six girls are to enter a dance with ten boys so that every girl is between two boys. what is the probability that some specified boy remains between two boys?

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  1. First let arrange the 10 boys.
    This can be done in 10! ways

    Now there are 9 places in this arrangement where we can place a girl, but we only have 6 girls.
    So we have to choose 6 of those 9 places to insert the single girls.
    There are C(9,6) or 84 ways of doing so
    But we can also arrange the girls in 6! ways to fit into those 84 places,

    So I get 84(10!)(6!) ways of doing this
    = rather large , I get 2.1947 x 10^11

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  2. Thank you. But sorry the question is #six girls are to enter a dance with ten boys to form a ring...#

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  3. The ten boys created 10 successive intervals (arcs). 6 girls can choose six of the intervals in C(10, 6) ways.

    If a particular boy is to be between 2 boys, the number of intervals (arcs) reduces by 2 to 8 and the choice now is C(8, 6) ways
    So, probability =C(8, 6)/C(10, 6)=2/15

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