Is there an alternative? Part 2

First, let's find the answer to the second question, as it will help us.

If there are 30 identical chocolates in a line, instead of picking them up, we can put them in some order: put the 6 chocolates you want to eat (written X ) and then put the 2 mandatory chocolates (written o ) in between each X .

_Xoo_Xoo_Xoo_Xoo_Xoo_X_

You are left with $30-16=14$ chocolates, that you have to put in 7 spaces (written _ ): between the X 's, at the far left, or at the far right. As you can put several times some chocolate in any of these spaces, that's a multicombination and there's $\Big(\!\!\Big(\begin{matrix}7\\14\end{matrix}\Big)\!\!\Big)={7+14-1 \choose 14}={20 \choose 14}=38'760$ ways, by stars and bars , to do that.

With similar steps, we have this generic formula for taking $p$ chocolates from a line of $n$ while keeping at least two chocolates in-between is: $\boxed{\Big(\!\!\Big(\begin{matrix}p+1\\n-p-2(p-1)\end{matrix}\Big)\!\!\Big)={n-2p+2 \choose p}}$

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Now, when the chocolates are in circles:

First, pick up on chocolate (30 ways to do that). You can pick up the remaining 5 chocolates from $30-1-2-2=25$ possible places, as you cannot take the first chocolate again, nor the $2\times2$ chocolates on each of its sides.

From the first part here, we now that there is ${25-2·5+2 \choose 5}={17 \choose 5}=6'188$ ways to pick 5 chocolates from 25 with at least 2 between them. We have then $30·6'188$ ways to pick 6 chocolates while deciding which one should be considered as the first one . But for all of the 6 choosen chocolates, each can be picked as first one , and therefore we have counted every possibility to pick up the chocolates 6 times. We should then divide our result by 6.

Answer is $\dfrac{30·6'188}{6}=\boxed{30'940}$ .

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Generic formula for $p$ chocolates picked from $n$ displayed in circle, while leaving at least $k$ chocolates in-between is $\frac{n}{p}{n-1-k·p \choose p-1}$

We generalize this for $n$ chocolates and atleast $p$ chocolates gap and choosing $r$ chocolates.

We firstly choose a chocolate ,

Now , call the number of chocolates between $1^{st} , 2^{nd}$ chosen chocolates as $a_1$ , similarly between $2^{nd} , 3^{rd}$ as $a_2$ , similarly define $a_i$ , where $i \in (1 , 2 , 3 \cdot \cdot \cdot r - 1)$ , $a_r$ is the number of chocolates between $r^{th} , 1^{st}$ chosen chocolates.

Now , we want that $a_1 + a_2 + a_3 + \cdot \cdot \cdot + a_r = n - r$ because there are $n - r$ chocolates remaining . With the condition that $a_i$ , where $i \in (1 , 2 , 3 \cdot \cdot \cdot r) \geq p$ because we desire $p$ chocolates between any two chosen chocolates.

By Stars And Bars Theorem , we have the number of solutions to be ${{n - r - rp + r - 1}\choose{r - 1}} = {{n - rp - 1}\choose{r - 1}}$ .

Now , for choosing the first chocolate , we have $n$ ways. But for any set of $r$ chocolates , call $b_1 , b_2 , b_3 , \cdot \cdot \cdot b_r$ , we are counting it $r$ times , as we can suppose any of the $r$ chocolates to be the first one.

Therefore , our final answer is $\boxed{{{n - rp - 1}\choose{r - 1}}\times{\frac{n}{r}}}$

Plugging in the values $n = 30 , r = 6 , p = 2$ , we get $\boxed{30940}$ as our answer.