Permutations and combinations

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  • Thread starter Kaushik
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Ok. I would start by 1) counting it as though exact seating mattered. 2) Then divide by cases where multiple seating arrangements will be considered equivalent.

1) Counting as though exact seating matters
There are 10 seats for 10 people. The people can be put in them in 10! ways.

2) Divide by cases where multiple seating arrangements will be considered equivalent.
The seating at the small table (4 seats) will be considered equivalent if they are just rotations of each other. ABCD = BCDA = CDAB = DABC. So divide by 4.
Similarly, the seating at the large table (6 seats) will be considered equivalent if they are just rotations of each other. So divide by 6.

Final answer: 10!/(4*6).

Note: This is the same as 10!/4! that some people got. I couldn't exactly follow the logic of the people who got 10!/4! in one step.
Note: If clockwise and counterclockwise are also considered equivalent, then divide by 8 and 12 rather than 4 and 6.
 
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Ok. I would start by 1) counting it as though exact seating mattered. 2) Then divide by cases where multiple seating arrangements will be considered equivalent.

1) Counting as though exact seating matters
There are 10 seats for 10 people. The people can be put in them in 10! ways.

2) Divide by cases where multiple seating arrangements will be considered equivalent.
The seating at the small table (4 seats) will be considered equivalent if they are just rotations of each other. ABCD = BCDA = CDAB = DABC. So divide by 4.
Similarly, the seating at the large table (6 seats) will be considered equivalent if they are just rotations of each other. So divide by 6.

Final answer: 10!/(4*6).

Note: This is the same as 10!/4! that some people got. I couldn't exactly follow the logic of the people who got 10!/4! in one step.
Note: If clockwise and counterclockwise are also considered equivalent, then divide by 8 and 12 rather than 4 and 6.
Thanks!
 
  • #28
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Question - Find the number of ways in which 6 gentlemen and 3 ladies can be seated around a table so that every gentleman may have a lady by his side.

I tried it using your approach.
Consider, GLGGLGGLG (G-> Gentleman L-> Lady)
1)The gentlemen can be seated in ##6!## ways.
2)The 3 ladies can sit in any of the 3 spots between the gentlemen in ##3!## ways.
3)The ladies and gentlemen can be switched.
4)They can be rotated in 9 ways which are all equivalent.
Hence, ##\frac{(6!)(3!)(2)}{9}##
Is this correct? If not, where did I commit my mistake?
 
  • #29
280
17
Question - Find the number of ways in which 6 gentlemen and 3 ladies can be seated around a table so that every gentleman may have a lady by his side.

I tried it using your approach.
Consider, GLGGLGGLG (G-> Gentleman L-> Lady)
1)The gentlemen can be seated in ##6!## ways.
2)The 3 ladies can sit in any of the 3 spots between the gentlemen in ##3!## ways.
3)The ladies and gentlemen can be switched.
4)They can be rotated in 9 ways which are all equivalent.
Hence, ##\frac{(6!)(3!)(2)}{9}##
Is this correct? If not, where did I commit my mistake?
When I wrote it down, I noticed that they are equivalent 6 times and not 9. So it must be ##\frac{(6!)(3!)(2)}{6}##. But am I supposed to write down everything and see or will I get used to it by time (problem solving).
 

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