In fact the answer is 2!!!!

You need a minimun of 2 weighs to find the bad coin IF YOU ARE LUCKY!

1- You take 2 coins, you weight them and you get an unbalanced situation.

2- You take one of them and you compare to an outside coin and you still find an unbalanced situstion ==> The bad coin is X and its 9g or 11g

So the **minimum ways** needed are 2.

First, I’d like to complement the clever redundant use of the factorial sign. (2!!!! = 2).

However, your answer is wrong. The original question was:

**“… how many minimum ways …”**. Whereas, you pointed out only one such way. Indeed, the

**minimum number of weighings **to determine counterfeit coin is 2, if you get lucky. But there are 27!/(25!*2!)=351 ways to compare 2 coins out of 27. And then there are 25*2 ways to compare one of those two to the 25 on the side. Or you could choose to drop 13 coins onto one side of the scale and 13 onto another, and get lucky finding that they weigh the same. And there are C(27,13)*C(14,13) ways, or 27!/(14!*13!)*14 of doing that. Then there are 26 ways to compare the counterfeit coin with one of the true ones. Simplified, the correct answer is: 27! / 24! + 27! * 26 / (13! * 13!). That seems like a very large number, which may not fit into the space allotted by this forum.

(Also, see my previous post for further discussion of the puzzle as it was meant.)