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Combination question

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Problem: How many ways are there for 15 boys and 15 girls at a dance to pair up into 15 heterosexual dance couples?

    2. Relevant equations
    This is probably somewhat helpful, but I couldn't figure out exactly how to implement it.

    nCr = [tex]\frac{n!}{k!(n-k)}[/tex]

    3. The attempt at a solution
    What I thought was if lets say you have all of the boys in a row 1 to 15, and you just rearrange the girls in every possible combination 15! and just match them up with each boy. then I multiply that number by 15 for the boys, and I get a number of like 19615115520000 which seems quite high...
    I also tried to do it using a smaller amount of numbers to get an easier idea of how to do it, but it didn't help me much because then the number seemed too small.
  2. jcsd
  3. Apr 21, 2008 #2
    In this problem, there are two events, choosing a boy to dance(E[tex]_{1}[/tex]) and choosing a girl to dance(E[tex]_{2}[/tex]).
    The number of combinations can be given by: E = E[tex]_{1}[/tex] [tex]\ast[/tex] E[tex]_{2}[/tex].
    E[tex]_{1}[/tex] can be written as [tex]_{15}[/tex]C[tex]_{1}[/tex]. This is the number of combinations of 1 person out of a group of 15.
    E[tex]_{2}[/tex] is the same: [tex]_{15}[/tex]C[tex]_{1}[/tex].
    Using the equation you provided, [tex]_{15}[/tex]C[tex]_{1}[/tex] = [tex]\frac{15!}{1!(15-1)}[/tex] = 15

    Therefore: E = E[tex]_{1}[/tex] [tex]\ast[/tex] E[tex]_{2}[/tex] = [tex]_{15}[/tex]C[tex]_{1}[/tex] [tex]\ast[/tex] [tex]_{15}[/tex]C[tex]_{1}[/tex] = 15 * 15 = 225
    So, there are 225 combinations.
  4. Apr 21, 2008 #3
    Actually, the equation should be:

    nCr = [tex]\frac{n!}{r!\ast(n-r)!}[/tex] = [tex]\frac{15!}{1!\ast(15-1)!}[/tex]
  5. Apr 21, 2008 #4

    Gib Z

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    Homework Helper

    I'm not sure if thats correct guys, but Welcome to PF both of you.

    Your idea was a good one - try the same problem with 2 girls and 2 boys. The first girl has 2 options. Then the second girl only has 1 option. So multiply it- 2 combinations.

    3 girls and 3 boys - The first girl can choose between 3 boys. When she chooses, the second girl only has 2 to choose from. Then the last one only has 1 to choose from. So thats 3! combinations.

    Extend this to 15 girls and 15 boys.
  6. Apr 21, 2008 #5


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    Staff Emeritus
    Science Advisor

    The way I would do this problem is this:

    First put the boys in a line- you can do that any way you like- their order does not matter.

    Now choose a girl to dance with the first boy. How many choices do you have?
    Once you have done that, choose a girl to dance with the second boy. How many choices do you have for that?
    Go through the line of boys like that, and, finally, using the "fundamental principal of counting", multiply the number of choices each time. The answer should be easy- and you should see immediately how that fits Gibz's suggestion.
  7. Apr 21, 2008 #6
    Sorry for giving the incorrect information. :eek:
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