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Math Help - Seating Arrangement Problem

  1. #1
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    Seating Arrangement Problem

    I need some help with two problems. The first, I don't even know how to approach...

    Three married couples are seated randomly in a row of six seats. What is the probability that no wife and husband are seating next to one another?



    The second problem...

    On average, how many chocolate chips should a cookie contain if the probability that a cookie contains at least one chip is to be at least .99?

    I'm pretty sure that this is a poisson problem
    <br />
e^{-\lambda}{\lambda^k \over{k!}}<br />
    where the entire expression is equal to .99, k = 1, and I have to solve for lambda. The problem is that I don't know how to solve for lambda in

    <br />
e^{-\lambda}{\lambda}<br />


    Can anyone help me?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by valosn View Post
    I need some help with two problems. The first, I don't even know how to approach...

    Three married couples are seated randomly in a row of six seats. What is the probability that no wife and husband are seating next to one another?
    the total number of seating arrangements is 6!

    it is easier to find the arrangement where every husband does sit next to his wife. to do this, we can consider each couple as one object with a permutation of 2! (since the wife can sit on the right or left of the husband). Thus we have 3 objects, each with a permutation of 2!. thus the number of seating arrangements where each husband sits next to his wife is 2!3!

    so what is the probability of each husband sitting next to his wife? Hence, what is the probability that no husband sits next to his wife?
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  3. #3
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    Allow C_i to denote the case couple i is seated together. Then \overline {C_i } is the complement of that event. Now you want P\left( {\overline {C_1 }  \cap \overline {C_2 }  \cap \overline {C_3 } } \right) = 1 - P\left( {C_1  \cup C_2  \cup C_3 } \right) = \frac{{\sum\limits_{k = 0}^3 {\left( { - 1} \right)^k {{6} \choose {k}} 2^k \left( {6 - k} \right)!} }}{{6!}}.
    Last edited by Plato; December 4th 2007 at 02:33 PM.
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