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    identical round tables

    There are 2 identical round tables.At least 1 person must sit around
    every table. In how many ways 5 persons may sit?

    This is the solution:
    C(5,1).0!.3!+C(5,2).1!.2!

    How do we get 0! and 1!?
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    Re: identical round tables

    The number of ways that N people can sit at a round table is (N-1)!, assuming that the only important things is the relatie position of each person. So for example for 3 people there are only 2 arrangements: either B sits to A's left and C to A's right, or vice vesa. One way to think of it us that the first person can sit anywhere - it doesn't matter in which sea - then the 2nd person can sit in any one of (N-1) remaining seats, the third in (N-2) remaining seats, etc.

    The solution as shown considers that there are two possible cases: 1 person sits at one table and 4 at the other, or 2 people sit at one table and 3 at the other. It does not matter which table is which, so if person A sits at table 1 and persons B, C, D and E at table 2 it's the same as person A sitting at table 2 and persons B, C, D and E sitting at table 1. For the case of 1 person at one of the tables there are C(5,1) was to choose who that 1 person is, and (N-1)! = 0! ways to arrange that one person at his table. Then for the remaining 4 people there are (N-1)!=3! ways to arrange them at the other table. For the case of 2 people at one of the tables there are C(5,2) ways to pick those two people, and (N-1) = 1! ways to arrange them at that table, and then 2! ways to arrange the other three at the other table. Hope this helps!
    Last edited by ebaines; March 13th 2014 at 08:43 AM.
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    Re: identical round tables

    many thanks.
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    Re: identical round tables

    Hello, kastamonu!

    ebaines is correct.
    Let me give it a try . . .


    There are 2 identical round tables. At least 1 person must sit at each table.
    In how many ways 5 persons may sit?

    This is the solution:
    C(5,1)\!\cdot\!0!\!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!

    How do we get 0! and 1!?

    As he noted: n people can sit around a round table in (n-1)! ways.


    There are two cases to consider.


    [1] One at one table and Four at the other.
    . . .There are C(5,1) ways to choose the One person.
    . . .He can sit at his table in 0! = 1 way.
    . . .The other Four people can sit at their table in 3! ways.
    Hence: . C(5,1)\!\cdot\!0!\!\cdot\!3!\text{ ways.}

    [2] Two at one table and Three at the other.
    . . .There are C(5,2) ways to choose the Two people.
    . . .They can sit at their table in 1! way.
    . . .The other Three people can sit at their table in 2! ways.
    Hence: . C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}


    Therefore: . C(5,1)\!\cdot\!0!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}
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    Re: identical round tables

    Many Thanks
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