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  1. #1
    Senior Member pankaj's Avatar
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    Find number of ways

    There are 8 students from 4 different schools(2 from each school).In how many ways can these students be put in 4 different rooms(2 students in each room) so that no room contains students of the same school.
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  2. #2
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    Re: Find number of ways

    Quote Originally Posted by pankaj View Post
    There are 8 students from 4 different schools(2 from each school).In how many ways can these students be put in 4 different rooms(2 students in each room) so that no room contains students of the same school.
    There are 4! to assign the students from school X to a different room.
    What about the students from school Y~?
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  3. #3
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    Re: Find number of ways

    Hello, pankaj!

    There are 8 students from 4 different schools (2 from each school).
    In how many ways can these students be put in 4 different rooms (2 students in each room)
    so that no room contains students of the same school.

    Suppose the students are: . A,A,B,B,C,C,D,D
    In how many ways can they partitioned into four unmatched pairs?

    This is derangement of four objects.
    There are 9 derangements. **

    Then the four pairs can be assigned rooms in 4! ways.

    Answer: . 9\cdot4! \:=\:216 ways.


    ** You can count them if you like . . . \begin{array}{cccc}A&B&C&D \\ \hline B&A&D&C \\ C&A&D&B \\ D&A&B&C \\ B&D&A&C \\ C&D&A&B \\ D&C&A&B \\ B&C&D&A \\ C&D&B&A \\ D&C&B&A \end{array}
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  4. #4
    Senior Member pankaj's Avatar
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    Re: Find number of ways

    @Soroban.Please check my solution.

    Number of ways to put the 8 students in 4 rooms such that there are 2 students in every room =\dbinom{8}{2}\times\dbinom{6}{2}\times\binom{4}{2  }=28\times15\times6=2520.

    Let A_i denote the property that the  ith room contains students of the same school.
    n(A_1\cup A_2\cup A_3\cup A_4)=\sum n(A_{i})-\sum n(A_{i}\cap A_{j})+\sum n(A_{i}\cap A_{j}\cap A_{k})- n(A_{1}\cap A_{2}\cap A_{3}\cap A_{4}).

    n(A_{1})=\binom{4}{1}\times\big( \binom{6}{2}\times \binom{4}{2}\times 1\big)=4\times 90=360.

    Therefore, \sum n(A_{i})=360\times 4=1440

    n(A_{1}\cap A_{2})=\dbinom{4}{2} \times 2! \times \dbinom{4}{2} =72. Therefore, \sum n(A_{i}\cap A_{j})=\dbinom{4}{2} \times 72=432

    n(A_{1}\cap A_{2}\cap A_{3})=\dbinom {4}{3}\times 3!=24.Therefore, \sum n(A_{i}\cap A_{j}\cap A_{k})=24=24

    n(A_{1}\cap A_{2}\cap A_{3}\cap A_{4})=4!=24

    n(A_1\cup A_2\cup A_3\cup A_4)=1440-432+24-24=1008



    Thus required number of ways= 2520-n(A_1\cup A_2\cup A_3\cup A_4)=2520-1008=1512
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  5. #5
    Senior Member pankaj's Avatar
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    Re: Find number of ways

    I want to make a slight correction

    \sum n(A_{i}\cap A_{j}\cap A_{k})=4\times 24=96

    Thus \ n(A_{1}\cup A_{2}\cup A_{3}\cup A_{4})=1440-432+96-24=1080

    Finally,required number of ways= 2520-1080=1440
    Last edited by pankaj; January 21st 2014 at 05:13 AM.
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