Number of copies

Something very interesting happened yesterday. One of my friends was going to give a presentation in front of a group of people (24 in total), he prepared a work with other three guys (a group of 4 people).
They thought that it would be nice if everyone in the audience has a copy of the presentation slides (so they can take notes, for example). The presentation had 30 slides, my friend went to the copy center and asked for 24 copies. The guy who was there told him that he had a lot of people in the queue and he couldn't do all the copies for him, he could do only 10 as much (it was late, the most probable thing is that he was tired and didn't want to work). My friend left the copy center without any copy (really upset), told what had happened to his classmates. As they needed the copies, they decided to go to the copy center and ask for 6 copies each (so they will have the 24 copies and they avoid any problem with the problematic guy). They resolved their problem and gave the presentation, but they left me a doubt in my head. In how many ways they could distributed the number of copies so they can have 24 copies in total?, forget about the money....

So, bassically the problem is,

\begin{equation}
x_1 + x_2 + x_3 + x_4 = 24\;\;\;where\;\;\;x_i\leq 10\;\;for\;\;0\leq i \leq 4
\end{equation}

I now have an equation, I feel relaxed...


Consider only the equation, forget about the $x_i\leq 10$ restriction, counting the number of solutions of that equation is like thinking that we have 24 balls and we want to distribute the 24 balls between 4 children.

So, the solution of the equation is a known problem, it is a selection, with repetition, of size 24 from a collection of size 4. What?!?!, yes, this is one of the mathematical things that are simpler to see the formula than the description.

we can say then that the number of solutions of the equation is $|S|=\displaystyle\binom{4+24-1}{24}$.

So we have the solution of the equation, which was not what I wanted, but we are quite close. If I do $|S|$ (number of solutions) minus all the solutions in $S$ that does not satisfy the condition $x_i\leq10$ I will have the number of ways the guys could had distributed their copies. (see the inclusion and Exclusion Principle)


I need to find the solution of the equation,

\begin{equation}
x_1 + x_2 + x_3 + x_4 = 24\;\;\;where\;\;\;x_i\geq 11\;\;for\;\;0\leq i \leq 4
\end{equation}

we say that solution $x_1,x_2,x_3,x_4$ satisfies condition ci, where $1\leq i\leq 4$, if $x_i\geq 11$. So the answer of our initial problem is $N(\overline{c_1}\;\overline{c_2}\;\overline{c_3}\;\overline{c_4})$ (find solutions don't satisfy $c_1$ and $c_2$ and $c_3$ and $c_4$).

$N(\overline{c_1})$ is equal to find,

\begin{equation}
(x_1 + 11) + x_2 + x_3 + x_4 = 24\;\;\;where\;\;\;x_i\geq 0\;\;for\;\;0\leq i \leq 4
\end{equation}

which is equal to:

\begin{equation}
x1 + x2 + x3 + x4 = 13\;\;\;where\;\;\;x_i\geq 0\;\;for\;\;0\leq i \leq 4
\end{equation}

That is the same as what we solved before!!

So, $N(c_1) =\displaystyle\binom{4+13-1}{13}$, as it is a simmetric problem we have $N(c_1) = N(c_2) = N(c_3) = N(c_4)$, so $|S_1| = \displaystyle\binom{4}{1} \displaystyle\binom{4+13-1}{13}$ (the total number of solutions individually)

We need also to calculate $N(c_i,c_k)$ where $0\leq i,k\leq 4$ and $i \neq k$, which is also easy,

\begin{equation}
(x_1 + 11) + (x_2+11) + x_3 + x_4 = 24\;\;\;where\;\;\;x_i\geq 0\;\;for\;\;0\leq i \leq 4
\end{equation}


\begin{equation}
x1 + x2 + x3 + x4 = 2\;\;\;where\;\;\;x_i\geq 0\;\;for\;\;0\leq i \leq 4
\end{equation}


So $N(c_1,c_2) = \displaystyle\binom{4+2-1}{2}$, as it is a simmetric problem $|S_2| = \displaystyle\binom{4}{2} \displaystyle\binom{4+2-1}{2}$ ( the total number of solutions of (2) where $0\leq i,k\leq 4$ and $i \neq k$)

$N(c_i,c_j,c_k) = 0$ and $N(c_i,c_j,c_k,c_t) = 0$ (do the same thing as we did before and you get a negative number on the right side of the equation)

So now we have all the calculations we need

$N(\overline{c_1}\;\overline{c_2}\;\overline{c_3}\;\overline{c_4}) = |S| - |S1| + |S2| - |S3| + |S4| $

$N(\overline{c_1}\;\overline{c_2}\;\overline{c_3}\;\overline{c_4}) = \displaystyle\binom{4+24-1}{24} - \displaystyle\binom{4}{1} \displaystyle\binom{4+13-1}{13} + \displaystyle\binom{4}{2} \displaystyle\binom{4+2-1}{2} - 0 + 0 $

$N(\overline{c_1}\;\overline{c_2}\;\overline{c_3}\;\overline{c_4}) = \frac{27!}{24! 3!} - \frac{4!}{1! 3!} \frac{16!}{13! 3!} + \frac{4!}{2! 2!} \frac{5!}{2! 3! }= 745 $

Again, who can ever thing that maths are awful!