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 February 26th, 2013, 01:48 PM #1 Newbie   Joined: Feb 2013 Posts: 2 Thanks: 0 Permutation I am going through some permutation tutorials, one of them has an exercise asking in a 15-puzzle, if we assume the top right corner is 2, how many configurations satisfy this? I think the answer is 14! Am I right? My sincere thanks
 February 26th, 2013, 01:52 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,946 Thanks: 1137 Math Focus: Elementary mathematics and beyond Re: Permutation What about the empty space. Doesn't that make it 15!?
February 26th, 2013, 02:07 PM   #3
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Re: Permutation

Quote:
 Originally Posted by greg1313 What about the empty space. Doesn't that make it 15!?
I think you are correct

 February 27th, 2013, 07:15 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Permutation If you are allowed to move each square independently then there are 15! ways to place the other 15 squares. However, in a typical "15 puzzle", you cannot do that. You can only move into configurations that are even permutations of the original positions (which is, typically, $\begin{bmatrix}1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & \end{bmatrix}$).
February 27th, 2013, 12:11 PM   #5
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Re: Permutation

Quote:
 Originally Posted by HallsofIvy If you are allowed to move each square independently then there are 15! ways to place the other 15 squares. However, in a typical "15 puzzle", you cannot do that. You can only move into configurations that are even permutations of the original positions (which is, typically, $\begin{bmatrix}1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & \end{bmatrix}$).
The question is how many permutations, not how many can you get from a given start. So your note is not relevant for this question.

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