers, $1$ portugese, $1$ chinese and they are allowed to sit around a circular table so that no two people of same nationality sit side by side. Answer is $3148$. I ve tried it many a times but...answer given in book is 3148
combinatorics permutations
combinatorics permutations
asked 2 days ago
Harshit Raj
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Harshit Raj
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked 2 days ago
Harshit Raj
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Harshit Raj
92
asked 2 days ago
Harshit Raj
92
92
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Harshit Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, KReiser, Lord Shark the Unknown, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago
add a comment |
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
The Americans must be placed: A _ _ A _ A _
There are $3!$ ways to do that.
There are four remaining people to put in the four seats. This can be done in $4!$ ways, two of which have the British in the two adjacent slots; these cases must be eliminated.
Hence $4! - 2 = 22$ cases where we do not care where the first person is seated (i.e., we can rotate the table). If we do care about orientation, we multiply by 6.
Answer: $3! (4!-2) 6/2 = 396$.
answered 2 days ago
David G. Stork
9,67621232
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The Americans must be placed: A _ _ A _ A _
There are $3!$ ways to do that.
There are four remaining people to put in the four seats. This can be done in $4!$ ways, two of which have the British in the two adjacent slots; these cases must be eliminated.
Hence $4! - 2 = 22$ cases where we do not care where the first person is seated (i.e., we can rotate the table). If we do care about orientation, we multiply by 6.
Answer: $3! (4!-2) 6/2 = 396$.
answered 2 days ago
David G. Stork
9,67621232
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
add a comment |
The Americans must be placed: A _ _ A _ A _
There are $3!$ ways to do that.
There are four remaining people to put in the four seats. This can be done in $4!$ ways, two of which have the British in the two adjacent slots; these cases must be eliminated.
Hence $4! - 2 = 22$ cases where we do not care where the first person is seated (i.e., we can rotate the table). If we do care about orientation, we multiply by 6.
Answer: $3! (4!-2) 6/2 = 396$.
answered 2 days ago
David G. Stork
9,67621232
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
add a comment |
The Americans must be placed: A _ _ A _ A _
There are $3!$ ways to do that.
There are four remaining people to put in the four seats. This can be done in $4!$ ways, two of which have the British in the two adjacent slots; these cases must be eliminated.
Hence $4! - 2 = 22$ cases where we do not care where the first person is seated (i.e., we can rotate the table). If we do care about orientation, we multiply by 6.
Answer: $3! (4!-2) 6/2 = 396$.
answered 2 days ago
David G. Stork
9,67621232
The Americans must be placed: A _ _ A _ A _
There are $3!$ ways to do that.
There are four remaining people to put in the four seats. This can be done in $4!$ ways, two of which have the British in the two adjacent slots; these cases must be eliminated.
Hence $4! - 2 = 22$ cases where we do not care where the first person is seated (i.e., we can rotate the table). If we do care about orientation, we multiply by 6.
Answer: $3! (4!-2) 6/2 = 396$.
answered 2 days ago
David G. Stork
9,67621232
edited 2 days ago
answered 2 days ago
David G. Stork
9,67621232
answered 2 days ago
David G. Stork
9,67621232
answered 2 days ago
David G. Stork
9,67621232
9,67621232
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
add a comment |
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
Ooopsss... yes... another factor of 2. Thanks.
– David G. Stork
2 days ago
1
1
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
Sorry, that was not what I meant. What I was trying to say is that once you seat the Americans, there are $4$ bad cases to be eliminated from the $4!$ ways you can seat the other four people since for each of the two ways the British could be seated together, there are two ways to seat the Portugese and Chinese people in the remaining two positions.
– N. F. Taussig
2 days ago
add a comment |
Welcome to MathSE. When you pose a question here, you should show what you have attempted so that users can detect any errors you may have made. Explaining what you know and where you are stuck is also helpful. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
2 days ago