
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. (6)(4)(2^5)=61440" /> Correct answer is 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 Then permutate each couple and children so total number of ways 
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 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 (61)! ways. Thus, 16 x 5! = 16 x 120 = 1920 
February 1st, 2012, 06:53 PM  #3  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 2  Re: Permutation and combinations Quote:
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= 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  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 2  Re: Permutation and combinations Quote:
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?  

Tags 
combinations, permutation 
Search tags for this page 
four married couples attend a wedding dinner. one of the couples brought along,if 2 must sit together permutate,how to arrange couple in permutation and combination,danis way at couple are not sitting together
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
PERMUTATION AND COMBINATIONS  maths4computing  Advanced Statistics  4  July 21st, 2013 03:06 PM 
Combinations /Permutation question  calh27  Advanced Statistics  4  June 11th, 2012 05:48 PM 
Permutation and combinations (seating)  Punch  Algebra  15  January 31st, 2012 10:48 PM 
Complex Combinations Within Combinations Problem?  bugrocket  Advanced Statistics  2  January 23rd, 2011 05:02 PM 
Permutation or Combinations  lowlyfe97  Real Analysis  3  August 4th, 2009 08:43 AM 