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Permutations in a conference

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    At a conference of 5 powers,each deligation consists of 3 members. If each delegation sits together, with the leader in the middle, in how many ways ca the members be arranged at a round table?



    2. Relevant equations

    No. of ways of arranging n objects around a circle=(n-1)!
    P(n,r)=n!/(n-r)!


    3. The attempt at a solution

    I understand that 15 people go around the table
    => no. of ways to arrange=14!
    But again in their different delgations,each can be arranged 2!/(2-2)!=2ways
    => There are 2(15) ways to arrange within the delegations
    .'. there are 14!(30) ways
    IS THIS CORRECT?
     
  2. jcsd
  3. Nov 6, 2009 #2
    We can not consider 15 people as a whole since we do not have complete flexibility regarding their arrangement.
    Since there are 5 groups which always individually sit together, there are 4! ways in which the groups can be organized on the table.
    Now, each group can internally have 2 possible arrangements. => 2^5 arrangements in all.
    i.e. Total no. of permutations = 4! * 2^5
     
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