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Probability - Couples seated round a table

  1. Mar 11, 2010 #1
    [SOLVED] Probability - Couples seated round a table

    As per my textbook: (Ross, 8th ed.), the probability of 10 couples being seated around a table, where every guy's with his girl, or guy, is:


    Where any:

    [tex]P(E_{i_{1}}, E_{i_{2}}, E_{i_{3}}, E_{i_{4}}....E_{i_{n}}) = \frac{2^{n}(19 - n)!}{19!}[/tex]

    The book explains that it considers each of the 10 couples a single entity, and therefore calculates all possible outcomes of placing these entities around the table.

    Yet why is it that for all the [tex]E_{i}[/tex] intersections in the above equation, n is subtracted from 19, instead of 10?? When n = 1, there are now 18 other people that can be arranged in whatever way, which makes sense to me.

    But when n = 2, we're permuting 17 others, when 2 couples, 4 people, have been removed from the table?? Shouldn't it be 16 for n = 2, 14 for n= 3... ?

    Last edited: Mar 11, 2010
  2. jcsd
  3. Mar 11, 2010 #2
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