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Unformatted text preview: MATH 681 Exam #2 1. (10 of 12 students attempted this) A necklace consists of 6 gems; the gems can be garnets, tourmaline, or zircons. Two necklaces are considered to be identical if one can be obtained by rotating or flipping the other. (a) (5 points) Find a pattern inventory for all such necklaces. You need not alge braically expand the pattern inventory. Let us consider the pattern inventory of invariants under each permutation in D 6 ; we may then use P´olya’s Theorem to get the pattern inventory for the equivalnece classes. Every sequence of six gems is invariant under the identity permutation e ; this will yield pattern inventory ( g + t + z ) 6 . Under the oneelement rotation r (or its inverse r 5 ), all gems are in one big cycle, so to be invariant they must all be the same type: our pattern inventory for such invariants is ( g 6 + t 6 + z 6 ) (which we shall double in our final accounting, since it is the invariant inventory for two different group elements). The rotation r 2 (and similarly r 4 cycles elements in collections of three, yielding pattern inventory ( g 3 + t 3 + z 3 ) 2 . The rotation r 3 swaps opposite pairs, so its invariants have pattern inventory ( g 2 + t 2 + z 2 ) 3 . Likewise, any of the three flips which move every gem devolve into 3 swaps, with invariant pattern inventory ( g 2 + t 2 + z 2 ) 3 . The three flips which use an axis passing through two gems, on the other hand, have two fixed points and two swaps, leading to pattern inventory ( g + t + z ) 2 ( g 2 + t 2 + z 2 ) 2 . Assembling all these in the style of P´ olya’s theorem, we get: ( g + t + z ) 6 + 2( g 6 + t 6 + z 6 ) + 2( g 3 + t 3 + z 3 ) 2 + 4( g 2 + t 2 + z 2 ) 3 + 3( g + t + z ) 2 ( g 2 + t 2 + z 2 ) 12 (b) (5 points) Either using your pattern inventory or by other means, determine the number of necklaces which use two of each gemstone. We are looking for coefficients of the form g 2 t 2 z 2 . Considering each of the terms in the pattern inventory discovered above in turn, we find that ( g + t + z ) 6 has the term ( 6 2 , 2 , 2 ) g 2 t 2 z 2 , by the multinomial theorem. Neither ( g 6 + t 6 + z 6 ) nor ( g 3 + t 3 + z 3 ) 2 have a g 2 t 2 z 2 term. ( g 2 + t 2 + z 2 ) 3 has the term ( 3 1 , 1 , 1 ) g 2 t 2 z 2 , by the multinomial theorem. Lastly, ( g + t + z ) 2 ( g 2 + t 2 + z 2 ) 2 may also be identified to have the term ( 3 1 , 1 , 1 ) g 2 t 2 z 2 : after selecting the two g 2 + t 2 + z 2 factors to each contribute one of g 2 , t 2 , or z 2 , there is a unique factor which can be contributed in a unique way by the remaining terms. Thus, the coefficient ofin a unique way by the remaining terms....
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.
 Fall '09
 WILDSTROM
 Math

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