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    "Two punts can each hold 6 people. A party of 10 wishes to use these punts. In how many ways can the party be divided?"

    The answer is 462, but I can't for the life of me work out how to do it.

    Help, please?
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    You can split it into two cases, and sum the number of solutions for each case to arrive at the answer.

    As a hint, the first case would be where the party is divided equally into 5 people per punt. For this case, in how many ways can the party be divided?
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    You also need to treat the punts as indistinguishable. Though it's not clear from the question that you need to do that.
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    Thanks.
    I've wasted too much time on this question and I'm getting nowhere though, can somebody please just show me how to reach the answer?
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    (Original post by TheMan100)
    Thanks.
    I've wasted too much time on this question and I'm getting nowhere though, can somebody please just show me how to reach the answer?
    2 cases:

    a) 5 in each punt.

    So, how many choices for the first punt? 10C5, and the 2nd punt takes whatever is left in each case.

    b) 6 in one punt and 4 in the other.

    Have a go.

    Then add the two values.
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    (Original post by ghostwalker)
    2 cases:

    a) 5 in each punt.

    So, how many choices for the first punt? 10C5, and the 2nd punt takes whatever is left in each case.

    b) 6 in one punt and 4 in the other.

    Have a go.

    Then add the two values.
    FINALLY

    Thanks... at one point I was close, but I was doing 10C5 twice for some reason.
 
 
 
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