I am getting the answer as $4676$, someone please check what you are getting. Here's what I did, I don't know if it's correct. Let $a_1,a_2,a_3,a_4,a_5,a_6$ denote the number of sectors of each colour. Then, $a_1+a_2+a_3+a_4+a_5+a_6=12$ By PHP, or observation/logic rather, each $a_i\leq 6$, else there will be 2 adjacent sectors of same colour. Now, it's just the coefficient of $x^{12}$ in $\frac{(1-x^7)^6}{(1-x)^6}$

I tried to use recurrence but messed up.... Got nothing...

Umm, anyone else? Even those who didn't appear for it, please try the question. I think there will be a generalised theorem

Try this way: Let the number of colouring methods be An,when there are 2n+2 equal sectors. Just consider the bottom sector and the two sectors beside it. If the colour of the two are the same,then there are 5 colouring methords for the bottom one.Mark it as Sn(Same) If different,then 4 methods.Mark it as Dn.(Different) The total number of methods is 4Dn+5Sn.Mark it as An. Imagine that you know Sn Dn already. Then easy to konw Sn+1=5Sn+4Dn Dn+1=20Sn+21Dn

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And then An=25^(n+1)+5(=Sn+1) So the answer is 25^6+5=2444140630 (Have no idea about whether it is correct.)

Math-wiz wrote: I am getting the answer as $4676$, someone please check what you are getting. Here's what I did, I don't know if it's correct. Let $a_1,a_2,a_3,a_4,a_5,a_6$ denote the number of sectors of each colour. Then, $a_1+a_2+a_3+a_4+a_5+a_6=12$ By PHP, or observation/logic rather, each $a_i\leq 6$, else there will be 2 adjacent sectors of same colour. Now, it's just the coefficient of $x^{12}$ in $\frac{(1-x^7)^6}{(1-x)^6}$ I also did it exactly the same way but got it wrong

Karsa wrote: Try this way: Let the number of colouring methods be An,when there are 2n+2 equal sectors. Just consider the bottom sector and the two sectors beside it. If the colour of the two are the same,then there are 5 colouring methords for the bottom one.Mark it as Sn(Same) If different,then 4 methods.Mark it as Dn.(Different) The total number of methods is 4Dn+5Sn.Mark it as An. Imagine that you know Sn Dn already. Then easy to konw Sn+1=5Sn+4Dn Dn+1=20Sn+21Dn

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And then An=25^(n+1)+5(=Sn+1) So the answer is 25^6+5=2444140630 (Have no idea about whether it is correct.) Opppps,I forgot to

There are $6^{12}$ total colorings of the disc without restriction. We try to count the number of colorings that at least two adjacent sectors are the same color. Between every two consecutive sectors it exists one border and now we count number of colorings that at least one border , i.e. two adjacent sectors, is the same color. Let $A_{1},A_{2},...,A_{12}$ be the sets of colorings of the disc where the borders are the same color. Therefore, by the PIE we get \begin{align*} 6^{12}-\left |A_{1}\cup A_{2}\cup ...\cup A_{12} \right | &=6^{12}-\left [ \binom{12}{1}6^{11}-\binom{12}{2}6^{10}+\binom{12}{3}6^{9}-\binom{12}{4}6^{8}+\binom{12}{5}6^{7} -\binom{12}{6}6^{6}+\binom{12}{7}6^{5}-\binom{12}{8}6^{4}+\binom{12}{9}6^{3}-\binom{12}{10}6^{2}+\binom{12}{11}6^{1}-6 \right ]\\&=\left ( 6-1 \right )^{12}-1+6=5^{12}+5\\&=\boxed {244140630} \end{align*}============ If a coloring that can be obtained from another by rotation is considered identical then we must apply Burnside's lemma to this particular problem we get the formula: $g(n,k)=\frac{1}{n}\sum_{d|n}\phi (d)((k-1)^{n/d}(-1)^{n/d}(k-1))$ where $d\in \mathbb{N} $ sectors in each cycle and $\phi (d)$ is the Euler Totient Function. With $n=12, k=6$: \begin{align*} g(12,6)&=\frac{1}{12}(\phi (1)(5^{12}+5)+\phi (2)(5^{6}+5)+\phi (3)(5^{4}+5)+\phi (4)(5^{3}-5)+\phi (6)(5^{2}+5)+\phi (12)(5^{1}-5))\\&=\frac{1}{12}((5^{12}+5)+(5^{6}+5)+2(5^{4}+5)+2(5^{3}-5)+2(5^{2}+5))\\&=\frac{1}{12}\left ( 244140630+15630+1260+240+60 \right )\\&=\boxed {20346485} \end{align*}

Taiharward wrote: There are $6^{12}$ total colorings of the disc without restriction. We try to count the number of colorings that at least two adjacent sectors are the same color. Between every two consecutive sectors it exists one border and now we count number of colorings that at least one border , i.e. two adjacent sectors, is the same color. Let $A_{1},A_{2},...,A_{12}$ be the sets of colorings of the disc where the borders are the same color. Therefore, by the PIE we get \begin{align*} 6^{12}-\left |A_{1}\cup A_{2}\cup ...\cup A_{12} \right | &=6^{12}-\left [ \binom{12}{1}6^{11}-\binom{12}{2}6^{10}+\binom{12}{3}6^{9}-\binom{12}{4}6^{8}+\binom{12}{5}6^{7} -\binom{12}{6}6^{6}+\binom{12}{7}6^{5}-\binom{12}{8}6^{4}+\binom{12}{9}6^{3}-\binom{12}{10}6^{2}+\binom{12}{11}6^{1}-6 \right ]\\&=\left ( 6-1 \right )^{12}-1+6=5^{12}+5\\&=\boxed {244140630} \end{align*} I'm sorry but you didn't turn the disk.It is a circle,and if you can turn one colouring method into another,then they are the same. the same mistake as the one I made.

Math-wiz wrote: I am getting the answer as $4676$, someone please check what you are getting. Here's what I did, I don't know if it's correct. Let $a_1,a_2,a_3,a_4,a_5,a_6$ denote the number of sectors of each colour. Then, $a_1+a_2+a_3+a_4+a_5+a_6=12$ By PHP, or observation/logic rather, each $a_i\leq 6$, else there will be 2 adjacent sectors of same colour. Now, it's just the coefficient of $x^{12}$ in $\frac{(1-x^7)^6}{(1-x)^6}$ Won't the answer be $6*5^{10}*4$

@Karsa It depends on whether the disc has labelled sectors or unlabelled .The above answer is correct for the first situation. If a coloring that can be obtained from another by rotation is considered identical then we must apply Burnside’s Lemma, the formula giving rise to the cycle index of a combinatorial necklace.

@Taiharward As we have solved the first situation,we can take time to think about the second one.

OK. D'accord, Monsieur.

I am getting $6*5^{10}*9$ I firstly coloured a sector in 6 ways.We'll not consider the choosing of 1 out of 12 sectors as all sectors are identical. We colour the next 10 sectors one by one by choosing colors from remaining 5 colors ( as adjacent sector has a color ), which can be done in $5^{10}$ ways.The last sector to be colored lies $bet^n$ 2 colored sectors. Case 1: Both sectors are of same color. Then the last sector to be colored must be colored in 5 ways. Case 2:The 2 sectors are of different color. Then can be colored in 4 ways. In total #ways=$6*5^{10}*(5+4)$