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

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 February 1st, 2012, 01:56 AM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 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
 February 1st, 2012, 02:06 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 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: $N=3!\cdot5\cdot3!\cdot4!\cdot10=43200$
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.

$\text{Duct-tape the first family together: }\:\fbox{\text{Man & parents}}$
[color=beige]. . [/color]$\text{They can be permuted in }3!\text{ ways.}$

$\text{Duct-tape the second family together: }\:\fbox{\text{Woman & parents}}$
[color=beige]. . [/color]$\text{They can be permuted in }3!\text{ ways.}$

$\text{Seat the first family:\:10 choices.}$

$\text{Seat the second family:\:5 choices.}$
[color=beige]. . [/color]$\text{(Think about it.)}$

$\text{Then the last four can be seated in }4!\text{ ways.}$

$\text{The number of seating arrangements is: }\3!)(3!)(10)(5)(4!) \:=\:43,200" />

 February 1st, 2012, 09:46 PM #4 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 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
 February 1st, 2012, 10:34 PM #5 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 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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### in how many ways can a family of 5 sit round a table

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