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If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.

My solution with in-ex (Wrong): $$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} = 25$$ $,1\le x_{i} \ge 4 $ $$x_{i}= 1 +z_{i} $$ $$z_{1}+1 + z_{2}+1 + z_{3}+1 + z_{4}+1 + z_{5} +1+ z_{6}+1 + z_{7}+1 +z_{8} +1+ z_{9}+1 + z_{10}+1 = 25$$

$$z_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$

${sticks+plus} \choose sticks $$=$${24} \choose 15$. That is all cominations when everyone gets at least one.

The forbidden combinations, $z_{i} \ge 4 $,:$$z_{i}=4+u_{i}$$ $$4+u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$

$$u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 11$$

${sticks+plus} \choose sticks $$=$${20} \choose 11$, and this can be done in ten ways. So my answer is: ${24} \choose 15$$-10$${20} \choose 11$

According to the book this is wrong.

When doing it with GF I get: $X^{10}\frac{(1-x^{4})^{10}}{(1-x)^{10}}$. And then I don'tknow what to do. How to now which coefficient to find?

callculus42
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theUbuntuNewbie
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2 Answers2

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You have $[ x^{10} ]\cdot\frac{(1-x^4)^{10}}{(1-x)^{10}}$, where $[ x^{k} ]$ denote the coefficient of operator.

$\frac{1}{(1-x)^{10}}$ can be written as $\sum_{j=0}^\infty \binom{-10}{j}(-x)^j$ (Binomial series).

$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{-10}{j}(-x)^j$$

Now we use that $\binom{-r}{s}=\binom{r+s-1}{r-1}\cdot (-1)^s$

$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$

The coefficient of operator $[ x^{10} ] $can be combined with $x^k, 0\leq k\leq 10$ as $[ x^{10} ]\cdot x^k=[ x^{10-k} ]$

$$\left([ x^{10} ]\cdot \binom{10}{0} -[ x^{6} ]\cdot \binom{10}{1}+[ x^{2} ]\cdot \binom{10}{2}-\ldots \right)\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$

Finally we choose the corresponding coefficients

$${19 \choose 9 }-10\cdot {15 \choose 9} +45\cdot {11 \choose 9}=44,803$$

callculus42
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  • I'm assuming you are using $1-x^4$ instead of $1-x^5$ since there is a minimum of 1 cookie we are essentially looking at 15 cookies with 3 cookie maximum, but why are you solving for $[x^{10}]$ and not $[x^{15}]$? – Older Amateur Dec 14 '23 at 06:06
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I find it a little clearer not to skip the first step, which is writing the generating function. you just write down a series $A(x)$ of formal variables $x^k$, where k denotes how many cookies each child becomes. As specified k must be between 1 and 4. Thus $$ A(x) = x+x^2+x^3+x^4 $$

As you have 10 children, the generating functions becomes, $$B(x) = \mathrm{A(x)}^{10} = \mathrm{(x+x^2+x^3+x^4)}^{10} = \frac{(1-x^4)^{10}}{ (1-x)^{10}} $$ Then extracting coefficients using convolution.

user729424
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Fakhreddine Milouchi
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