By Joyner W.D.

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N}. 2 Prove this using the multiplication counting principle from the section on counting in the previous chapter. Example 47 The matrix of the permutation f given by 1 2 3 2 1 3 f= is 0 1 0 P (f ) = 1 0 0 0 0 1 Note that matrix multiplication gives 0 1 0 1 2 1 0 0 2 = 1 0 0 1 3 3 from which we can recover the 2 × 3 array. Theorem 48 If f : T → T is a permutation then (a) P (f ) 1 2 .. n = f (1) f (2) .. f (n) Furthermore, the inverse of the matrix of the permutation is the matrix of the inverse of the permutation: (b) P (f )¡1 = P (f ¡1 ), and the matrix of the product is the product of the matrices: (c) P (f g) = P (f )P (g).

Bt } are disjoint. 42 CHAPTER 3. PERMUTATIONS Lemma 51 If f and g are disjoint cyclic permutations of T then f g = gf . proof: This is clear since the permutations f and g of T affect disjoint collections of integers, so the permutations may be performed in either order. ✷ Lemma 52 The cyclic permutation (a1 a2 ... ar ) has order r. , f r¡1 (a1 ) = ar , f r (a1 ) = a1 , by definition of f . , r, we have f r (ai ) = ai . ✷ Theorem 53 Every permutation f : T → T is the product of disjoint cyclic permutations.

S. , 15 - no working model could be supplied, so his patent was denied. 2. DEVIL’S CIRCLES (OR HUNGARIAN RINGS) 55 The possible moves are R = Ru,r,d,l = (r 16) = swap r and 16, L = Lu,r,d,l = (l 16) = swap l and 16, U = Uu,r,d,l = (u 16) = swap u and 16, D = Du,r,d,l = (d 16) = swap d and 16. 2 Verify that the five defining properties of a permutation puzzle are satisfied by this example. We shall call the 15 puzzle a planar puzzle since all its pieces lie on a flat board. 2 Devil’s circles (or Hungarian rings) This is a planar puzzle consisting of two or more interwoven ovals, each of which has several labeled (by colors or numbers) pieces, some of which may 56 CHAPTER 4.