How many different circular necklaces containing ten beads can be made using beads of at most two colors?
I know I need calculate situation with 2 colors, 8 second color, 3 first color, 7 second color, 4 first color, 6 second color and 5 first color, 5 second color.
But I have problems with my calculations:
$$\frac{1}{20}\left(\frac{10!}{8!2!} + 3\times 5 + 1\times 5 + 3\times 15\right) = \frac{110}{20}. $$
Please show to me how to solve this problem correctly.
The answer to this problem is the Necklace polynomial. For a necklace of length $n$ utilising $a$ different colours, where rotating the necklace so that the "fist" bead changes is counted as the same necklace, but flipping the necklace so that the order of the beads is reversed is counted as a different necklace, the total number of necklaces is given by $$\frac{1}{n}\sum_{d|n}\phi\left(\frac{n}{d}\right)a^d$$ The notation $d|n$ means "$d$ divides $n$", and the sum is taken for all values of $d$ between $1$ and $n$ inclusive for which $d$ is a factor of $n$. $\phi$ is Euler's totient function.
In this case, $n=10$ and $a=2$. The divisors of $10$ are $1,2,5,10$, so $d$ takes each of these values in the sum. The answer is $$\frac{1}{10}\sum_{d\in\{1,2,5,10\}}\phi\left(\frac{10}{d}\right)2^d=\frac{1}{10}\left[2^1\cdot\phi(10)+2^2\cdot\phi(5)+2^5\cdot\phi(2)+2^{10}\cdot\phi(1)\right]=\frac{1}{10}\left[2\cdot4+4\cdot4+32\cdot1+1024\cdot1\right]=108$$
Using Polya's Theorem : Zd10=(1/|D10|)(p1^10+4p5^2+4p10+6p2^5+5p1^2p2^4)=(1/20)(20 a^10 + 20 a^9 * b + 100 a^8 * b^2 + 160 a^7 * b^3 + 320 a^6b^4 + 320 a^5*b^5 + 320 a^4 * b^6 + 160 a^3 * b^7 + 100 a^2 * b^8 + 20 a * b^9 + 20 * b^10)=1560/20=78 , where p1=a+b , p2=a^2+b^2 ......
