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Math Help - easy probability?

  1. #1
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    easy probability?

    herbert and his two chums , along with five of herberts doting aunties,
    have to sqeeze onto the backseat of his fathers bently , en en rout to
    royal ascot .

    given that herbert does not sit at either end , and that the seating order
    is otherwise random , find the probabilty of herbert having his best chumbs either side of him

    trying to aswer this 1 is like banging my head against a brick wall
    can any one help



    tx
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  2. #2
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    Quote Originally Posted by jickjoker View Post
    herbert and his two chums , along with five of herberts doting aunties,
    have to sqeeze onto the backseat of his fathers bently , en en rout to
    royal ascot .

    given that herbert does not sit at either end , and that the seating order
    is otherwise random , find the probabilty of herbert having his best chumbs either side of him

    trying to aswer this 1 is like banging my head against a brick wall
    can any one help



    tx
    If we temporarily disregard the restriction that Herbert does not sit at the end, there are 8! arrangements of the passengers. Of these, 7! have Herbert sitting at the left end, and 7! have him sitting at the right end. So there are 8! - 2\times 7! possible arrangements, which we will assume are all equally likely.

    How many of these arrangements have Herbert between his pals? Duct tape Herbert and his pals into a single unit with Herbert in the middle. This can be done in 2 ways depending on which pal sits on the left. There are then 6! ways to arrange the group of pals plus 5 aunts.

    So the probability is \frac{2 \times 6!}{8! - 2\times 7!}.
    Last edited by awkward; August 1st 2009 at 09:49 AM. Reason: wording
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  3. #3
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    Quote Originally Posted by jickjoker View Post
    herbert and his two chums , along with five of herberts doting aunties,
    have to sqeeze onto the backseat of his fathers bently , en en rout to
    royal ascot .
    given that herbert does not sit at either end , and that the seating order is otherwise random , find the probabilty of herbert having his best chumbs either side of him
    Suppose X~\&~Y are his two chums and that A,B,C,D,E are his aunts.
    The outcome you want is an arrangement of the string \boxed{XHY},A,B,C,D,E that is six items.
    We could also have the string \boxed{YHX},A,B,C,D,E.
    So there are 2\cdot 6! ways to get the outcome you want.

    Given that Herbert will not sit at either end, how may total ways could 8 people sit?
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