Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Combinatorics problem! Need help

  1. Sep 7, 2014 #1
    Hi, this is the problem: Delegates from 10 countries, including Russia, France, England, and the United States, are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and U.S. delegates are not to be next to each other.

    So apparently the answer is 564 480 combinations, I come close to that, but not quite. Can anyone please correct my way of doing it which is as follows:

    18 x (8!-14x(6!)) = 544 320 combinations

    To justify my answer, the first 18 is for the 18 different combinations where England and France are sitting next to each other, then I tried to do the rest of the problem as if there were only 8 countries so if anyone sat next to anyone it would be 8! but - 14 (for all the different ways Russia and the US could be sitting next to each other) x 6! for all the different sitting patterns of the remaining countries! Obviously my way is wrong (unless a mistake in the answer key) so can anyone explain to me what I am doing wrong and what method would be better!!!! Thank you very much in advance!
     
  2. jcsd
  3. Sep 7, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If Russia and the US sit next to each other at the 8-party-table, it could still be a valid combination if France and England are in between in the full setup.
     
  4. Sep 7, 2014 #3
    Hi, not sure if I get your answer, but do you mean if Russia and the US were only separated by France and England? How can I bring that into my equation? Thank you!
     
  5. Sep 7, 2014 #4
    Hi! Figured it out thanks to your helpful answer! Thank you very much!!
     
  6. Sep 7, 2014 #5

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Getting their answer: Consider France and England as one unit. Ignoring the restriction on Russia and the U.S., calculate the number of permutations. That is 9!. But it is doubled because France and England can be in either order. So 2*9!. Similarly, count how many have France and England together and also have Russia and the U.S. together: 2*2*8! Subtract to get 2*9! - 2*2*8! =564480.
    This is 18*8! - 16*14*6!

    Your answer: 18 x (8!-14x(6!)) = 18*8! - 18*14*6!

    The error: You can see that your calculation of the term -18*14*6! has the factor of 18 because you are allowing the combination of France and England to split Russia and the U.S. That is wrong. Russia and the U.S. are always together in this term. So there should be a 2*8 = 16 factor for it, not your 2*9=18 factor
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Combinatorics problem! Need help
  1. Combinatorics help (Replies: 1)

  2. Combinatorics problem (Replies: 3)

  3. Combinatorics problem (Replies: 8)

  4. Combinatorics problem (Replies: 4)

Loading...