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Find the number of ways n different games can be divided

  1. Dec 14, 2006 #1
    1. The problem statement, all variables and given/known data
    Find the number of ways n different games can be divided between n different children so that every time, exactly one child gets no game.


    2. Relevant equations



    3. The attempt at a solution

    Here, since exactly one child gets no games, n games are distributed among (n-1) children which gives one child 2 games. Now, since a child can get 2 games out of n in [tex]^n C_2 [/tex] and the children can be arranged in n! ways, to total ways to distribute the games is [tex]^n C_2 * n![/tex]

    I was wondering if the following solution is also correct:
    [tex]x_1 + x_2 +x_3 +... x_n =n[/tex] such that exactly one [tex]x_i =0[/tex] and exactly one [tex] x_j =2[/tex] where i and j are not the same elements and all other elements are equal to 1.

    Therefore, the solution should be:

    Coeff. of [tex]x^n[/tex] in [tex] ( (x^0)(x^1+x^2)(x^(n-2) )^n [/tex]

    Thats coeff. of [tex]x^n[/tex] in ( (x^0)(x^1+x^2)(x^(n-2) )^n

    Is this correct?
     
    Last edited: Dec 14, 2006
  2. jcsd
  3. Dec 14, 2006 #2

    AlephZero

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    As an independent check, here's a different argument.

    Pick one child to recieve no games. n choices.

    Pick the child who recieves 2 games. (n-1) choices.

    Pick the two games to be recieved by that child. n(n-1) choices.

    Give each remaining child one game chosen at random. (n-2)! choices.

    The above are independent, so the number of combinations = n^2(n-1)^2(n-2)! = n(n-1)n!
     
  4. Dec 14, 2006 #3
    btw, how would you find the coeff. of x^n in that equation though? And thank you Azero, thats the actual given solution.
     
  5. Dec 17, 2006 #4
    That would be n(n-1)/2 .

    Chaoseverlasting, I can't quite see how you came up with the
    [tex]x_1 + x_2 +x_3 +... x_n =n[/tex] to solve the problem, since that approach would only give you the number of combinations if the games were not distinct. In this case however, even the games are distinct and thereby increases the no. of combinations.

    Cheers
     
  6. Dec 17, 2006 #5

    AlephZero

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    Oops - you are right, of course. :blushing:
     
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