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Math Help - Diner Seating

  1. #1
    Member billym's Avatar
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    Diner Seating

    (Sorry not sure if this is basic or university level)

    A diner has 16 two-seat booths.

    26 guys walk in at random.

    what is the probability that at least one booth is empty?

    ...

    Is this right:

    Total seating arrangements = P( 32, 26) = 3.655 * 10^32

    ...

    Arrangements with 3 empty booths = 26! * P( 16, 3 ) = 1.355 * 10^30

    Arrangements with 2 empty = P( 28, 26) * P( 16, 2 ) = 3.659 * 10^31

    Arrangements with 1 empty = P( 30, 26) * 16 = 1.768 * 10^32

    ...

    (seating arrangements with empty booths) / (total seating arrangements)

    = ( 2.148 * 10^32 ) / ( 3.655 * 10^32 ) = 0.588

    ...

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  2. #2
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    Quote Originally Posted by billym View Post
    (Sorry not sure if this is basic or university level)

    A diner has 16 two-seat booths.

    26 guys walk in at random.

    what is the probability that at least one booth is empty?

    ...

    Is this right:

    Total seating arrangements = P( 32, 26) = 3.655 * 10^32

    ...

    Arrangements with 3 empty booths = 26! * P( 16, 3 ) = 1.355 * 10^30

    Arrangements with 2 empty = P( 28, 26) * P( 16, 2 ) = 3.659 * 10^31

    Arrangements with 1 empty = P( 30, 26) * 16 = 1.768 * 10^32

    ...

    (seating arrangements with empty booths) / (total seating arrangements)

    = ( 2.148 * 10^32 ) / ( 3.655 * 10^32 ) = 0.588

    ...

    Hi billym,

    I think you have some problems with over-counting. Your arrangements with 1 empty booth have been included in the count of 2 empty booths, etc.

    Are you familiar with the Principle of Inclusion / Exclusion, also known as the Sieve Principle? This might be a good place to apply it.
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  3. #3
    Member billym's Avatar
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    Quote Originally Posted by awkward View Post
    Hi billym,

    I think you have some problems with over-counting. Your arrangements with 1 empty booth have been included in the count of 2 empty booths, etc.

    Are you familiar with the Principle of Inclusion / Exclusion, also known as the Sieve Principle? This might be a good place to apply it.
    Okay so I was forgetting to subtract when then 2 empty seats circumstance yielded an event when the guys arranged themselves in a way that left another empty seat. And in the 1 empty seat circumstance there would be the events of both one extra empty seat and two extra empty seats that would need to be subtracted. (???)

    So: (16!/13!)*26! + (30!/4!) + (28!/2!)[(16!/14!) - 28)] = 4.472 * 10^31


    (4.472 * 10^31) / (3.655 * 10^32) = 0.122

    Does that look more reasonable? Am I doing anything right?
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  4. #4
    Member billym's Avatar
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    Can somebody please show me how to do this problem? I've tried it a billion ways and I keep getting a different answer. All it says is "focus on the empty seats."

    please?!!!
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  5. #5
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    Quote Originally Posted by billym View Post
    Can somebody please show me how to do this problem? I've tried it a billion ways and I keep getting a different answer. All it says is "focus on the empty seats."
    Let’s reframe the problem. “How many ways can we put twenty-six identical balls into sixteen different boxes if the boxes can not hold more two balls?”
    Let d stand for double occupancy, s stand for single occupancy and e stand for empty.
    Now we have four cases;
    \begin{array}{ccc}<br />
   d & s & e  \\ \hline<br />
   {13} & 0 & 3  \\<br />
   {12} & 2 & 2  \\<br />
   {11} & 4 & 1  \\<br />
   {10} & 6 & 0  \\ \end{array}.
    To find the number of ways place the balls is the sum of those four cases.
    For the second case: “how many ways can we arrange a string of 12 d’s, 2 s’s and 2 e’s?”
    We notice that only one of those cases do we have no empty box.
    Last edited by Plato; September 29th 2009 at 05:03 AM.
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  6. #6
    Member billym's Avatar
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    My brain is so fried right now i only kind of understand. I get the chart bit. Is it a simple as the probability of having at least one empty box is 0.75?

    What is the purpose of "For the second case: how many ways can we arrange a string of 12 ds, 2 ss and 2 es?

    Could you actually write down the correct answer so I'll know when I've done it right?
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  7. #7
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    The total number of ways to fill those boxes is 41328.
    The total number of ways to get the last row is 8008
    So the probability of having no empty booth is \frac{{8008}}{{41328}} = 0.194.
    The to the question is 1-0.194=0.806

    It will help if you go to this website Wolfram|Alpha
    In the input window, type in this exact expression: expand (1+x+x^2)^16 (you can copy & paste)
    Click the equals bar at the right-hand end of the input window.
    In the expansion you will see the coefficient of x^{26} is 41328 that is no accident.

    Now ask it expand (x+x^2)^16.
    See the coefficient of x^{26} is 8008 that is also no accident.
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