There are eight coins in a row, all showing heads. In each move, we can flip over two adjacent coins provided that they are both showing heads or both showing tails. How many different patterns of heads and tails can we obtain after a number of moves?
My solution: We can replace each head with the number '1', and each tail with the number '0', we will then end up with a binary sequence with a starting decimal value of 255. Each move will add or subtract $3*2^n$, $(6 \geq n \geq 0)$, thus there will be $255 / 3 + 1 = 86 $ solutions.
There is already a solution on StackExchange here: How many different patterns of heads and tails can we obtain after a number of moves?
Although I understand the accepted solution for the above link, I still don't understand why my solution is wrong, can anyone help continue my solution or give me a hint?