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Math Help - inclusion-exclusion principle question

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    inclusion-exclusion principle question

    Determine how many integer solutions there are to a+b+c+d=19 if

    0<=a<=5,
    0<=b<=6,
    3<=c<=7,
    3<=d<=8
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  2. #2
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    Quote Originally Posted by jmedsy View Post
    Determine how many integer solutions there are to a+b+c+d=19 if 0<=a<=5, 0<=b<=6, 3<=c<=7, 3<=d<=8
    This is an extremely tedious problem to do with inclusion/exclusion!
    It is easy to do with generating functions.

    I will start you off with inclusion/exclusion.
    The number of ways for condition a to happen is \binom{22}{3}-\binom{16}{3} .

    The number of ways for conditions c to happen is \binom{19}{3}-\binom{14}{3} .

    The number of ways for conditions a & c to happen together is \binom{19}{3}-\binom{8}{3} .

    Now to finish you have to figure for each of the other conditions in all possible combinations.


    With generating functions we expand \left( {\sum\limits_{k = 0}^5 {x^k } } \right)\left( {\sum\limits_{k = 0}^6 {x^k } } \right)\left( {\sum\limits_{k = 3}^7 {x^k } } \right)\left( {\sum\limits_{k = 3}^8 {x^k } } \right).
    The coefficient of x^{19} will be the answer.
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