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Homework Help: Permutation and Combination Problem

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data

    How many way can a team lose 3 of their next 5 games?
    How many ways can a team lose 2 of their next 5 games?
    Why are the two answers the same?

    2. Relevant equations

    Permutation = nPr = n! / (n-r)!
    Combination = nCr = nPr / r!
    n, r are non negative integers and r<=n.
    r is the size of each permutation.
    n is the size of the set from which elements are permuted.
    ! is the factorial operator.

    3. The attempt at a solution
    I think this is a problem where the permutation equation would be used. After all, the team can LLLWW or WWLLL and those would be two different permutations that would satisfy the question's requirements. But I just don't understand how I would go about solving it. Any help is much appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 28, 2012 #2
    When I was studying for the SAT's and dealt with these equations the trick was to remember "n choose r", meaning that in the equations n is the number you are choosing from and r is the amount you are choosing.

    In the first example you are choosing 3 out of 5 so n=5 r=3

    Get it?
  4. Apr 28, 2012 #3
    But then if we did the permutation formula for both questions then one wouldn't receive the same answer for both.
  5. Apr 28, 2012 #4


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    Science Advisor

    If you write "L" for lose and "W" for win you can see that the problem, is the same as 'In how many ways can you write 2 "W"s and 3 "L"s?'

    If you imagine that all of the "L" are labeled, say "1", "2", "3", and all of the "W" are labeled "1", "2", then you have 5 distinguishable objects. There are 5! ways of arranging them. But because the two "W"s are not distinguishable, such an arrangement as [itex]W_1L_1L_2L_3W_2[/itex] is exactly the same as [itex]W_2L_1L_2L_3W_1[/itex]. That is, because there are 2!= 2 ways to permute just the "W"s, we have to divide by 2! to get the number or arrangements ignoring rearrangements of just the "W"s. Similarly, there are 3!=6 ways to rearrange just the different "L"s. We must also divide by 3! since those are not "different" arrangements.

    Now do the same with 2 "L"s and 3 "W"s.
  6. Apr 28, 2012 #5
    If you were looking for all three-length permutations of {1,2,3,4,5}, you would get 123, 132, 213, 231, 312, 321 as six different possibilities.
    Are all six of these different ways of winning three games?
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