"È$ß#È"ß$È#ß%È% This same mapping could also be represented in the form

Random Permutations A permutation of the objects "ß á ß defines a mapping. For example, the permutation 1 œ $ß "ß #ß % of the objects "ß #ß $ß % defines the mapping "È$ß#È"ß$È#ß%È% This same mapping could also be represented in the form "È$ß$È#ß#È" and %È% or more succinctly as "ß $ß # ß % The separate parts are referred to as cycles of the permutation. As argued in Riordan, An Introduction to Combinatorial Analysis, Chapter, The Cycles of Permutations, it is easy to see that every permutation can be uniquely represented by its cycles provided we adopt the convention that expressions such as "ß $ß #, 3ß 2ß 1, and 2ß 1ß 3, which represent the same cycle, are indistinguishable. We note that in the literature of cycles of permutations it is standard notation to write a cycle with its smallest element in the first position. Clearly x equals the total number of permutations of "ß á ß. It is well known that =ß>, the signless Stirling Number of the First Kind, counts the total number of permutations of "ß á ß with exactly > cycles. The article A Review of the Stirling Numbers, Their Generalizations and Statistical Applications, Charalambides, Ch. A.; Singh, Jagbir; Communications in Statistics, Theory and Methods, Vol. 1, No., 19, pages 2533--2595, is an excellent resource on Stirling Numbers. If a permutation of "ß á ß is selected uniformly at random from the set of all x permutations of "ß á ß, we will refer to this as a random permutation. If a permutation of "ß á ß is selected uniformly at random from the set of all = ß > permutations of "ß á ß with > cycles, we will refer to this as a random permutation with > cycles. We will need the following definitions.

: the infinite product space 0,1, á 0,1, á â : the set of all vectors s,s, á in such that "= #= á œ " # " # ß > : the set of all vectors s,s, " # á in such that " " "= #= # á œ = = á œ > # ß> ß> For any T define T œ T and T œ T We note that the condition that "=" #= # á œ implies that = œ! for all Þ Hence all vectors in T and Tß> are of the form + ", á, + ß!ß!ß á Þ For all T Á!ß!ßá, let be the collection of -dimensional vectors formed by taking each infinite-dimensional vector in and truncating after +ÞSo for example, T +, á, + ß!ß!ß á +, á, + " " Define ß> similarly. For notational consistency it is necessary to separate out the case T œ!ß!ßá. We will refer to a cycle with < elements as an r - cycle. A permutation of elements with 5 " - cycles, á, 5 - cycles is said to be of cycle class 5ßáß5Þ " " Let \ equal the number of - cycles in a random permutation of "ß á ß with > cycles. We can construct the set of all permutations of a set of elements into > cycles such that \" œ B" ßáß\ œ B in the following manner. Take any one of the x permutations of the elements and use the first B " elements of that permutation to fill the B " "- cycles, use the next #B# elements of that permutation to fill the B # #- cycles, and so on. In total we would use the "B" #B# á B œ elements to fill the B B á B cycles. " # This assignment yields x permutations but not all of these permutations are distinct. In particular this count would assume that rearranging the B -cycles amongst themselves leads to distinct permutations - which they do not. Furthermore this count would assume that all rearrangements of the elements within a cycle leads to distinct permutations - which they do not. Therefore it is necessary to divide the count of x by the number of ways to arrange the B

-cycles amongst themselves œ"ßáßand by the number of ways to arrange the elements in each cycle and not change the cycle. It follows that there are x " " â BxâBx " " B B " permutations of a set of "ß á ß with cycle class BßáßB " provided "B á B œ and B!ß áß a " and T \ " ß ß\ œ B" ß ßB œ! x " B" " B B" xâbx " â =ß> "B áb œ " B" ábœ> B!ß"ßáß a otherwise If we define [ as the number of - cycles in a random permutation of "ß á ß then it follows similarly that T [ " ß ß[ œ A" ß ßA œ! " " A " â" A "A" á Aœ AxâAx " " A!ß"ßáßa otherwise We note that it follows from the law of total probability that " " " " â œ " AxâAx " "A" áaœ A!ß"ßáßa " A A which is Cauchy's identity. Theorem 1.

". " E1 [ " ß ß[ œ E1 ] " ß] # ß x.- " - - œ! where 1 + " ß+ # ßá is any function and 1 + " ßáß+ œ1 + " ßáß+ ß!ß!ßáand for T Á!ß!ßá ". " T [ " ß ß[ œ T] " ß] # ßá T x.- " - -œ! where that ]ß]ßá " # is an infinite sequence of independent Poisson random variables such - C exp - T ] œ C œ C œ!ß "ß #ß á and œ "ß #ß á è Cx Theorem 2. > E 1 \ ß ß\ œ.. " " " E =ß> 1 ] " ß] # ß.-.) > "- >x ) -œ! ) œ! where 1 + " ß+ # ßá is any function and 1 + " ßáß+ œ1 + " ßáß+ ß!ß!ßáand for T Á!ß!ßá >.. " " T\ " ß ß\ ß> œ T " ß # ßá T =ß>.-.) > ] ] "- >x ) -œ! ) œ! where that ]ß]ßá " # is an infinite sequence of independent Poisson random variables such )- C exp )- T ] œ C œ C œ!ß "ß #ß á and œ "ß #ß á è Cx Now consider the set of all possible ways to distribute distinguishable keys onto > distinct key rings such that no key ring is left empty and where the position of keys on a key ring matters, but only up to circular shifts. Suppose we pick a distribution from this set uniformly at random.

Let Yßœ"ßáß> equal the number of keys on the key ring. Then, >2 TYß " ßY> œ?ß " ß? > œ! x?" â? > >x =ß>?" á? > œ? "ß#ßá a else Let Zß œ"ßáß equal the number of (distinguishable) key rings containing keys. Then, TZ" ß ßZ œ @ " ß ß@ œ! x " @ " @ " @" xâ@ x " â =ß> "@ á@ œ " @" á@ œ> @!ß"ßáß a otherwise We note that the above probability distribution is based on a model where the > key rings are distinguishable (e.g. different colors) but clearly the latter probability would be the same if the > key rings were the same color or not. However distributing distinguishable keys onto > like key rings is equivalent to forming a permutation of "ß á ß with > cycles. That is, for all ß> ß> TZ ß ßZ œ T \ ß ß\ " ß> " ß> where as before the \ ßœ"ßáß> equal the number of cycles with elements in a random permutation of "ß á ß with > cycles. We will need the following definitions. > À the > -dimensional product space ",#á, â ",#á, > > " > À the set of all vectors?, áß? in such that? á? œ " > > " ß> " > Define h as that set such that Z ß ßZ Í Y ß ßY h. Theorem 3.

T \ ß ß\ " d œ " T ] ] >x = ß > > " ß> " > ln ), á, h d) ) œ! where ]ßáß] " > are iid logarithmic series random variables with parameter ). That is, T ] œ C œ! C ) C ln") C "ß #ß á else è Problem 1. (a) The probability that a random permutation of "ß #ß á ß contains exactly 5 cycles of length equals " 35 3 x " x 5 3x 3 3œ5 This result was derived by Goncharov, V.L. (19), Some Factors from Combinatorics, Izv. Akad. Nauk SSSR, Ser. Mat., 3. Translated from Russian in Goncharov, V.L. (1962), On the Field of Combinatory Analysis, Translations from American Mathematical Society, 19, 1 6. (b) The probability that a random permutation of "ß #ß á ß into > cycles will contain exactly 5 cycles of length is > " 3 x = 3ß>3 " = ß> 35 5 3x3 3x 3œ5 (c) The probability that a random permutation of "ß #ß á ß into > cycles contains at least 5 cycles of length is

> " 3" x = 3ß>3 " = ß> 35 5" 3x3 3x 3œ5 and the probability that a random permutation of "ß #ß á ß contains at least 5 cycles of length is " 35 3" x " x 5 " 3x 3 3œ5 >2 (d) The < descending factorial moment of the number of cycles of length in a random permutation of "ß #ß á ß ß is "!Ÿ<Ÿ < >2 and the < descending factorial moment of the number of cycles of length in a random permutation of "ß #ß á ß with > cycles is x = <ß > < < x < = ß> <"Ÿ>Ÿ< " The first of these two results can be found in Riordan, J. An Introduction to Combinatorial Analysis, page, Problem 12. (e) The probability that a random permutation of "ß #ß á ß contains exactly 5 " cycles of length ",exactly 52 cycles of length 2, á, and exactly 5r cycles of length is r 3 3 3 " " á3< 5" á5 < " < â " â 5 5 3" 3< 3xâ3x â 3" ßáß3 < 3" 5" á3< 5< Ÿ 3 " 5" ßáß 3 < 5< " < " < " <

Problem 2. The descending factorial moment of the number of cycles in a random permutation of "ß #ß á ß is < >2 <x x = "ß<" This result can be found in Riordan, J. An Introduction to Combinatorial Analysis, page 1, equation (12). Problem 3. is The number of permutations of "ß #ß á ß which have 5 cycles, none of which an < cycle is < x " = <ß 5 x <x< œ! Riordan, J. An Introduction to Combinatorial Analysis, page 3, equation (1) is the special case <œ". In the case <œ" these numbers are referred to as the Associated Stirling Numbers of the First Kind. Riordan gives a recurrence relation for general < in Problem 16(a), page 5. Problem. (a) The probability that exactly 5 cycle lengths are multiples of in a random permutation of "ß #ß á ß with > cycles is

> > x 5 1 " = 3 ß = 3ß > =ß> 5 3x 3x œ5 3œ (b) The probability that every cycle length in a random permutation of "ß #ß á ß with > cycles is a multiple of (taking 5 œ > in above problem) simplifies to = ß> x " =ß> x > L. Carlitz, Set Partitions, Fibonacci Quarterly, Nov. 196, pages 32-32 gives the formula in (b) for the case œ#. Riordan, An Introduction to Combinatorial Analysis, Problem 1, pages 6-, gives a table of the values of the above for œ# and Ÿ)Þ (c) The number of permutations of "ß á ß where every cycle length belongs to the set =ß=ß=#ßá for some integers = and,!ÿ=is x d ' " = " d) "') œ! ) œ! where ' œ/ # 13Î. (d) The number of permutations of "ß á ß where every cycle length is a multiple of (special case of (c) with = œ!) equals x " " This result can be found in Sachkov, Probabilistic Methods in Combinatorial Analysis, Chapter 5, Random Permutations, page 151.

Notes: Goulden and Jackson, Combinatorial Enumeration, Wiley-Interscience Series in Discrete Mathematics, 193, Problem 3.3.12(a), page 1 give the answer in the form x x " " œ" Bolker and Gleason, Counting Permutations, Journal of Combinatorial Theory, Series A, Vol. 29, 190, pages 236-22 give the answer in the form œ " divides ) where )! else œ" (e) The number of permutations of "ß á ß where every cycle length belongs to the set # ß # ß # #ß á for even (special case of (c) with = œ # and even ) equals # x " œ! # " " provided # divides. This result can be found in Sachkov, Probabilistic Methods in Combinatorial Analysis, Chapter 5, Random Permutations, page 151. However there is a misprint where the above sum starts at œ" instead of œ!. (f) The number of permutations of "ß á ß where no cycle is a multiple of is x " 3 3œ! 3 "

This result can be found in Goulden and Jackson, Combinatorial Enumeration, Wiley-Interscience Series in Discrete Mathematics, 193, Problem 3.3.12(b), page 1. (g) The probability that a random permutation of "ß #ß á ß contains an even number of cycles all of which have odd length equals " Î# # This result can be found in Wilf, generatingfunctionology, 2nd edition, page. Problem 5. / 9 Let - (- ) equal the number of even (odd) permutations of "ß #ß á ß and let / 9 0 0 equal the number of even (odd) permutations of "ß #ß á ß with 5 ß5 cycles. ß5 (a) / - œ " œ " 9 - œ! œ " x and x # # # # This result can be found in Riordan, John, An Introduction to Combinatorial Analysis, Problem 20, page -. Note: The standard proof wherein one demonstrates a bijection by switching the position of elements " and in any permutation and noting that the parity changes is a much simplier proof but the present approach illustrates another aspect of Theorem 1. (b)

0 / ß5 œ =ß5 =ß5 # =ß5 5 œ is even! 5 is odd and 9 0 ß5 œ =ß5 =ß5 #! 5 is even œ =ß5 5is odd This result can be found in Sachkov, Probabilistic Methods in Combinatorial Analysis, page 15. Note: Applying Theorem 2 is a useful demonstration but a simplier proof follows from the observation that for fixed and > such that "B" á B œ and B á B œ > with B!ß"ßáßa, then " B# B% B' á is even Í > is even. It follows from this observation that 0 / ß5 œ x " " â BxâBx " "B" ábœ B" ábœ> B# B% B' áœeven B!ß"ßáßa " B B " x " B" " B Bx " âb x " â > œ even "B" ábœ œ B" ábœ> B!ß"ßáßa! > œ odd =ß> œ >œeven! > œ odd Problem 6. (a) How many permutations of "ß #ß á ß are there for which the longest run of

"-cycles is less than or equal to 5? (b) How many permutations of "ß #ß á ß with > - cycles are there for which the longest run of "-cycles is less than or equal to 5? (c) How many permutations of "ß #ß á ß are there which have exactly < runs of length 5 of "- cycles? (d) How many permutations of "ß #ß á ß with > -cycles are there which have exactly < runs of length 5 of "- cycles? Problem. if Define [ œ " \ œ @! else if Define X œ " \!! else (a) How many permutations of "ß #ß á ß are there for which exactly < of the values in the cycle class 5 ßáß5 equal @? i.e. R [ á [ œ< " " (b) E[ á [ " < (c) TX á X œ< " (d) EX á X " < Problem. How many permutations of "ß #ß á ß are there for which all cycle lengths are between 6 and? inclusive? i.e. \ á\ \ á\ œ! " 6"?"

Problem 9. (a) Suppose a permutation of "ß #ß á ß is picked at random form the set of all x permutations of "ß #ß á ß and from this permutation a cycle is picked at random. Let [ represent the length of this cycle. Find T [ œ A and E[. < (b) Suppose a permutation of "ß #ß á ß is picked at random form the set of all permutations of "ß #ß á ß with > cycles and from this permutation a cycle is picked at random. Let [ represent the length of this cycle. Find T [ œ A and E[ < (c) Suppose a cycle is picked at random from the set of all cycles. (explain). Let [ represent the length of this cycle. Find T [ œ A and E[ < (d) Suppose a permutation of "ß #ß á ß is picked at random and an element is picked at random from that permutation. Let [ represent the length of this cycle containing the randomly picked element. Find T [ œ A and E[ < Goulden & Jackson, page 190, Problem 3.3.19. Show that the number of permutations of "ß #ß á ß in which the cycle containing has length, is " x, for any œ "ßáßÞ Lovasz, page 29, problem 3. Shows that the probability that the cycle containing " 1 has length 5 is for 5 œ"ß#ßáßþ Hence expectation follows immediately. Grusho, A.A. Properties of random permutations with constraints on the maximum cycle length, Probabilistic Methods in Discrete Mathematics, (Petrozavodsk, 1992), pages 60-63 considers this problem with the additional constraint that no cycle can have length greater than -.

Also look at other problems similar to this covered in section on Random Set Partitions Problem 10. T \ \ \ á œ< # % ' Problem 11. Determine the number of permutations of "ß #ß á ß for which the number of < - cycles equals the number of = - cycles. This problem is discussed in Riordan, John, An Introduction to Combinatorial Analysis, Problem 15(b), page -5. Problem 12. Determine the number of permutations of "ß #ß á ß which have no - cycles for any #Þ This problem is discussed in Riordan, John, An Introduction to Combinatorial Analysis, Problem 1, page 5-6. Problem 13. Determine the number of even (odd) permutations of "ß #ß á ß which have no "- cycles.

This problem is discussed in Riordan, John, An Introduction to Combinatorial Analysis, Problem 21, page -9. Problem 1. Show that the number of permutations of "ß #ß á ß which have 5 cycles, none of which is a " cycle, # cycle, á, or < cycle is Howard, F. T. refers to these numbers as the <-associated Stirling Numbers of the First Kind in Fibonacci Quarterly, Associated Stirling Numbers, Vol 1, no., 190, pages 303-315. Problem 15. Record values. Goldie, Charles (199). Records, permutations and greatest convex minorants, Mathematical Proceedings of the Cambridge Philosophical Society, 106, no. 1, pp. 169-1. Let C be a random element of W, the set of permutations of, all x elements of W being equally likely. C may be written as a product of cycles. Let us say that 3 is a new-cycle index if 3does not belong to the cycles containing "ß á ß 3 "Þ The random set of new-cycle indices is denoted V. It always contains 1. Stam, theorem 3) has shown that the events 3 V are independent, with respective probabilities "Î3Þ ÒStam, A.J. (193). Cycles of random permutations, Ars Combinatoria, 16, pages 3-. Ó

Let Y be the set of record times. That is, Y À œ Ö3 À \ œ max Ð\ ß á ß \ Ñ 3 " 3 Let e be the set of record values. That is, ÐÑ " e À œ Ö\ 3 À 3 Y ÐÑ " Let \ ßáß\ be the order statistics of \ ßáß\ Theorem 3.1 The events Ö e ß 3 œ "ß á ß are independent with probabilities \ 3 ÐÑ T e œ \ 3 ÐÑ " "3 Proof ÐÑ ÐÑ ÐÑ ÐÑ 3 3 3" ÐÑ ÐÑ \" ßáß\ \ 3" á \ ÐÑ "ÎÐ " 3Ñ \ 3 ÐÑ ÐÑ \ 3" á \ ÐÑ "ÎÐ " 3Ñ \ 3 \ e if and only if \ occurs before \, á, \ in the finite sequence. Whatever order,, occur in among themselves, there is probability that occurs earlier. Thus given information on which of,, are record values, there is always probability that is a record value. (?) I give this argument here because I wonder if this argument is (1) rigorous and (2) can be used to prove the hook length formula. Note that in case where ÐÑ \ ßáß\ is a permutation of 1, á,, then \ œ. Let " ÐÑ M œ " \ e! else Consider TM œ"±m œ"ßáßm œ" " ÐÑ ÐÑ We are given that \ ", á,\ are record values so we know that ÐÑ ÐÑ ÐÑ " # " \ occurs before \, á,\ in the finite sequence \ ßáß\

and and ÐÑ ÐÑ ÐÑ # $ " \ occurs before \, á,\ in the finite sequence \ ßáß\ ÐÑ " ÐÑ " \ occurs before \ in the finite sequence \ ßáß\ conditioning but (as the argument goes) that tells us nothing about whether ÐÑ " ÐÑ " before \, á,\ in the finite sequence \ ßáß\. The ÐÑ " \ ÐÑ occurs only tells us about the position of \, á,\ relative to each other. ÐÑ Therefore TM œ"±m" œ"ßáßm œ" œtm œ" and subsequently TM" œ"ßm# œ"ßáßm œ" œtm œ"t M" œ"±m œ" âtm" œ"±m# œ"ßáßm œ" œtm œ"t M œ" âtm œ" " " which shows independence. Karamata-Stirling laws. The OW probability law is defined to be that of ^ á ^ where ^ ß á ß ^ " are independent and T ^ œ " œ "Î3ß T ^ œ! œ " "Î3Þ " 3 3 Explicitly, T^" á ^ œ5 œ =ß5 x Ð5œ!ß"ßáßÑ OW is the distribution of ( i) the number of cycles in a random permutation of n objects ( ii) the number of records in n exchangeable unequal r.v.s. ( iii) the number of sides in the gem of an n-step random walk with

Problem 16. exchangeable rationally independent increments