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

Punch

New member
Jan 29, 2012
23
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
 

PaulRS

Member
Jan 26, 2012
37
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.
 

grgrsanjay

New member
Jan 28, 2012
21
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
 

soroban

Well-known member
Feb 2, 2012
409
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

New member
Jan 28, 2012
21
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??
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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.