Polya's Counting TheoryDate: 09/12/2003 at 15:55:29 From: Frederic Subject: 6-sided dice Playing with dice with my daughter, I wondered how many different 6-sided dice can be made using the numbers 1, 2, 3, 4, 5 and 6. All of them have to be used. Each of them can be used only once. In other words, each die has "1" on one face, "2" on another face, "3" on another face and so on. Also, I consider that the orientation of the written numbers doesn't matter. In other words "4" and "4 upside down" are considered to be the same. For the first face I have 6 possibilities, for the second one I have 5 choices etc ... for the sixth face I have one choice. Then I consider cyclic permutation: a die with 1, 2, 3, 4 going around and 5 and 6 on side faces is the same as a die with 2, 3, 4, 1 and 5 and 6 on side faces etc .... So I come up with N= 180 different dice (N= 6x5x4x3x2 / 4). But I'm not sure dividing by 4 is enough. Date: 09/12/2003 at 19:35:01 From: Doctor Tom Subject: Re: 6-sided dice Hello Frederic, I assume that two proposed numberings of the faces of a die are to be considered equivalent if one of them could be rotated in space to exactly align with the other. In other words, to check if two numberings were the same, first you'd set the dice with the same number up, and then you'd verify that the number opposite it was the same. If not, they're different. If the top and bottom are the same, then leaving the top on top, you'd pick any number around the middle and line it up with the corresponding number on the other. If, by doing that, all the numbers matched up, then they'd be the same. If not, then you've got two distinct numberings. So my interpretation of your question is that given this method to determine if two numberings are the same, then how many different numberings are there? If it's right, here's how to work it: You know that both dice have the number 1, so put that face up in both cases, and in reality, you only need to find different numberings of the other faces. You can clearly number the bottom in 5 different ways, and once you've selected the bottom number, you've got four numbers to rearrange around the middle. But since that die can be rotated, leaving the top on top, to four different positions, you'll have to divide the total number of those rearrangements by 4. There are 4x3x2x1 = 24 rearrangements, but when you divide by 4, you obtain 6. That, times the 5 possible numbers opposite the number 1, make 30. (Of course there's nothing special about choosing 1 to lie on top, but whatever number you selected to begin your comparison would leave the same number of slots to fill in and you'd again obtain the number 30.) You have actually stumbled across a VERY interesting area of combinatorics that are solved using a method known as "Polya's counting theory". It relates the number of arrangements to the number of operations on the whole object that you consider to be "the same". There are plenty of practical applications of this. For example, in chemistry, suppose you've got a benzene ring where two of the 6 hydrogens are replaced by a chlorine and a bromine atom (chloro-bromo-benzene). How many different molecules are there, given that the molecules can freely rotate in space? This one is as easy as your dice problem, but with not much more complexity in the problem, the analysis can become seeming completely intractable. I have presented talks to bright high-school kids about this counting theory and have prepared a handout for them that tries to give the basic ideas, and hopefully gets them excited about some areas of math that they've never seen. It's available on the web. Go to http://www.geometer.org/mathcircles and download the paper "polya.pdf" (Acrobat format) or "polya.ps" (PostScript format). - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
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