Mixing
Number of Ways to seat Delegates around a round table $A, B,C, D$ such that $A$ and $B$ sit together and $C$ and $D$ do not sit together. [on hold]
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Delegates from $9$ countries including $A, B, C$ and $D$ are to be seated on a roundtable such that $A$ and $B$ sit together and $C$ and $D$ do not sit together. How many combinations are possible in accommodating all 9 delegates on a round table?
combinatorics permutations
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
asked Aug 3 at 16:25
Namay verma
111
put on hold as off-topic by Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig Aug 3 at 20:15
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig
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show 2 more comments
up vote
-2
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Delegates from $9$ countries including $A, B, C$ and $D$ are to be seated on a roundtable such that $A$ and $B$ sit together and $C$ and $D$ do not sit together. How many combinations are possible in accommodating all 9 delegates on a round table?
combinatorics permutations
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
asked Aug 3 at 16:25
Namay verma
111
put on hold as off-topic by Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig Aug 3 at 20:15
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig
2
Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
1
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Delegates from $9$ countries including $A, B, C$ and $D$ are to be seated on a roundtable such that $A$ and $B$ sit together and $C$ and $D$ do not sit together. How many combinations are possible in accommodating all 9 delegates on a round table?
combinatorics permutations
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
asked Aug 3 at 16:25
Namay verma
111
Delegates from $9$ countries including $A, B, C$ and $D$ are to be seated on a roundtable such that $A$ and $B$ sit together and $C$ and $D$ do not sit together. How many combinations are possible in accommodating all 9 delegates on a round table?
combinatorics permutations
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
asked Aug 3 at 16:25
Namay verma
111
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
edited Aug 3 at 20:14
N. F. Taussig
37.9k93053
37.9k93053
asked Aug 3 at 16:25
Namay verma
111
asked Aug 3 at 16:25
Namay verma
111
asked Aug 3 at 16:25
Namay verma
111
111
put on hold as off-topic by Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig Aug 3 at 20:15
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig
put on hold as off-topic by Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig Aug 3 at 20:15
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Adrian Keister, amWhy, Leucippus, N. F. Taussig
2
Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
1
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
2
Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
1
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago
2
2
Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
1
1
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago
 |Â
show 2 more comments
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Welcome to Math.SE. It's assumed that you would like an answer if you're posting a question. Please (a) make the title more descriptive of the problem you're having, and (b) show what you've tried to solve the problem.
– John
Aug 3 at 16:28
The question is unclear. Are you asking only about the number of ways of accommodating A,B,C,D (ignoring distinctions in the seating of the other five delegates), or are you asking about all combinations including different ways of seating E,F,G,H,I as well?
– Keith Backman
Aug 3 at 17:12
1
@KeithBackman we have to find the no of ways all 9 can be accommodated (keeping in mind the conditions)
– Namay verma
Aug 3 at 17:24
Place the delegates in a line that starts and ends with $A$ and $B$. Then we can close the line to a circle and $A$ and $B$ will be together. There are $2$ options for $A,B$. Two options for $C$ and $D$ for which one appears first in the line. Now, separate in case for how many people are going to be between $C$ and $D$. There can be $1,2,3,4$, or $5$ people between them. If $5$ people sit between $C$ and $D$ there are $5!$ ways to sit them in a line. That is $2cdot2cdot5!$.
– spiralstotheleft
2 days ago
If only $4$ people sit between $C$ and $D$, then the last person can sit either in the second spot or in the second to last. That is $2cdot2cdot2cdot4!$.
– spiralstotheleft
2 days ago