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Thread: 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;
    $\displaystyle \begin{array}{ccc}
    d & s & e \\ \hline
    {13} & 0 & 3 \\
    {12} & 2 & 2 \\
    {11} & 4 & 1 \\
    {10} & 6 & 0 \\ \end{array}$.
    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; Sep 29th 2009 at 04:03 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 $\displaystyle 41328$.
    The total number of ways to get the last row is $\displaystyle 8008$
    So the probability of having no empty booth is $\displaystyle \frac{{8008}}{{41328}} = 0.194$.
    The to the question is $\displaystyle 1-0.194=0.806$

    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 $\displaystyle x^{26}$ is $\displaystyle 41328$ that is no accident.

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