Nine light bulbs are equally spaced around a circle. When the power is turned on, each of the nine light bulbs turns blue or red, where the color of each bulb is determined randomly and independently of the colors of the other bulbs. Each time the power is turned on, the probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $\dfrac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n$. (PurpleComet 2023, Middle School, Problem 20) Click here for the solutions.
Solution 1: Overall, there are total $2^9=512$ combinations when the light bulbs are turned blue or red randomly. The problem can be divided into following cases, depending on the number of light bulbs turning blue, for satisfying the requirement that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle.
Case 1: The number of the blue bulbs is $0$. There is only $1$ way to arrange the bulbs to satisfy the requirement:
Case 2: The number of the blue bulbs is $1$. There is $0$ way to arrange the bulbs to satisfy the requirement, as the color of the two adjacent bulb of the blue bulb is red.
Case 3: The number of the blue bulbs is $2$. These two blue bulbs must be next to each other, and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 4: The number of the blue bulbs is $3$. These three blue bulbs must be next to each other, and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 5: The number of the blue bulbs is $4$. This can be divided into the following sub-cases:
Case 5.1: If the $4$ blue bulbs are arranged to be next to each other and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 5.2: If there are only $3$ blue bulbs are arranged to be next to each other, There is $0$ way to arrange the bulbs to satisfy the requirement, as the color of the two adjacent bulb of the remaining blue bulb is red.
Case 5.3: If the $4$ blue bulbs are arranged to $2$ groups, and each group contains $2$ blue blubs that are arrange to be next to each other. There must be $2$ or $3$ red blubs in-between these $2$ groups. There are $9$ ways to arrange these $5$ red bulbs in order to satisfy the requirement:
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Case 6: If there are $5$, $6$, $7$, $8$, and $9$ blue bulbs, there are $4$, $3$, $2$, $1$, and $0$ red bulbs, which would be same for case 1, 2, 3, 4, and 5.
Therefore, the number of arrangement of $9$ light bulbs satisfying the requirement is $$2\cdot(1+9+9+9+9)=74$$
The probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $$\dfrac{74}{512}=\dfrac{37}{256}$$ Therefore $m=37$, $n=256$, and $$m+n=37+256=\boxed{293}$$
Solution 2: Let $n$ be the number of light bulbs on the circle. There will be $2^n$ combinations when the light bulbs are turned blue or red randomly. We try to examine combinations $f(n)$ satisfying the requirement that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle.
If a red blub is assigned as bit $0$, and a blue blub as bit $1$. There will be $2^n$ $n$-bit binary numbers. In order to satisfy the requirement, the binary number meets all the following conditions:
1. It cannot contain a sequence of $101$, nor $010$
2. It cannot start with $01$ and end up with a $1$
3. It cannot start with $10$ and end up with a $0$
4. It cannot starts with a $1$ and end up with $10$
5. It cannot starts with a $0$ and end up with $01$
$$ \begin{array}{|c|c|c|c|} \hline n & 2^n & f(n) & f(n-1)+f(n-2) & g(n)=f(n)-(f(n-1)+f(n-2)) \\ \hline 1 & 2 & 2 & & \\ \hline 2 & 4 & 2 & & \\ \hline 3 & 8 & 2 & 4 & -2 \\ \hline 4 & 16 & 6 & 4 & 2 \\ \hline 5 & 32 & 12 & 8 & 4 \\ \hline 6 & 64 & 20 & 18 & 2 \\ \hline 7 & 128 & 30 & 32 & -2 \\ \hline 8 & 256 & 46 & 50 & -4 \\ \hline 9 & 512 & 74 & 76 & -2 \\ \hline 10 & 1024 & 122 & 120 & 2 \\ \hline 11 & 2048 & 200 & 196 & 4 \\ \hline 12 & 4096 & 324 & 322 & 2 \\ \hline 13 & 8192 & 522 & 524 & -2 \\ \hline 14 & 16384 & 842 & 846 & -4 \\ \hline 15 & 32768 & 1362 & 1364 & -2 \\ \hline 16 & 65536 & 2206 & 2204 & 2 \\ \hline 17 & 131072 & 3572 & 3568 & 4 \\ \hline 18 & 262144 & 5780 & 5778 & 2 \\ \hline 19 & 524288 & 9350 & 9352 & -2 \\ \hline 20 & 1048576 & 15126 & 15130 & -4 \\ \hline \end{array} $$Therefore
$$ f(n)=\left\{ \begin{array}{crl@{\qquad}l} 2 & n=1 & \\ 2 & n=2 & \\ f(n-1)+f(n-2)+g(n) & n\ge 3 \\ \end{array} \right. \ \ \ \ \ \ g(n)=\left\{ \begin{array}{crl@{\qquad}l} -2 & n=1 & \\ -4 & n=2 & \\ g(n-1)-g(n-2) & n\ge 3 \\ \end{array} \right. $$ $$ Or \ \ \ g(n)=\left\{ \begin{array}{rcl@{\qquad}l} -2 & (n-1)\pmod{6}\equiv 0 & \\ -4 & (n-1)\pmod{6}\equiv 1 & \\ -2 & (n-1)\pmod{6}\equiv 2 & \\ 2 & (n-1)\pmod{6}\equiv 3 & \\ 4 & (n-1)\pmod{6}\equiv 4 & \\ 2 & (n-1)\pmod{6}\equiv 5 & \\ \end{array} \right. $$The probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $$p(n)=\dfrac{f(n)}{2^n}$$