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February 1st, 2012, 01:56 AM   #1
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No. of ways to seat round a table (numbered seats)

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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February 1st, 2012, 02:06 AM   #2
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Re: No. of ways to seat round a table (numbered seats)

Permutate the first family, there are five ways the second family may be seated, then permutate them, then permutate the 4 remaining people, then permutate the ten ways each arrangement may be rotated or permutated:

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February 1st, 2012, 05:17 AM   #3
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Re: No. of ways to seat round a table (numbered seats)

Hello, Punch!

Quote:
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.


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3!)(3!)(10)(5)(4!) \:=\:43,200" />

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February 1st, 2012, 09:46 PM   #4
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Re: No. of ways to seat round a table (numbered seats)

There's been a lot of round tables in this forum recently. So where's King Arthur? :P
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February 1st, 2012, 10:34 PM   #5
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Re: No. of ways to seat round a table (numbered seats)

First, my guess is that you'll find King Arthur and Queen Guenevere having a romantic dinner at a square table, probably.
(You've got to suspend your disbelief a little bit in this.)

Let me try another way:
Group all the members in the first and second family together and that will generate two distinct groups.
To permutate each of these two groups, we have 3! ways.

Now, to arrange a total of (2 other men + 2 other women + 2 groups that we formed above) in a round table, we have (6-1)! way.

Hence, the number of possible arrangements if the members of the same family are seated together=3!*3!* (6-1)! ways.

But we're told that the seats are numbered,
so the number of possible arrangements if the members of the same family are seated together=3!*3!* (6-1)!*10 ways=43200 ways
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