Double Derangement Permutations

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1 Ope Joural of iscrete Matheatics, 206, 6, Published Olie April 206 i SciRes ouble erageet Perutatios Pooya aeshad, Kayar Mirzavaziri 2, Madjid Mirzavaziri 3 Ferdowsi Uiversity of Mashhad, Iteratioal Capus, Mashhad, Ira 2 Natioal Orgaizatio for evelopet of Exceptioal Talets (NOET I, Mashhad, Ira 3 epartet of Pure Matheatics, Ferdowsi Uiversity of Mashhad, Mashhad, Ira Received 28 Septeber 205; accepted 9 April 206; published 2 April 206 Copyright 206 by authors ad Scietific Research Publishig Ic This work is licesed uder the Creative Coos Attributio Iteratioal Licese (CC BY Abstract Let be a positive iteger A perutatio a of the syetric group S of perutatios of [ ] = {, 2,, } is called a derageet if a( i for each i [ ] Suppose that x ad y are two arbitrary perutatios of S We say that a perutatio a is a double derageet with respect to x ad y if a( x( ad a( y( for each i [ ] I this paper, we give a explicit for- x, y, the uber of double derageets with respect to x ad y Let 0 k ad ula for ( let { i,, i k } ad { a, a }, k be two subsets of [ ] with ij a ad = j { i,, ik} { a,, ak} Suppose that ( k,, deotes the uber of derageets x such that x( ij aj result, we show that if 0 ad z is a perutatio such that z( i for i z( = i for i k >, the ( ez = ( ( k ( i ik ( i,, ik = { i,, ik} { z( i,, z( ik } Keywords,,,,,, where k= 0 i < < ik Syetric Group of Perutatios, erageet, ouble erageet = As the ai ad Itroductio Let be a positive iteger A derageet is a perutatio of the syetric group S of perutatios of [ ] = {, 2,, } such that oe of the eleets appear i their origial positio The uber of derageets of S is deoted by or A siple recursive arguet shows that How to cite this paper: aeshad, P, Mirzavaziri, K ad Mirzavaziri, M (206 ouble erageet Perutatios Ope Joural of iscrete Matheatics, 6,

2 P aeshad et al ( ( = + 2 The uber of derageets also satisfies the relatio ( i ( is explicitly deteried by! = + It ca be proved by the iclusio- exclusio priciple that This iplies that li 0 i!! = e These facts ad soe other results cocerig derageets ca be foud i [] There are also soe geeralizatios of this otio The problèe des recotres asks how ay perutatios of the set [ ] have exactly k fixed poits The uber of such perutatios is deoted by ad is give by k, k, = k k k, Thus, we ca say that li! = ke! Soe probabilistic aspects of this cocept ad the related otios cocerig the perutatios of S is discussed i [2] ad [3] Let e be the idetity eleet of the syetric group S, which is defied by ei ( = i for each i [ ] We ca say that a perutatio a of [ ] is a derageet if ai ( ei ( for each i [ ] We deote this by a e Thus, is the uber of a with a e If c is ay fixed eleet of S the the uber of a S with a x is also, sice a x if ad oly if ax e I the preset paper, we exted the cocept of a derageet to a double derageet with respect to two fixed eleets x ad y of S 2 The Result I the followig, we assue that is a positive iteger ad the idetity perutatio of the syetric group S of perutatios of [ ] is deoted by e Moreover, for two perutatios a ad b of S, the otatio a b eas that a( b( for each i [ ] We also deote the uber of eleets of a set A by A efiitio Suppose that x ad y are two arbitrary perutatios of S We say that a perutatio a is a double derageet with respect to x ad y if a x ad a y The uber of double derageets with respect to x ad y is deoted by ( xy, Propositio Let 0 k ad let { i,, } ad { a,, be two subsets of [ ] with ij a ad j = { i,, ik} { a,, ak} The ( k,,, the uber of derageets x such that x( ij, is deteried by ( k,, Proof Let ar { i,, ik} { a,, ak} a Case a {,, s i } Let as it oly if the derageet x of the set [ ] { } k k ( + ( k+ if k ad 2k 0 i ( k + = k if k = 0 otherwise Thus r = is for soe s r Now there are two cases: = I this case a derageet x satisfies the coditio x( ij if ad \ i t satisfies the coditio x ( ij = a j for all j t, where a j = a for j s ad a j s = at This provides a oe to oe correspodece betwee the derageets x of [ ] with x( ij for the give sets { i,, } ad { a,, with eleets i their itersectios, ad the derageets x of [ ] \ { i t} with xi = a j j for the give sets { i,, i } \ k { it} ad { a,, a } \ k { a t} with eleets i their itersectios Case 2 a {,, s i } I this case a derageet x satisfies the coditio x( ij if ad oly if the derageet x of the set [ ] \ { a s} satisfies the coditio x ( ij for all j s This provides a oe to oe correspodece betwee the derageets x of [ ] with x( ij for the give sets { i,, } ad { a,, with eleets i their itersectios, ad the derageets x of [ ] \ { a s} with x ( ij for the give sets { i,, i } \ k { is} ad { a,, a } \ k { as} with eleets i their itersectios k,, =, k, Iteratig this arguet, we have These cosideratios show that ( ( ( k = ( k = ( k = = ( k,,,, 2, 2, 2,, 0 We ca therefore assue that = 0,0,0 = For k, we clai that have ( We thus evaluate ( k,,0, where 2k For k = 0, we obviously 00

3 P aeshad et al For a derageet x satisfyig x( ij ( k ( k ( k,, 0 =,, 0 + 2,, 0 there are two cases: x( a = i or x( a i i a for with 0 eleets i their itersectios The uber is equal to If the first case occurs the we have to evaluate the uber of derageets of the set [ ] \{, } the give sets { i,, 2 } ad { a,, 2 ak} ( 2, k,0 If the secod case occurs the we have to evaluate the uber of derageets of the set [ ] \ { } give sets { i,, 2 } ad { a,, 2 ak} (, k,0 We ow use iductio o k to show that For k = we have a for the with 0 eleets i their itersectios The uber is equal to k k ( + ( k+ ( k,,0 =, 2 2 k 0 i ( k+ (,,0 = (,0,0 + ( 2,0,0 = + 2 = Now let the result be true for k We ca write ( k,, 0 (, k, 0 ( 2, k, 0 = + k 2 k 2 k 2 ( 2 k + i k ( ( k + = + 0 i ( ( k + 0 i ( 2 ( k + k 2 k 2 k ( ( k k ( k+ i = + 0 i ( k+ i ( ( k+ i k 2 k 2 = + k 2 ( 2 2 ( ( + k k k + k+ i ( + ( 2k = k i i ( k + ( 2k k 2 ( k k 2 ( ( k ( + ( 2k = + + k i ( k + ( 2k = k 2 k 2 ( + k ( + ( k+ ( + ( 2k ( + ( k+ + + k i ( k+ ( 2k i ( k+ k 0 ( + ( k + i k i ( k+ Corollary Let k be a positive iteger The k 0 k i k i k+ i = k! Proof Let = 2k, ij = j ad aj = k + j for j =,, k The a derageet x satisfies the coditio x( ij if ad oly if x defied by x ( = x( k + for i [ k] is a perutatio of [ k ] The uber of such perutatios x is k! The followig Table gives soe sall values of ( k,,0 The followig lea ca be easily proved Lea Let x ad y be two arbitrary perutatios ad 0 be the uber of i s for which x( y( The there is a perutatio z such that z( i for i ad z( = i for i > ad ( xy, = ( ez, Theore 2 Let 0 ad let z be a perutatio such that z( i for i ad z( = i for i > The k ( = ( ( ( k ez, k,, i,, i, k= 0 i < < ik 0

4 P aeshad et al Table Values of ( k,,0 for 0 ad 2k \k , ,329 8, where ( i,, ik = { i,, ik} { z( i,, z( ik } Proof Let E i be the set of all derageets x for which x( z( ( ez, E i =, where i The = We use the iclusio-exclusio priciple to deterie E i For each 0 k ad i < < ik we have E E = k,, i,, i, i ( ( ik k where ( i,, ik = { i,, ik} { z( i,, z( ik } This iplies the result Our ultiate goal is to fid a explicit forula for evaluatig (, we eed to state two eleetary euerative probles cocerig subsets A of the set [ ] ec for a arbitrary cycle c Prior to that with k eleets ad exactly cosecutive ebers Lea 2 Let S( k,, be the uber of subsets A= { a,, of [ ] for which the equatio r = s+ has exactly solutios for r ad s i A If 0 < k the k + k S( k,, = k Moreover, S(,0,0 = ad S( k,, = 0 for other values of k,, Proof We ca restate the proble as follows: We wat to put k oes ad k zeros i a row i such a way that there are exactly appearace of oe-oe To do this we put k zeros ad choose k places of k + the k + possible places for puttig k blocks of oes i ways Let the uber of oes i k the i-th block be ri We the ust have r + + rk = k The uber of solutios for the latter equatio is k Now suppose that we write, 2,, aroud a circle We thus assue that is after ad so, is also assued to be cosecutive Uder this assuptio we have the followig result Lea 3 Let C( k,, be the uber of subsets A= { a,, of [ ] for which the equatio r s+ (od has exactly solutios for r ad s i A If 0 < k < the k k C( k,, = k k C,0,0 C,, C k,, = 0 for other values of k,, Moreover, ( = ( = ad ( 02

5 P aeshad et al Proof Siilar to the above arguet, we wat to put k oes ad k zeros aroud a circle i such a way that there are exactly appearaces of oe-oe At first, we put the i a row There are four cases: Case There is o block of oes before the first zero ad after the last zero I this case we put k zeros k ad choose k places of the k possible places for puttig k blocks of oes i k ways Let the uber of oes i the i-th block be ri We the ust have r + + r k = k The uber of k solutios for the latter equatio is Case 2 There is o block of oes before the first zero but there is a block after the last zero I this case we put k zeros ad choose k places of the k possible places for puttig k blocks of k oes i ways Let the uber of oes i the i-th block be ri We the ust have k k r + + rk = k The uber of solutios for the latter equatio is Case 3 There is a block of oes before the first zero but there is o block after the last zero This is siilar to the above case Case 4 There is a block of oes before the first zero ad a block of oes after the last zero I this case we ust have appearace of oe-oe i the row forat, sice we wat to achieve appearace of oe-oe i the circular forat Thus we put k zeros ad choose k ( 2 places of the k possible k places for puttig k ( 2 blocks of oes i ways Let the uber of oes i the i-th block k k be ri We the ust have r + + rk ( = k The uber of solutios for the latter equatio is These cosideratios prove that k k k k k k C( k,, = k k k A straightforward coputatio gives the result The followig Table 2 gives soe sall values of ( 0,, Table 2 Values of ( 0,, C k for k 0 ad k C k k\

6 P aeshad et al Theore 3 Let c be be a cycle of legth The k ( = ( ( ( ec, C k,, k,, 0 2k Proof Let c be the cycle defied by c ( j = j+ for j, c ( = ad c ( i + i The ( ec, = ( ec, Usig the otatios of Theore 2, ( i,, i k = if ad oly if the subset A= { i,, } of [ ] exactly solutios for the equatio r s+ (od for rs, i A Thus the uber of { i,, ik } property ( i,, i k = is C( k,, Applyig Theore 2, we have the result Exaple We evaluate 5( ec, 5 ad 5( ec, 3 Applyig Theore 3 with = 5 we have 5( ec, 5 = C( 5,0,0 ( 5,0,0 C( 5,,0 ( 5,,0 + C( 5,2,0 ( 5,2,0 + C( 5, 2, ( 5, 2, C( 5,3, ( 5,3, C( 5,3, 2 ( 5,3, 2 + C( 5, 4,3 ( 5, 4,3 C( 5,5,5 ( 5,5,5 = C( 5,0,0 ( 5,0,0 C( 5,,0 ( 5,,0 + C( 5,2,0 ( 5,2,0 + C( 5, 2, ( 4,, 0 C( 5,3, ( 4, 2, 0 C( 5,3, 2 ( 3,, 0 + C ( 5, 4,3 ( 2,, 0 C ( 5,5,5 ( 0, 0, 0 = = 3, ad ( (, ( 2, ( 3, ( 4, ( 5 ( 5, 4,, 2,3,( 5, 4,,3, 2,( 5, 4, 2,,3 x x x x x for the 3 double derageets x with respect to e ad c 5 are Applyig Theore 3 with = 3 we have 3,,5, 2, 4, 3, 4,5,, 2, 3,5,, 2, 4, 3,5, 2,, 4, 4,,5, 2,3, 4,,5,3, 2, 4,5,, 2,3, 4,5,,3, 2, 4,5, 2,,3, 5,, 2,3, 4, 5 3 ( ( ( ( C( 3, 2, ( 5, 2, C( 3, 3, 3 ( 5, 3, 3 ( ec, = C 3,0,0 5,0,0 C 3,,0 5,,0 + = = 9, ad ( (, ( 2, ( 3, ( 4, ( 5 ( 3,,4,5,2,( 5,,2,3,4,( 4,,2,5,3,( 3,,2,5,4 x x x x x for the 9 double derageets with respect to e ad c 3 are 3,4,5,,2, 3,5,4,,2, 3,4,5,2,, 3,5,4,2,, 4,5, 2,,3, 5, 4, 2,,3, 4,5, 2,3,, 5, 4, 2,3,, 4,,5, 2,3, 5,, 4, 2,3, 4,,5,3,2, 5,,4,3,2, 3,5,2,,4, 3,4,2,5,, 3,,5,2,4, = for has with the The above exaple shows that how ca we evaluate (, forula for evaluatig ( ez, for ay perutatio z Applyig Lea, we ca copute (, ay perutatios x ad y Refereces ec for a cycle c Moreover, Theore 2 gives a xy for [] Graha, RL, Kuth, E ad Patashik, O (988 Cocrete Matheatics Addiso-Wesley, Readig [2] Pita, J (997 Soe Probabilistic Aspects of Set Partitios Aerica Matheatical Mothly, 04, [3] Kopfacher, A, Masour, T ad Wager, S (200 Records i Set Partitios The Electroic Joural of Cobiatorics, 7, R09 04