{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers


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1 . Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets, we quicly oted that a oto fuctio must map two elemets of the domai oto a sigle elemet of the rage. The rest of the mappig was a  fuctio of the remaig three domai elemets to the remaiig three rage elemets. Thus, there are &! = 0 differet oto fuctios from a domai of size five to a rage of size four. For a domai of six elemets ad a rage of four elemets, we ote that two cases are possible whe costructig a oto fuctio. Either three elemets of the domai map to a sigle elemet of the rage or two pairs of elemets of the domai are mapped to two separate eleemets of the rages. The umber of ways to do this is left as a exercise for the reader. Most importat to ote is that as the differece of the sizes of the domai ad rages grows, so does the umber of differet ways to costruct a oto fuctio. The heart of the problem is determiig the umbers of ways the domai ca be partitioed ito a umber of oempty subsets equal to the size of the rage. This umber is ow as the Stirlig umber of the secod id. There are two types of Stirlig umbers. We will iitially examie Stirlig umbers of the secod id due to the fact that they occur more frequetly. The Stirlig umber of the secod id, or S(, ), is the umber of partitios of a elemet set ito oempty subsets. The symbol,, is geerally read as subset. For example, = because there exist seve differet ways to partitio a set S ={,,, } ito two oempty subsets as listed i Table... {{,, }, {}} {{, }, {, }} {{,, }, {}} {{, }, {, }} {{,, }, {}} {{, }, {, }} {{}, {,, }} Table.. The Stirlig umber of the secod id below i Table... =. The six differet partitios are give {{, }, {}, {}} {{}, {, }, {}} {{, }, {}, {}} {{}, {, }, {}} Table.. {{, }, {}, {}} {{}, {}, {, }}
2 Let A ={,,, } ad B ={, }. How may differet oto fuctios f : A d B exist? For the fuctio f to be oto B we must partitio A ito two differet oempty subsets. This ca be doe i = differet ways. Next we must assig each of those subsets a uique image i B. This ca be doe i! = ways. Thus, there are & = differet possible oto fuctios f : A d B. For A={,,, } ad B ={,, }, how may differet oto fuctios f : A d B exist? This would be &! = differet oto fuctios f : A d B. The values of four particular pairs of parameters are simple to determie which is ot true i the geeral case. For all positive itegers, = 0 as it is impossible to partitio a 0 elemet set ito o subsets. Liewise, = 0 for all >. For all positive itegers, = due to the fact that the oly partitio of a set ito oe oempty subset is the set itself. Similarly, for all positive itegers, =. The oly partitio of a elemet set ito subsets is the collectio of all sigle elemet subsets of the origial set. The value is also easy to compute. The oly partitio of a elemet set ito oempty subsets will cosist of oe subset of size two ad all other subsets will be sigle elemet subsets. Hece, selectio of the uordered pair determies the partitio ad =. Next cosider. I order to partitio a elemet set ito two oempty subsets, we eed just to select a proper, oempty subset from our elemet set. The partitio will cosist of this subset ad its complemet. It is obvious why this iitial subset must ot be empty. It must also be proper, otherwise its complemet would be empty. There are ways to select this subset. But, we have couted each partitio twice. Selectig the subset A geerates the partitio {A, A}. However selectig A also geerates the same partitio. Thus, = =. Theorem..: The geeral formula for computig the Stirlig umber of the secod id is =.! ( ) i ( i) i=0 i
3 With the formula i Theorem.. it is possible to determie the Stirlig umber of the secod id S(, ) without geeratig all possible partitios. For example, =! ( ) i ( i) = i=0 i ( )0 0 + ( ) + ( ) 0 = [ + 0] =. Next let A ={,,,,,, } ad B ={,,, }. How may differet oto fuctios f : A d B exist? For the fuctio f to be oto B we must partitio A ito four differet oempty subsets. This ca be doe i = 0 differet ways. Next we must assig each of those subsets a uique image i B. This ca be doe i! = ways. Thus, there are 0 & =, 00 differet possible oto fuctios f : A d B. Suppose a group of twelve college studets eeds to head to three differet locatios i order to gather items for a scaveger hut. They decide to split ito three differet groups i order to maximize efficiecy. How may differet ways ca they split ito three differet groups? Clearly, we wish to partitio the twelve studets ito three oempty subsets. This ca be doe i =, differet ways. If we wish to also decide the locatio that each group will visit we eed to compute the umber of differet ways to select a group for each locatio. There are! differet possible such assigmets. Thus, the umber of ways to partitio the groups ad decide their destiatios is &! = 9,. As with biomial coefficiets, there exists a simple recursive formula for Stirlig umbers of the secod id. Theorem..: For itegers m, = & +. The careful reader will otice that this equality ad subsequet proof are both very similar to the idetity = + ad its combiatorial proof of Theorem... Proof: Let A =,,...,. O oe had, the umber partitios of A ito oempty subsets is. O the other had, let A be the collectio of all partitios B of A ito oempty subsets such that {} c/ B ad let A be the collectio of all partitios B of A ito oempty subsets such that {} cb. Clearly A A = π. Every partitio of A either cotais the sigle elemet set {} or it does ot. Hece, every partitio of A is cotaied i either A or A. By the sum rule, the umber of partitios of A ito oempty subsets is A + A.
4 For A, the subset {} must be icluded i the partitio. Thus, we must partitio {,,..., } ito oempty subsets ad A =. For A, we must partitio A without icludig the sigle elemet subset {}. This forces the elemet to be part of a larger subset ad there must be oempty subsets without regard to the elemet. There are ways to partitio {,,..., } ito oempty subsets. I order to have a partitio of A ito oempty subsets all we must do is select oe of those subsets ad uio it with the elemet. This ca be doe i differet ways. Hece, A = & ad = & +. Usig this formula, we ca ow more easily compute S(, ). Usig the Theorem.. ad some basic properties of Stirlig umbers of the secod id, = & + = & (. )+ = + = Stirlig umbers of the secod id cotribute to the eumeratio of oto fuctios. Stirlig umbers of the first id also assist i coutig a particular class of fuctios, amely permutatios of a special form. Whe D = R =, there exist exactly! differet permutatios from D to R. However, we ow wish to examie the permutatios ad partitio them accordig to a represetatio as a uio of disjoit cycles. A commo way to write a permutatio of objects is i cycle form. The permutatio or represets the fuctio f where f()=, f()=, f()=, f()=ad f()=. A shorter way to defie f is to represet the cycles of the fuctio. The fuctio f is represeted by the cycles ()() because is mapped to which i tur is mapped to which is mapped to which maps bac to. This is a cycle of legth four. The poit is fixed ad cosists of a cycle of legth. Graphically, the cycles of a permutatio ca be represeted by a directed graph. The permutatio ()() is represeted by the directed graph with two compoets i Figure... This is a case where loops must be allowed i order to represet a fixed poit of a permutatio. Figure..
5 The Stirlig umber of the first id, or s(, ), is the umber of permutatios of elemets defied by disjoit cycles. The symbol,, is geerally read as cycle. For example, = because there are eleve differet ways to write the permutatios of S ={,,, } ito two disjoit cycles. ()() ()() ()() ()() ()() ()() ()() ()() Table.. ()() ()() ()() 0 As with there are some simple formulae for some values of. As with, 0 = 0 because a permutatio caot be writte without at least oe cycle. For >, = 0 because a uique simple form is used for cycles ad does ot allow repetitio of poits. If repetitio of poits is allowed i cycle otatio the a uique represetatio of a permutatio does ot exist. There is oly oe permutatio of elemets ito cycles, amely the idetity fuctio that fixes every poit ad =. The umber of permutatios of elemets ito oe cycle is just a variatio of the travelig salesma problem ad =( )!. The value is also easy to compute. The oly permutatio of poits ito cycles will cosist of oe cycle of legth two ad all other cycles will be fixed poits. Note that the cycle (ab) is the same as the cycle (ba). Selectio of the uordered pair determies the permutatio ad =. Much lie = & umbers of the first id. +, there is a recursive formula for Stirlig
6 Theorem..: homewor. =( ) & +. The proof of this idetity is left for the Homewor. Compute the followig Stirlig umbers of the secod id. i. ii. iii. iv. v. vi.. Compute the followig Stirlig umbers of the secod id. 0 9 i. ii. 9 iii. iv. v. vi List all the partitios of {,,,, } ito two oempty subsets.. List all the partitios of {,,,, } ito four oempty subsets.. Let A ={a, b, c, d, e, f, g, h} ad ={,,,, }. How may fuctios f : A d exist? How may oto fuctios f : A d exist?. Let A ={a, b, c, d, e, f, g, h, i} ad ={, ƒ,, }. How may fuctios f : A d exist? How may oto fuctios f : A d exist?. Let D = ad R = where is a fiite iteger. How may oto fuctios f : D d R exist.. Let D = ad R = where is a fiite iteger. How may oto fuctios f : D d R exist. 9. A project egieer has four systems to maitai ad six egieers i his group to assig to the differet systems. How may differet ways ca the project egieer assig employees to the systems if i. each system must be maitaied by at least oe employee?
7 ii. each system must be maitaied by at least oe employee ad each employee wors o exactly oe system? 0.. Prove = Compute by listig all the appropriate permutatios i cycle form. 0. Compute usig the recursio formula.. Prove Theorem..: =( ) & +.. Prove = +.. Costruct the followig table with etries for Costruct the followig table with etries for.
8 Let D = ad R = where is a fiite iteger. How may oto fuctios f : D d R exist. Let D = ad R = where is a fiite iteger. How may oto fuctios f : D d R exist.
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