HMMT 二月 2005 · COMB 赛 · 第 9 题

英文解析

9. Eight coins are arranged in a circle heads up. A move consists of flipping over twoadjacent coins. How many different sequences of six moves leave the coins alternatingheads up and tails up?
   Solution: 7680
   Imagine we flip over two adjacent coins by pushing a button halfway between them.
   Then the outcome depends only on the parities of the number of times that each buttonis pushed. To flip any coin, we must push the two buttons adjacent to that coin a totalof an odd number of times. To flip every other coin, the parities must then progressaround the circle as even, even, odd, odd, even, even, odd, odd. There are 4 ways toassign these parities. If we assume each button is pressed either once or not at all,
   this accounts for only four presses, so some button is also pressed twice more. Supposethis button was already pushed once. There are 4 of these, and the number of possiblesequences of presses is then 6! / 3! = 120. Suppose it has not already been pressed.
   There are 4 of these as well, and the number of possible sequences is 6! / 2! = 360. Thetotal number of sequences is then 4(4 · 120 + 4 · 360) = 7680.
   2004 2004