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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- Feb 7th 2010, 06:58 PMjmedsyinclusion-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 - Feb 8th 2010, 06:39 AMPlato
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 $\displaystyle \binom{22}{3}-\binom{16}{3} $.

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

The number of ways for conditions a & c to happen together is $\displaystyle \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 $\displaystyle \left( {\sum\limits_{k = 0}^5 {x^k } } \right)\left( {\sum\limits_{k = 0}^6 {x^k } } \right)\left( {\sum\limits_{k = 3}^7 {x^k } } \right)\left( {\sum\limits_{k = 3}^8 {x^k } } \right)$.

The coefficient of $\displaystyle x^{19}$ will be the answer.