No. of ways to seat round a table (numbered seats)

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Discussion Overview

The discussion revolves around the problem of determining the number of ways to arrange two families at a round table with numbered seats, ensuring that members of the same family sit together. The conversation includes various approaches to the problem, calculations, and considerations regarding the arrangement of seats.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the arrangements as 6!(3!)(3!) = 25920 but notes that the correct answer is 43200.
  • Another participant emphasizes the importance of considering the numbering of seats and how it affects the arrangement, particularly noting that seats 10, 1, and 2 are consecutive.
  • A different participant proposes that when ignoring the seat numbers, the formula for arrangements is (n-1)!, leading to 5!(3!)(3!) = 4320, and then adjusts for numbered seats to arrive at 43200.
  • One participant suggests a detailed breakdown of the arrangement process, including the seating choices for each family and the remaining individuals, ultimately confirming the answer as 43200.
  • Another participant reiterates the calculation of 43200 but questions whether their logic applies to similar problems.
  • A later reply affirms the correctness of the logic but suggests that once seats are numbered, the circular nature of the table is no longer relevant, implying a simplification of the problem.

Areas of Agreement / Disagreement

Participants generally agree on the final answer of 43200 arrangements, but there is disagreement regarding the necessity of considering the circular nature of the table once the seats are numbered. Some participants question the complexity of their approaches.

Contextual Notes

There are unresolved considerations regarding the implications of numbering the seats on the circular arrangement and whether the logic applies universally to similar problems.

Punch
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Two families are at a party. The first family consists of a man and both his parents while the second familly consists of a woman and both her parents. The two families sit at a round table with two other men and two other women. Find the number of possible arrangements if the members of the same family are seated together and the seats are numbered.

What I did was to consider the 2 families, the 2 woman and 2man as 6 groups of people.
6!(3!)(3!)=25920
but correct answer is 43200
 
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If we number the seats 1,2,3,...,10 . Note that, for example, seats 10, 1 and 2 are consecutive seats because we are working with a round table!

So you have to consider the cases in which seats 10 and 1 correspond to the same family too.
 
First,i am ignoring the numbers on the seat,

this is a round combination

So, formula is (n-1)!
no.of.ways is 5!(3!)(3!)= 4320

Now the seat are numbered,
then i can more these combinations 1 seats,2seata,...9 seats apart from the original one

so,number of ways is 43,200
 
Hello, Punch!

Two families are at a party.
The first family consists of a man and both his parents
. . while the second familly consists of a woman and both her parents.
The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if the members of the same family
. . are seated together and the seats are numbered.

Answer: 43,200
Duct-tape the families together.

We have: .$\text{(M, P, P)}$ . . . and they have $3!$ possible orders.
We have: .$\text{(W, P, P)}$ . . . and they have $3!$ possible orders.

We also have: .$m,\:m,\:w,\:w$$\text{M}$ has a choice of $10$ seats.
When he is seated, he and his family occupy three seats.
Among the remaining seven seats, $\text{(W, P, P)}$ has $5$ choices for seating.
. . (Think about it.)
Then the remaining four people can be seated in $4!$ ways.Therefore: .$(3!)(3!)(10)(5)(4!) \:=\:43,200$ arrangements.
 
grgrsanjay said:
First,i am ignoring the numbers on the seat,

this is a round combination

So, formula is (n-1)!
no.of.ways is 5!(3!)(3!)= 4320

Now the seat are numbered,
then i can more these combinations 1 seats,2seata,...9 seats apart from the original one

so,number of ways is 43,200

I Wanted to know whether my logic holds good for every similar problem??
 
grgrsanjay said:
I Wanted to know whether my logic holds good for every similar problem??
Your logic is correct. But why complicate matters?
Once the seats are numbered, we no longer have a circular table.
So there is no need for that.
 

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