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 January 28th, 2012, 10:32 PM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 Permutation and combinations Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together. I need to know the logic and thinking process behind how the answer is derived. What I tried is: First person has 10 seats to choose, second person 8 seats to choose and so on. Each couple can then seat on different sides. $10((6)(4)(2^5)=61440" /> Correct answer is $1920$ Another way of thinking I have is this: Consider each couple and the 2 children as each individual groups. Total number of ways of arranging the 5 groups in a round table is $(5-1)!=24$ Then permutate each couple and children$=2^5$ so total number of ways $= (2^5)24=768$
 January 29th, 2012, 07:40 AM #2 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 Re: Permutation and combinations I think 1920 is correct. There is nothing that says the kids have to sit together or even with their parents. So there are $2^4$ ways that the 4 married couples can be ordered (husband is on the left or right of the wife). Now consider the 6 objects - the 4 couples and the 2 kids. They can be seated around the table in (6-1)! ways. Thus, 16 x 5! = 16 x 120 = 1920
February 1st, 2012, 06:53 PM   #3
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Re: Permutation and combinations

Quote:
 Originally Posted by mrtwhs I think 1920 is correct. There is nothing that says the kids have to sit together or even with their parents. So there are $2^4$ ways that the 4 married couples can be ordered (husband is on the left or right of the wife). Now consider the 6 objects - the 4 couples and the 2 kids. They can be seated around the table in (6-1)! ways. Thus, 16 x 5! = 16 x 120 = 1920
Hi, I'm thinking along the line where to arrange each of the 4 married couples (since if each couple must sit together), we have 4*(2!) way, then to arrange these 4 married couples (which already became 4 units) and two children to sit around a round table will be (6-1)!, by using the Product Rule of probability, the number of ways in which these ten people can be seated round a table if each couple must sit together = 4*(2!)* (6-1)!=960, could you please tell me what did i do wrong?

Thanks.

 February 1st, 2012, 08:03 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Permutation and combinations Just the way I see it: kids sit together (between 2 couples): 8 ways not together : 12 ways Total: 20 ways couples arrangements: 96 ways 20*96 = 1920
 February 1st, 2012, 09:29 PM #5 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 2 Re: Permutation and combinations Argh...... Finally I know where it went wrong. "to arrange each of the 4 married couples (since if each couple must sit together), we have 2!+2!+2!+2!=4*(2!) way" is not correct because it means I only considered the first couple must sit together OR the second couple must sit together OR the third and fourth couple must sit together. It turned out that the setting didn't meet the requirement. It should look like 2*2*2*2=$2^4$ ways, i.e. the first couple must sit together AND the second couple must sit together AND the third and fourth couple must sit together. By the way, Denis, I must thank you for giving me another insight to tackle this problem. THANKS!
March 30th, 2012, 09:36 PM   #6
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Re: Permutation and combinations

Quote:
 Originally Posted by Denis Just the way I see it: kids sit together (between 2 couples): 8 ways not together : 12 ways Total: 20 ways couples arrangements: 96 ways 20*96 = 1920
Denis,
I'm really sorry to bother you on this one.
58 days have passed and still no fruitful attempt to even try to understand your concept.
I'm very interested to know your logic and reason behind your work...
Would you mind to explain it to me, please?

,

,

,

### danis way at couple are not sitting together

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