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Combinatorics problems

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    12 subjects (6 male, 6 female) are put on 3 lines, 4 in each. In how many ways can this be done, if one of the males and one of the females want to be on the same line?

    2. Relevant equations

    None.

    3. The attempt at a solution

    I thought it like this. I can pick the first 10 spots and leave the lovebirds behind. That can be done in 2 * (4C2) * 4! * 4! = 6912. 2*(4C2) because I'm picking spots for the two on the line where the lovebirds are missing, and they can be switched places so I multiply by 2. 4! * 4! is for the normal lines.

    When I add the lovebirds to this, I get 6912 * 3 * 2 = 41472 because the unfinished line could be any of the 3, and multiply by 2 because, for each possibility in 4C2 above, I could have just put the couple there and switched them inwards.
     
  2. jcsd
  3. Sep 22, 2013 #2

    vela

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    If you didn't have the requirement that the lovebirds sit in the same line, there'd be 12!=479001600 ways to arrange the subjects, right? It seems that requiring two of them sit together shouldn't result in such a drastic reduction in possibilities.

    I don't understand the factor of 3. Any of the three what?

    Don't you need to account for who sits in which rows still?
     
  4. Sep 22, 2013 #3
    Edit: vela has a better response than me and I can't seem to delete posts any more for some reason.
     
  5. Sep 22, 2013 #4

    vela

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    I got 130636800 ways. I might be off by a factor of 3, though.

    It's okay to give out the answer in a problem like this. It's just not okay to show explicitly how you got the answer. It's the OP's job to figure out how to get the answer.
     
  6. Sep 22, 2013 #5
    Edit: I've gotten so confused with this problem that I want to edit this post even though vela has already responded. My fault.

    Vela started his argument by saying there are 12! ways to arrange these people, but that does not try to eliminate double-counting configurations such as:
    (A B C D) (E F G H) (I J K L) and (E F G H) (A B C D) (I J K L)
    So this problem is ambiguous about whether or not those two cases are equivalent. So, OP, do you know whether the lines are "distinguishable" or not?
     
    Last edited: Sep 22, 2013
  7. Sep 22, 2013 #6

    vela

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    I didn't consider those equivalent. Why do you assume they are?
     
  8. Sep 22, 2013 #7
    Well I guess there is ambiguity in the question. It doesn't say "Line A, Line B, Line C" so you could assume they are "indistinguishable" lines.
     
    Last edited: Sep 22, 2013
  9. Sep 22, 2013 #8

    vela

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    Could be. I guess the OP can tell us.
     
  10. Sep 22, 2013 #9
    Alright I think I figured out how you got your answer vela, and I do believe it is correct for the distinguishable lines case.

    You said there are 12! possible orderings for the whole group of people, and then you multiplied this by the probability that the couple is randomly placed in the same line. [You would calculate this by assuming the male lovebird is sitting in one spot, and the female could go into any of the other 11 spots].

    To take care of the indistinguishable lines case, if we treat the different permutations of lines as equivalent, then we'd get 21772800.
     
    Last edited: Sep 22, 2013
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