My discrete math textbook has the following solution to the problem using generating functions (the solution has to use generating functions):


Once we start developing the formal power series we get the following:

$$x^{10}(1-x^6)^{10}\frac{1}{(1-x)^{10}}$$ Because of $x^{10}$ we now need to find the coefficient of only $x^{15}$: $$\binom{10}{0}\binom{10-1+15}{15}-\binom{10}{1}\binom{10-1+9}{9}+\binom{10}{2}\binom{10-1+3}{3}$$ I completely understand the first two expressions but why are we using the third one: $$\binom{10}{2}\binom{10-1+3}{3}$$ Where does the $3$ come from and why this isn't enough: $$\binom{10}{0}\binom{10-1+15}{15}-\binom{10}{1}\binom{10-1+9}{9}$$ ?

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    $\begingroup$ Can you explain your understanding of the first two terms? $\endgroup$ – Henning Makholm Aug 4 '16 at 19:35
  • $\begingroup$ As an aside, you have a typo in the first line. You write $(\color{red}{x}+x^1+x^2+\dots+x^6)$ where it should instead have been $(x^1+x^2+\dots+x^6)$. $\endgroup$ – JMoravitz Aug 4 '16 at 19:35
  • $\begingroup$ you're right corrected $\endgroup$ – Yos Aug 4 '16 at 19:44

The coefficient of $x^{25}$ in $(x+x^2+\ldots+x^6)^{10}$ equals the coefficient of $x^{15}$ in $$ (1+x+\ldots+x^5)^{10} = (1-x^6)^{10}\cdot\frac{1}{(1-x)^{10}} $$ so the answer is given by $$ [x^{15}] \left(\sum_{k=0}^{10}\binom{10}{k}(-1)^k x^{6k}\right)\cdot\left(\sum_{j\geq 0}\binom{9+j}{j}x^j\right) $$ and it is enough to consider just the terms associated with $k=0,1,2$, since $x^{6k}$ has an exponent greater than $15$ for any $k>2$. By computing the Cauchy product between the above series it follows that the answer is given by: $$ \underbrace{\binom{10}{0}\binom{24}{15}}_{k=0,\;j=15}-\underbrace{\binom{10}{1}\binom{18}{9}}_{k=1,\;j=9}+\underbrace{\binom{10}{2}\binom{12}{3}}_{k=2,\;j=3}.$$

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    $\begingroup$ thanks, I understood that I need to go back to sigma notation for better understanding how to solve such problems, thanks! $\endgroup$ – Yos Aug 4 '16 at 20:03

As opposed to generating functions, I personally learned to solve this question via inclusion exclusion.

We were looking for the number of integer solutions to the system: $\begin{cases} x_1+x_2+\dots+x_{10}=25\\ 1\leq x_i\leq 6&\text{for all}~i\end{cases}$

Through a change of variable, we have instead the system $\begin{cases}y_1+y_2+\dots+y_{10}=15\\ 0\leq y_i\leq 5\end{cases}$

Approaching via inclusion-exclusion on the events that the respective upper bound conditions are violated. Let $S$ be where we ignore any upper bound conditions. Let $A_i$ be the set where the $i$'th upper bound condition is violated. The total with no upperbound conditions violated is:

$$|S|-\sum\limits_{i=1}^{10}|A_i|+\sum\limits_{i=1}^{10}\sum\limits_{j=i+1}^{10}|A_i\cap A_j| - \sum\limits_{i=1}^{10}\sum\limits_{j=i+1}^{10}\sum\limits_{k=j+1}^{10}|A_i\cap A_j\cap A_k|+\dots$$

if we have no upper bound conditions violated, there are $\binom{10+15-1}{15}$ solutions. If one upper bound condition is violated, pick which one. Each of which will have a total of $\binom{10+9-1}{9}$ solutions for a total of $\binom{10}{1}\binom{18}{9}$ needing to be removed from the count. If two upperbound conditions are violated, pick which two they are. Each of which will have a total of $\binom{10+3-1}{3}$ solutions for a total of $\binom{10}{2}\binom{12}{3}$ needing to be added back. Three upperbound conditions being simultaneously violated is impossible.

This gives, again the solution of:


The explanation for the final term being with this understanding, that you had subtracted the same "bad outcomes" from your total too many times otherwise. Why we use a $3$ would be from how we solve the question of finding how many violate the $i$'th and $j$'th upperbound conditions. Suppose $i=1$ and $j=2$, we have then the system:

$\begin{cases} y_1+y_2+\dots+y_{10}=15\\ 6\leq y_1\\ 6\leq y_2\\ 0\leq y_3\\ \vdots \\ 0\leq y_{10}\end{cases}$

After a change of variable, we have the system

$\begin{cases} z_1+z_2+\dots+z_{10}=3\\ 0\leq z_i\end{cases}$ which is of a known form.

  • $\begingroup$ thanks, I agree that the problem can be solved via inclusion/exclusion principle but the problem I posted had to be solved via generating functions. $\endgroup$ – Yos Aug 4 '16 at 20:04
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    $\begingroup$ My goal was to provide an alternate explanation to why the final term appeared the way it did is it may be more intuitive when viewed from a different angle. $\endgroup$ – JMoravitz Aug 4 '16 at 20:36

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