1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
    where,
    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Permutation and Combination Problem
Loading...