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Counting problem

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    100 hundred people are to be divided into 10 discussion groups with 10 people in each group
    how many ways can this be done.
    3. The attempt at a solution
    So if we think of it as people on a 10 by 10 grid their are 100! ways of populating the grid and then 10! ways or rearranging the columns and 10! ways of rearranging the rows.
    so would the answer be
    [itex] \frac{100!}{10!^{10}10!} [/itex]
    I have 10!^10 because there are ten columns and each of those 10 columns can be arranged 10! ways. And the other 10! on the bottom because I could rearrange those rows 10! ways. And I divide 100! by them because we are over counting. If we switch the people around in the group it is still a distinct group.
     
  2. jcsd
  3. Jun 28, 2011 #2

    lanedance

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    didn't totally follow the re-arranging of the cols, so as another way for the first group we choose 10 from 100 without order
    [tex] \frac{100!}{10!90!} [/tex]

    for the 2nd group we choose 10 from 90 without order
    [tex] \frac{90!}{10!80!} [/tex]

    so putting these together for both groups we get
    [tex] \frac{100!}{10!90!} \frac{90!}{10!80!} = \frac{100!}{10!^280!} [/tex]

    and so on, once finished counting the groups we need to account for the different ways to arrange ten groups which represents the repeated counts -

    but its starting to look pretty similar to your answer...
     
    Last edited: Jun 28, 2011
  4. Jun 28, 2011 #3

    lanedance

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    also another check would be to see if you formula generalise to n^2 people in n groups and check for n = 2 and n= 3
     
  5. Jun 28, 2011 #4

    lanedance

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    ok - so i get your idea now, but think it needs a little tweek.. the rows got me

    100! ways to populate a grid, where the columns represent a group
    - 10! ways to rearrange within a single column and 10 columns gives a total (10!)^10 ways to rearrange within columns
    - there is also 10! different ways to arrange the columns themsleves
    So in all there is (10!)^11 ways to rearrange the grid for a given combination that leads to the same group structure
     
    Last edited: Jun 28, 2011
  6. Jun 28, 2011 #5
    thanks for your responses, ya i think my answer makes sense.
     
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