Distributing $\textbf{m\&m's}$

We notice that there is recursion of demands that comes in pairs.

For instance,

the $\textbf{1st}$ child wants no more than $\textbf{1}$ $\textcolor{#D61F06}{red}$ $\textbf{m \& m's}$ and the $\textbf{7th}$ child wants no more than $\textbf{2}$ $\textcolor{#D61F06}{red}$ $\textbf{m \& m's}$ ,

the $\textbf{2nd}$ child wants no more than $\textbf{1}$ $\textcolor{#EC7300}{orange}$ $\textbf{m \& m's}$ and the $\textbf{8th}$ child wants no more than $\textbf{2}$ $\textcolor{#EC7300}{orange}$ $\textbf{m \& m's}$ ,

...

With that being said, we shall focus on computing the pair of requests from one of the colors, say $\textcolor{#D61F06}{red}$ and compute the total number of ways of distribution by rule of product .

Let $x_{1}$ be the number of chocolates received by the $\textbf{1st}$ child and $x_{7}$ be the number of chocolates received by the $\textbf{7th}$ child.

The only possible solutions for $x_{1},x_{7}$ are:

$x_{1}=0,x_{7}=0$

$x_{1}=1,x_{7}=0$

$x_{1}=0,x_{7}=1$

$x_{1}=1,x_{7}=1$

$x_{1}=0,x_{7}=2$

$x_{1}=1,x_{7}=2$

When $x_{1}=0,x_{7}=0$ ,

$x_{2}+x_{3}+...+x_{6}+x_{8}+...+x_{12}=3$

By Stars and Bars Theorem , there are $12\choose9$ non-negative integer solutions to the equation.

When $x_{1}=1,x_{7}=0$ and $x_{1}=0,x_{7}=1$ ,

$x_{2}+x_{3}+...+x_{6}+x_{8}+...+x_{12}=2$

By Stars and Bars Theorem , there are $11\choose9$ non-negative integer solutions to the equation.

But since there are two possible solutions for $x_{1},x_{7}$ , we have $11\choose9$ $\cdot2$ solutions here.

When $x_{1}=1,x_{7}=1$ and $x_{1}=2,x_{7}=0$ ,

$x_{2}+x_{3}+...+x_{6}+x_{8}+...+x_{12}=1$

By Stars and Bars Theorem , there are $10\choose9$ non-negative integer solutions to the equation.

But since there are two possible solutions for $x_{1},x_{7}$ , we have $10\choose9$ $\cdot2$ solutions here.

When $x_{1}=2,x_{7}=1$ ,

$x_{2}+x_{3}+...+x_{6}+x_{8}+...+x_{12}=0$

There is a only $1$ non-negative integer solution here.

$\therefore$ , for one single color, there is a total of $12\choose9$ $+$ $11\choose9$ $\cdot2$ $+$ $10\choose9$ $\cdot2$ $+1$ $=351$

By rule of product , the number of ways of distribution $=351^6=(3^3\cdot13)^6=3^{18}\cdot13^6$

$\therefore p_{1}+n_{1}+p_{2}+n_{2}=3+18+13+6=40$

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$\textbf{Bonus:}$

$12\choose9$ $+$ $11\choose9$ $\cdot2$ $+$ $10\choose9$ $\cdot2$ $+1$

$=$ $12\choose9$ $+$ $11\choose9$ $+$ $10\choose9$ $+1$ $+$ $11\choose9$ $+$ $10\choose9$

$=$ $12\choose9$ $+$ $11\choose9$ $+$ $10\choose9$ $+$ $10\choose10$ $+$ $11\choose9$ $+$ $10\choose9$

$=$ $12\choose9$ $+$ $11\choose9$ $+$ $11\choose10$ $+$ $11\choose9$ $+$ $10\choose9$

$=$ $12\choose9$ $+$ $12\choose10$ $+$ $11\choose9$ $+$ $10\choose9$

$=$ $13\choose10$ $+$ $11\choose9$ $+$ $10\choose9$

$=$ $13\choose10$ $+$ $11\choose9$ $+$ $10\choose9$ $+1$ $-1$

$=$ $13\choose10$ $+$ $11\choose9$ $+$ $10\choose9$ $+$ $10\choose10$ $-1$

$=$ $13\choose10$ $+$ $11\choose9$ $+$ $11\choose10$ $-1$

$=$ $13\choose10$ $+$ $12\choose10$ $-1$

$=$ $13\choose3$ $+$ $12\choose2$ $-1$

$=$ $\cfrac{13\cdot12\cdot11}{3\cdot2\cdot1}$ $+$ $\cfrac{12\cdot11}{2\cdot1}$ $-1$

$=$ $286+66-1$

$=$ $351$

The identities I utilized in my computation above are $n \choose r$ $+$ $n+1 \choose r$ $=$ $n+1 \choose r+1$ and $n \choose r$ $=$ $n \choose n-r$

Detailed proofs can be read here .