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Probability involving combinations

  1. Feb 15, 2007 #1
    Surely there's an easier way to do this question than the method given?

    Q: Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

    A: Yes. Let's call the first couple C and c, the second couple K and k, and the single person S. Let's seat S in different places and figure out the possible ways to have no couples sit together. If S sits in the first seat, any of the remaining four people could sit next to S. However, only two people could sit in the next seat: the two who don't form a couple with the person just seated). For example, if we have S K so far, C or c must sit in the third seat. Similarly, we have only one choice for the fourth seat: the remaining person who does not form a couple with the person in the third seat. Because we have seated four people already, there is only one choice for the fifth seat; the number of ways is 4 × 2 × 1 × 1 = 8. Because of symmetry, there are also 8 ways if S sits in the fifth seat. Now let's put S in the second seat. Any of the remaining four could sit in the first seat. It may appear that any of the remaining three could sit in the third seat, but we have to be careful not to leave a couple for seats four and five. For example, if we have C S c so far, K and k must sit together, which we don't want. So there are only two possibilities for the third seat. As above, there is only one choice each for the fourth and fifth seats. Therefore, the number of ways is 4 × 2 × 1 × 1 = 8. Because of symmetry, there are also 8 ways if S sits in the fourth seat. This brings us to S in the third seat. Any of the remaining four can sit in the first seat. Two people could sit in the second seat (again, the two who don't form a couple with the person in the first seat). Once we get to the fourth seat, there are no restrictions. We have two choices for the fourth seat and one choice remaining for the fifth seat. Therefore, the number of ways is 4 × 2 × 2 × 1 = 16. We have found a total of 8 + 8 + 8 + 8 + 16 = 48 ways to seat the five people with no couples together; there is an overall total of 5! = 120 ways to seat the five people, so the probability is = . Whew!

    Thanks for your help.
     
  2. jcsd
  3. Feb 15, 2007 #2
    Try the other way round, ie how many ways are there so that the two couples always sit as a couple.
     
  4. Feb 15, 2007 #3
    That's not much easier. First of all, the reverse would be, at least one couple sits as a couple. Second of all, isn't there a way to do this using combinations?
     
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