Eight coins are arranged in a circle heads up. A move consists of flipping over two adjacent coins. How many different sequences of six moves leave the coins alternating heads up and tails up?
I need help finding where I went wrong; I know there is another way to do this, but I want to know where this is wrong! Here is my approach:
Label the 8 coins from 1 to 8. Either the even coins will be all tails or the odd coins will be all tails. We'll find the answer for the even coins being all tails, and then double our answer.
There are 2 cases:
One of the even coins is flipped 3x (turns to tails, heads, then back to tails), and the other 3 even coins is flipped 1x (turn to tails). There are 4 choices for which flip is done 3x. We can arrange the sequence in 6!/3! = 120 ways. So 120x4 = 480 sequences.
All even coins are flipped 1x (turn to tails), and one odd coin is flipped 2x (such that it remains a heads). There are 4 choices for which coin is flipped twice. We can arrange sequence in 6!/2! = 360 ways. So 360x4 = 1440.
Add up both cases: 1440 + 480 = 1920. Double our answer to count the ways that the odd coins end up tails, and we get 1920x2 = 3840.
The correct answer however, is 2 times this, or 7680. What am I missing?