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Math Help - Round Table Problem

  1. #1
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    Round Table Problem

    There are six chairs at a round table. A seating arrangement will be considered the same if everyone at the table has the same neighbor to the left and to the right.

    a) How many different ways can we seat six people, assuming that one person sits in each chair?
    b) How many different ways can we seat four people, assuming that two chairs are left empty?
    c) How many different wats can we seat three couples if every person is seated next to their date?

    a) is pretty obvious, which is 5!. However I'm not quite sure how to approach b and c. Any ideas? Thanks
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  2. #2
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    Re: Round Table Problem

    Quote Originally Posted by jpicks23 View Post
    There are six chairs at a round table. A seating arrangement will be considered the same if everyone at the table has the same neighbor to the left and to the right.
    a) How many different ways can we seat six people, assuming that one person sits in each chair?
    b) How many different ways can we seat four people, assuming that two chairs are left empty?
    c) How many different wats can we seat three couples if every person is seated next to their date?
    a) is pretty obvious, which is 5!. CORRECT
    How many ways can we have two empty chairs at a round table?
    In is not three: next to each other, one chair apart, two apart?
    Now the table is ordered. Seat the four in the non-empty seats?

    How many ways can couple A be seated at the table? Is it now ordered?

    Please tell us the answers.
    Last edited by Plato; April 12th 2012 at 12:57 PM.
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    Re: Round Table Problem

    Quote Originally Posted by Plato View Post
    How many ways can we have two empty chairs at a round table?
    In is not three: next to each other, one chair apart, two apart?
    Now the table is ordered. Seat the four in the non-empty seats?

    How many ways can couple A be seated at the table? Is it now ordered?

    Please tell us the answers.
    There're 6 ways you can put 2 empty chairs next to each other, 6 for a chair apart, and 6 for 2 chairs apart. Anything past 2 will be repetitive. But now how do you determine the number of ways the 4 people can be seated? Is it (6 x 4!) + (6 x 4!) + (6 x 4!)?

    As for c), would 3! = 6 be correct?
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    Re: Round Table Problem

    Quote Originally Posted by jpicks23 View Post
    There're 6 ways you can put 2 empty chairs next to each other, 6 for a chair apart, and 6 for 2 chairs apart. Anything past 2 will be repetitive. But now how do you determine the number of ways the 4 people can be seated? Is it (6 x 4!) + (6 x 4!) + (6 x 4!)?
    NO indeed. There are only three ways to pick the two empty chairs period.
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    Re: Round Table Problem

    Quote Originally Posted by Plato View Post
    NO indeed. There are only three ways to pick the two empty chairs period.
    So it's 3 x 4! for the total number of ways to seat 4 people, correct?
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  6. #6
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    Re: Round Table Problem

    Quote Originally Posted by jpicks23 View Post
    So it's 3 x 4! for the total number of ways to seat 4 people, correct?
    No that is not correct. The answer is 60.
    Look at the diagram.Round Table Problem-untitled.gif
    Think how many ways to seat four people for each of those three cases.
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