Permutations by number of fixed points and anti-excedances

Afr. Mat. (2012) 23:121–133DOI 10.1007/s13370-011-0025-y

Permutations by number of fixed pointsand anti-excedances

Fanja Rakotondrajao

Received: 5 November 2010 / Accepted: 12 April 2011 / Published online: 29 April 2011© African Mathematical Union and Springer-Verlag 2011

Abstract In this paper we study the bivariated distribution of anti-excedances and fixedpoints on the symmetric group and give its generating function. We also study the samedistribution on the alternating group and its complement, and prove combinatorially a rela-tion between the number of odd permutations and the number of even permutations havinga given number of anti-excedances and a given set of fixed points.

Keywords Permutations · Anti-excedances · Fixed points · Parity · Generating function

Mathematics Subject Classification (2000) 05A15 · 05A19

1 Introduction

The distribution of fixed points and the so called ‘Eulerian statistics’ over the symmetricgroup have been studied in depth but always separately [2–4,6,11]. Roberto Mantaci andthe author (2003) have introduced and studied the numbers an,k that count the derangementsover n objects having k anti-excedances. In the first section of this paper we study the fixedpoint distribution over symmetric group which generalizes the properties of derangementsor permutations without fixed points and give a combinatorial interpretation of the general-ization of the famous relation of derangement numbers [1,8,10–12]

dn = ndn−1 + (−1)n . (1)

The author was supported by the ‘Soutien aux Activités de Recherche en Informatique et Mathématiques enAfrique’ (SARIMA) project and by LIAFA during her stay at the University of Paris 7, France as invited‘Maître de conférences’.

F. Rakotondrajao (B)Département de Mathématiques et Informatique,Université d’Antananarivo, 101 Antananarivo, Madagascare-mail: [email protected]; [email protected]

123

122 F. Rakotondrajao

In the second section, we will study the bivariated distribution of the Eulerian statisticanti-excedances and of the statistic of fixed points on the symmetric group. We will studyin the paper the numbers an,k,m that count the permutations over n objects having k anti-excedances and m fixed points. We will give an exponential generating function for thesenumbers as well as recursive relations defining them. We will also study the distribution ofthe bivariate distribution of anti-excedances and fixed points on the alternating group and itscomplement and prove that the number of odd permutations and the number of even permu-tations having k anti-excedances and having the same given set F of fixed points differ by 1for all integers n and k and for all subset F of [n]. We will give a combinatorial proof of thisresult.

Let us denote by [n] the interval {1, 2, . . . , n}, by σ a permutation of the symmetric groupSn and by An the alternating group of rank n, that is, the group of all even permutations.

We will use the cycle notation to represent a permutation, that is, we will express thepermutation as a product of cycles corresponding to the orbits of the permutation. We willdenote by cyc(i) the length of the cycle which contains the integer i . We will say that i ∈ [n]is a fixed point for σ if σ(i) = i , that is, the integer i is the only integer in its cycle. Wewill say that σ presents an anti-excedance (resp. an excedance) in i ∈ [n] if σ(i) ≤ i (resp.σ(i) > i). In this case we will say that i is an anti-exceedant (resp. an exceedant) of thepermutation σ . We will denote by Fix(σ ) the set of the fixed points of σ , by F I X (σ ) theinteger |Fix(σ )| and by AX (σ ) the number of the anti-excedances of σ . We will denote byFn,m the set of permutations over n objects having m fixed points and by fn,m the cardinalityof this set, by Sn,k the set of permutations over n objects having k anti-excedances and byAn,k the cardinality of this set. The numbers An,k are the classical Eulerian numbers [2,11].These numbers satisfy the following recursive relation

An,k = (n − k + 1)An−1,k−1 + k An−1,k (1 ≤ k ≤ n). (2)

The Eulerian polynomials An(t) =∑

σ∈Snt AX (σ ) have the following generating func-

tion [2]

A(t, u) =∑

n≥0

An(t)un

n! = (1 − t)

1 − t exp((1 − t)u). (3)

Notice that the number fn,m of permutations having m fixed points over n objects is equal

to

(n

m

)dn−m where dn−m denotes the number of permutations without fixed points [2,11].

On the other hand, the numbers fn,m satisfy the following recursive relation [9]

fn,m = fn−1,m−1 + (m + 1) fn−1,m+1 + (n − 1 − m) fn−1,m (0 ≤ m ≤ n), (4)

which could be combinatorially interpreted as the number of ways to obtain the permutationshaving m fixed points over n objects from the set of permutations over n − 1 objects byinserting the integer n inside a cycle or creating the new cycle (n):

1. Inserting the integer n in a cycle will decrease by one the number of fixed points if it isinserted in a cycle of a fixed point, but will not change the number of fixed points if it isinserted in a cycle of length greater or equal to two.

2. Creating the new cycle (n) will increase by one the number of fixed points.

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Permutations by number of fixed points and anti-excedances 123

The generating function of the fixed point distribution on the symmetric group is given by[2,11]

F(x, u) =∑

n≥0

∑

σ∈Sn

x F I X (σ ) un

n! = exp ((x − 1)u)

1 − u. (5)

2 Fixed point distribution over symmetric group

We will give in this paper a combinatorial interpretation of the relation

fn,m = n fn−1,m + (−1)n−m(

n

m

)(6)

directly over permutations having m fixed points. The number of derangements dn is a clas-sical number and it is a particular case of the number fn,m . It satisfies the recurrent relation

dn = (n − 1)(dn−1 + dn−2) (7)

from which we could deduce by a simple computation the other one

dn = ndn−1 + (−1)n . (8)

There are different interpretations of this last equation in the literature [1,8,10,12].

2.1 Combinatorial interpretation of fn,m = n fn−1,m + (−1)n−m(

n

m

)

Let M = {p1, . . . , pm} be a subset of the set [n] of size m. Let {b1 < b2 < · · · < bn−m} =[n]\M.

Definition 1 We say that a permutation is a critical permutation if it has the form σn,M =(b1, b2)(b3, b4) · · · (bn−m−1, bn−m)(p1) · · · (pm).

We denote by Cn,m the set of all critical permutations σn,M for all subsets M ⊂ [n] of size m.

Remark 2 1. Notice that Cn,m = ∅ if the integer n − m is odd.

2. If n − m is even, then |Cn,m | =(

n

m

).

Definition 3 Let us define the map ϕ from Fn,m\Cn,m onto [n] × Fn−1,m as follows. Letσ ∈ Fn,m and let [n]\F I X (σ ) = {b1 < b2 < · · · < bn−m}. We have to distinguish thefollowing three cases.

1. If cyc(n) ≥ 3, then the pair (i, σ ′) = ϕ(σ) is defined by i = σ−1(n) and the permutationσ ′ is obtained from the permutation σ by deleting the integer n from his cycle.

2. If cyc(n) = 2, then let p be the smallest integer (if such an integer does exist) such thatthe cycle (b2p+1, b2p+2) is not a cycle of the permutation σ and ϕ(σ) = (n, σ ′) whereσ ′ is defined as below

(a) If σ(n) = b2p+1, then σ ′ is obtained from the permutation σ by deleting the cycle(n, σ (n)) and inserting the integer σ(n) in the cycle which contains the integerb2p+2 just before b2p+2.

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(b) If σ(b2p+1) = b2p+2, then the permutation σ ′ is obtained from the permutation σ

by deleting the cycle (n, σ (n)), removing the integer b2p+1 from the cycle whichcontains it and then adding the new cycle (b2p+1, σ (n)).

(c) If σ(b2p+1) �= b2p+2 or σ(b2p+1) �= n, then the permutation σ ′ is obtained fromthe permutation σ by removing the cycle (σ (n), n) and inserting the integer σ(n)

in the cycle which contains the integer b2p+1 just before the integer b2p+1.

In these three cases, the integer p remains the smallest integer such that the cycle(b2p+1, b2p+2) is not a cycle of the permutation σ ′.

3. If the integer n is a fixed point of the permutation σ , we have to distinguish the followingcases.

(a) If cyc(bn−m) ≥ 3, then ϕ(σ) = (σ (bn−m), σ ′) where the permutation σ ′ isobtained from the permutation σ by removing the cycle (n) and the integer σ(bn−m)

from its cycle and adding the new cycle (σ (bn−m)).(b) If cyc(bn−m) = 2, then let p be the integer defined as in the [2] and ϕ(σ) =

(bn−m, σ ′) where the permutation σ ′ is defined as below:

i. If σ(bn−m) = b2p+1, then σ ′ is obtained from the permutation σ by remov-ing the cycles (n) and (b2p+1, bn−m) and inserting the integer b2p+1 in thecycle which contains the integer b2p+2 just before b2p+2 and adding the newcycle (bn−m).

ii. If σ(b2p+1) = b2p+2, then the permutation σ ′ is obtained from the permu-tation σ by removing the cycles (n) and (bn−m, σ (bn−m)) and the integerb2p+1 from the cycle which contains it and then adding the new cycles(b2p+1, σ (bn−m)) and (bn−m).

iii. If σ(b2p+1) �= b2p+2 or σ(b2p+1) �= bn−m , then the permutation σ ′ isobtained from the permutation σ by removing the cycles (n) and (σ (bn−m),

bn−m) and inserting the integer σ(bn−m) in the cycle which contains theinteger b2p+1 just before the integer b2p+1 and adding the new cycle (bn−m).

For each case, the permutation σ ′ is an element of the set Fn−1,m .

Proposition 4 The map ϕ is injective.

Proof Let the pair (i, σ ′) ∈ [n]×Fn−1,m and let {b′1 < · · · < b′

n−1−m} = [n −1]\F I X (σ ′).

1. If i = n, then the antecedent σ of the pair (i, σ ′), if it does exist, is entirely deter-mined by the case [2]. Let p be the smallest integer, if it does exist, such that the cycle(b′

2p+1, b′2p+2) is not a cycle of the permutation σ ′.

(a) If n − 1 − k is even and the permutation σ ′ is an element of the set Cn−1,m , thenthe pair (i, σ ′) does not have an antecedent because the integer p does not exist.

(b) If n − 1 − k is even and the permutation σ ′ /∈ Cn−1,m or if n − 1 − m is odd, thenthe integer p does exist. This case falls in [2a] or [2b] or [2c]. We will distinguishagain two cases in spite of the length of the cycle containing b′

2p+1.

i. If cyc(b′2p+1) = 2, then, by [2c], the permutation σ is obtained from the

permutation σ ′ by removing the cycle(b2p+1, σ

′(b2p+1)), and inserting the integer b′2p+1 in the cycle which con-

tains the integer b′2p+2just before the integer b′

2p+2 and creating the newcycle (n, σ ′(b′

2p+1)).ii. If cyc(σ ′) ≥ 3

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A. If σ ′(b′2p+1) = b′

2p+2, then the case [2a] implies that the permuta-tion σ is obtained from the permutation σ ′ by removing the integerb′

2p+1and creating the new cycle (b′2p+1, n).

B. If σ ′(b′2p+1) �= b′

2p+2, then the case [2b] implies that the permuta-tion σ is obtained from the permutation σ ′ by removing the integerσ ′(b′

2p+1) and creating the new cycle (σ ′(b′2p+1), n).

2. If i < n, then we have to distinguish again two cases whether the integer i is a fixedpoint of the permutation σ ′ or not.

(a) If i /∈ F I X (σ ′), then [1] implies that the permutation σ is obtained from thepermutation σ ′ by inserting the integer n in the cycle which contains the integer ijust after the integer i .

(b) If i ∈ F I X (σ ′), then, by [3], we have to distinguish again the following two caseswhether i < b′

n−1−m or i > b′n−1−m .

i. If i < b′n−1−m , then the permutation σ is obtained from the permutation σ ′

by removing the cycle (i) then inserting the integer i between b′n−1−k and

σ ′(b′n−1−m) and creating the new cycle (n).

ii. If i > b′n−m−1, then the antecedent σ of the pair (i, σ ′), if it does exist, is

completely determined by [3(b)i] or [3(b)ii] or [3(b)iii]. Let p be the smallestinteger such that the cycle (b′

2p+1, b′2p+2) is not a cycle of the permutation

σ ′, if a such integer does exist.

A. If n − 1 − m is even and the permutation σ ′ is an element of the setCn−1,m , the integer p does not exist as well as the permutation σ .

B. If n − 1 − m is even and the permutation σ ′ /∈ Cn−1,m or if the integern − 1 − m is odd, then the integer p does exist. We have to distinguishthe following cases in spite of the length of the cycle which containsthe integer b′

2p+1.

– If cyc(b′2p+1) = 2, then, by [3(b)ii], the permutation σ is obtained from

the permutation σ ′ by removing the cycles (b′2p+1, σ

′(b′2p+2)) and (i),

then inserting the integer b′2p+1 between σ ′−1(b′

2p+2) and b′2p+2 and

creating the new cycles (σ ′(b′2p+1), i) and (n).

– If cyc(b′2p+1) ≥ 3 and σ ′(b′

2p+1) = b′2p+2, then, by [3(b)i], the permu-

tation σ is obtained from the permutation σ ′ by removing the cycle (i)and the integer b′

2p+1 from the cycle which contains it and creating thenew cycles (n) and (b′

2p+1, i).– If cyc(b′

2p+1) ≥ 3, then, by [3(b)iii], the permutation σ is obtained from

the permutation σ ′ by removing the cycle (i) and the integer σ ′−1(b′2p+1)

from the cycle which contains it and creating the new cycles (n) and(σ ′−1(b′

2p+1), i). Corollary 5 If n − 1 − m is even, all pairs (i, σ ′) where the permutation σ ′ is an element ofthe set Cn−1,m and i > max{ j |σ ′( j) �= j} do not have an antecedent. The number of such

pairs is equal to

(n

m

).

Corollary 6 If n − m is even, then |Fn,m | −(

n

m

)= |[n] × Fn−1,m |.

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Corollary 7 If n − m is odd, then |Fn,m | = |[n] × Fn−1,m | −(

n

m

).

Example 8 For n = 5 and m = 2, we have

F4,2 = {(1)(2)(34); (1)(3)(24); (1)(4)(23); (2)(3)(14); (2)(4)(13); (3)(4)(12)}and f4,2 = 6.

F5,2 = {(1)(2)(345); (1)(3)(245); (1)(2)(354); (1)(3)(254); (1)(4)(235);(1)(4)(253); (1)(5)(234); (1)(5)(243); (2)(3)(145); (2)(3)(154);(2)(4)(135); (2)(4)(153); (2)(5)(134); (2)(5)(143); (3)(4)(125);(3)(4)(152); (3)(5)(124); (3)(5)(142); (4)(5)(123); (4)(5)(132)}

and f5,2 = 20. The following ten pairs do not have an antecedent by the map ϕ

(5, (1)(2)(34)); (5, (1)(3)(24)); (4, (1)(4)(23)); (5, (1)(4)(23)); (5, (2)(3)(14));(4, (2)(4)(13)); (5, (2)(4)(13)); (3, (3)(4)(12)); (4, (3)(4)(12)) and (5, (3)(4)(12))

We have also

C5,1 = {(1)(23)(45), (2)(13)(45), (3)(12)(45), (4)(12)(35), (5)(12)(34)}.

Remark 9 Notice that we could deduce the relation (6) from the relation (8) if we replace n

by n − m and

(n

m

)dn−m by fn,m .

3 The anti-excedance and fixed point distribution on the symmetric group

We will give a recursive relation for the distribution of the bivariate statistic (AX, F I X) onthe symmetric group Sn , and will give its generating function. We will denote by Fn,k,m theset of permutations in Sn,k having m fixed points (or equivalently, the set of permutations inFn,m having k anti-excedances) and denote an,k,m the cardinality of Fn,k,m .

Remark 10 Notice that

1. For any integer n ≥ 0, one has an,n,n = 1.2. For a fixed integer m and a fixed integer n, the numbers an,k,m are symmetric, in the

sense that an,k,m = an,n−k+m,m for all integers n ≥ 1 and m ≤ k ≤ n.

Remark 11 We have

1. an,k = an,k,0 [7],2. an,k,m = 0 if k < 0 or m > k.

Theorem 12 For all integers n, k and m such that 0 ≤ m ≤ k ≤ n, one has

an,k,m = an−1,k−1,m−1 + (m + 1)an−1,k,m+1

+(n − k)an−1,k−1,m + (k − m)an−1,k,m (9)

with the initial value a0,0,0 = 1.

123


Proof Notice that all permutation σ ′ in the symmetric group Sn is obtained from a permu-tation σ in Sn−1 by multiplying σ on the left by a transposition (i, n) for an integer i ∈ [n]and we suppose that the integer n is an exceedant for σ . Notice also that if the integer i is anexceedant (resp. an anti-exceedant) for σ , then when we multiply σ by the transposition (i, n)

on the left, we create (resp. do not create) a new anti-exceedant for σ ′. When it is created,this new anti-exceedant is the integer n itself. Now let us look for the various cases for theinteger i

1. If i = n, then the transposition (i, n) is not a transposition but the 1- cycle (n) and thepermutation σ ′ = (n)σ has a new anti-exceedant fixed point, which is the integer n itself.

2. If the integer i is an exceedant for σ , then the permutation σ ′ = (i, n)σ has a newanti-excedant, which is the integer n itself, but does not have any new fixed point.

3. If the integer i is a fixed point of the permutation σ , then the permutation σ ′ = (i, n)σ

does not have a new anti-exceedant but it has one fewer fixed points than the permuta-tion σ .

4. If the integer i is an anti-exceedant non fixed point of the permutation σ , then the per-mutation σ ′ = (i, n)σ has neither a new anti-exceedant nor a new fixed point.

It follows straightforwardly that we obtain all the permutations in the set Fn,m having k anti-excedances by considering all the permutation σ indicated in the following four cases andby multiplying them by the appropriate transposition.

1. σ ∈ Fn−1,m−1 having k − 1 anti-excedances and the only possibility for the choice of‘transposition’ by which multiply σ is the 1-cycle (n).

2. σ ∈ Fn−1,m+1 having k anti-excedances and there exist m +1 possibilities for the choiceof the transposition: the transpositions (i, n) where the integer i is a fixed point of thepermutation σ .

3. σ ∈ Fn−1,m having k − 1 anti-excedances and there exist n − 1 − (k − 1) possibilitiesfor the choice of the transposition: the transpositions (i, n) where the integer i is anexceedant of the permutation σ .

4. σ ∈ Fn−1,m having k anti-excedances and there exist k − m possibilities for the choiceof the transposition: the transpositions (i, n) where the integer i is an anti-exceedant nonfixed point of the permutation σ .

Remark 13 Notice that a fixed point of a permutation σ is an anti-exceedant of the permu-tation σ . Notice also that a permutation of Sn having m fixed points and k anti-excedancescould be considered as a derangement over [n − m] having k − m anti-excedances, so wehave the simpler result

an,k,m =(

n

m

)an−m,k−m . (10)

3.1 Generating function

Let us denote by A(t, x, u) the exponential generating function of the distribution of the bivar-iate statistic (AX, F I X) on the symmetric groups, that is, the function A(t, x, u) defined by

A(t, x, u) =∑

n≥0

∑

σ∈Sn

t AX (σ )x F I X (σ ) un

n! .

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Proposition 14 The generating function A(t, x, u) of the distribution of the bivariate sta-tistic (AX, F I X) on the symmetric groups satisfies the following differential equation

xt A = (1 − tu)∂ A

∂u+ t (t − 1)

∂ A

∂t+ (x − 1)

∂ A

∂x, (11)

with the initial conditions

A(t, 1, u) = 1 − t

1 − t exp ((1 − t)u)and A(1, 1, 0) = 0. (12)

Proof The differential equation can be easily derived from the recurrence relation given inTheorem 12. The initial condition

A(t, 1, u) = 1 − t

1 − t exp ((1 − t)u)(13)

is due to the fact that when we set x = 1 the resulting formal series is the well-knowngenerating function of the Eulerian polynomials and when we set x = 1, t = 1 and u = 0we obtain the first value of the numbers an,k,m . Theorem 15 The generating function

A(t, x, u) =∑

n≥0

∑

σ∈Sn


n! (14)

of the distribution of the bivariate statistic (AX, F I X) on the symmetric group has thefollowing closed form

A(t, x, u) = (1 − t) exp ((x − 1)tu)

1 − t exp ((1 − t)u). (15)

Proof Using MAPLE to solve the partial differential equation (11), we obtained the result.

The function(1 − t) exp ((x − 1)tu)

1 − t exp ((1 − t)u)satisfies the differential equation given in the previous

proposition, as well as the initial conditions.

4 Anti-excedances, fixed points and parity

Mantaci [4] introduced the numbers Pn,k and Dn,k that count respectively the cardinality ofthe set An,k of even permutations having k anti-excedances, and the cardinality of the setSn,k\An,k . These numbers satisfy the following relations

Pn,k = Pn−1,k−1 + k Dn−1,k + (n − k)Dn−1,k−1 (16)

Dn,k = Dn−1,k−1 + k Pn−1,k + (n − k)Pn−1,k−1 (17)

for any integers n ≥ 0 and 1 ≤ k ≤ n with P0,0 = 1 and D0,0 = 0.Let us denote by

– pn,m the cardinality of the set of even permutations in Fn,m ,– in,m the cardinality of the set of odd permutations in Fn,m .

Proposition 16 The numbers pn,m and in,m satisfy the following relations

pn,m = pn−1,m−1 + (m + 1)in−1,m+1 + (n − m − 1)in−1,m (18)

in,m = in−1,m−1 + (m + 1)pn−1,m+1 + (n − m − 1)pn−1,m (19)

for any integers n ≥ 0 and 0 ≤ m ≤ n with p0,0 = 1 and i0,0 = 0.

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Proof We will use the same idea as in Theorem 12. Suppose i �= n. If σ is an even (resp.odd) permutation of Sn−1, by multiplying σ ∈ Sn−1 by the transposition (i, n) we obtainan odd (resp. even) permutation σ ′ in Sn . When i = n, we obtain a permutation having thesame parity.

We denote by

– Pn,k,m the cardinality of An,k,m , the set of even permutations having k anti-excedancesand m fixed points,

– Dn,k,m the cardinality of (Sn,k⋂

Fn,m)\An,k,m , the set of odd permutations having kanti-excedances and m fixed points.

We have the following results.

Proposition 17 For all positive integers n, k and m such that 0 ≤ m ≤ k ≤ n, one has

Pn,k,m = Pn−1,k−1,m−1 + (m + 1)Dn−1,k,m+1

+ (n − k)Dn−1,k−1,m + (k − m)Dn−1,k,m (20)

Dn,k,m = Dn−1,k−1,m−1 + (m + 1)Pn−1,k,m+1

+ (n − k)Pn−1,k−1,m + (k − m)Pn−1,k,m (21)

with the initial conditions P0,0,0 = 1 and D0,0,0 = 0.

Proof The process described in the Theorem 12 to prove the recursive formula for the num-bers an,k,m allows to construct an odd permutation of Sn starting from an even one of Sn−1

and vice-versa, when i �= n. In the case i = n, the parity remains the same.

Let F be a subset of [n] such that |F | = m. We denote by

– Fn,F the set of permutations having F as set of fixed points,– Dn,k,F the cardinality of the set Fn,F

⋂(Sn,k\An,k)

– Pn,k,F the cardinality of the set Fn,F⋂

An,k ,– in,F the cardinality of the set (Sn\An)

⋂Fn,F

– pn,F the cardinality of the set An⋂

Fn,F

We have the following theorem.

Theorem 18 For all positive integers n, k and m such 0 ≤ m ≤ k ≤ n and for all subsetF ⊂ [n] of size m, one has

Dn,k,F − Pn,k,F = (−1)n−m . (22)

Proof A combinatorial proof of this result is the subject of the next separate section of thispaper.

Remark 19 This result could be deduced also from Mantaci and Rakotondrajao’s [7] result,but we will give in this section an elegant proof which is more general than the proof by case.

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Corollary 20 For all integers n, k and m with 1 ≤ m + 1 ≤ k ≤ n − 1 and for all subsetF ⊂ [n] of size m, one has

in,F − pn,F = (−1)n−m(n − m − 1) (23)

in,m − pn,m = (−1)n−m(n − m − 1)

(n

m

)(24)

Dn,k,m − Pn,k,m = (−1)n−m(

n

m

)(25)

Dn,k − Pn,k = (−1)n−k+1(

n − 1k − 1

)(26)

Proof Notice that

1.

Fn,F =n−1⋃

k=m+1

Sn,k

⋂Fn,F . (27)

2.

Fn,m =⋃

F⊂[n]|F |=m

Fn,F . (28)

3.

Sn,k,m =⋃

F⊂[n]|F |=m

Fn,F

⋂Sn,k . (29)

4.

Sn,k =k−1⋃

m=0

Fn,m

⋂Sn,k (30)

and

k−1∑

m=0

(−1)n−m(

n

m

)= (−1)n−k+1

(n − 1

k − 1

). (31)

The last identity is a simple combinatorial exercise. 4.1 Generating functions of the numbers Pn,k,m and Dn,k,m

Proposition 21 The generating function

P(t, x, u) =∑

n≥0

∑

σ∈An


n! (32)

of the distribution of the bivariate statistic (AX, F I X) on the set of even permutations hasthe closed form

P(t, x, u) = 1

2

{texp ((x − 1)tu) − exp ((xt − 1)u)

1 − t+ (1 − t) exp ((x − 1)tu)

1 − t exp ((1 − t)u)

}. (33)

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The generating function

D(t, x, u) =∑

n≥0

∑

σ∈Snσ �∈An


n! (34)

of the distribution of the bivariate statistic (AX, F I X) on the set of odd permutations hasthe closed form

D(t, x, u) = 1

2

{texp ((x − 1)tu) − exp ((xt − 1)u)

t − 1+ (1 − t) exp ((x − 1)tu)

1 − t exp ((1 − t)u)

}. (35)

Proof The series∑

n≥0∑n

m=0∑n

k=m+1(−1)n−m(

n

m

)tk xm un

n! has the closed form

texp ((xt − 1)u) − exp ((x − 1)tu)

1 − t. (36)

Therefore, the two functions P(t, x, u) and D(t, x, u) are solutions of the system⎧⎪⎨

⎪⎩

P(t, x, u) + D(t, x, u) = A(t, x, u) = (1 − t) exp ((x − 1)tu)

1 − t exp ((1 − t)u)

D(t, x, u) − P(t, x, u) = texp((xt − 1)u) − exp ((x − 1)tu)

1 − t

5 Combinatorial interpretation of Dn,k,F − Pn,k,F = (−1)n−m

Definition 22 (Critical permutations) Let F = { f1, f2, . . . , fm} be a subset of [n] and{d1, d2, . . . , dn−m} = [n]\F with d1 < d2 < · · · < dn−m . We define the ‘critical permuta-tions’ by

�n,F,1 = ( f1)( f2) · · · ( fm)(d1d2 · · · dn−m−1 dn−m)

�n,F,i+1 = ( f1)( f2) · · · ( fm)(d2d1d3 · · · dn−m−i dn−m dn−m−1 · · · dn−m−i+1)

for i = 1, . . . , n − m − 2. We will denote by Kn,F the set of the critical permutations and byKn,m = ⋃

F⊂[n]|F |=mKn,F

Example 23 For n = 9 and F = {1, 3, 4, 8}. We have m = 4 and

– �9,F,1 = (1)(3)(4)(8)(2 5 6 7 9)

– �9,F,2 = (1)(3)(4)(8)(5 2 6 7 9)

– �9,F,3 = (1)(3)(4)(8)(5 2 6 9 7)

– �9,F,4 = (1)(3)(4)(8)(5 2 9 7 6)

Remark 24 For all positive integer i and for all subset F ⊂ [n] of size m, one hasAX (�n,F,i ) = m + i .

Definition 25 Let � be a map of Fn,F\Kn,F onto itself defined as follows.For each permutation σ , let d1 < d2 < · · · < dn−m be the non fixed points of σ .

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1. If σ(d1) �= d2 and σ(d2) �= d1, then the permutation σ ′ = �(σ) is obtained by mul-tiplying the permutation σ on the right by the transposition (d1, d2). In other terms,σ ′ is obtained from σ by exchanging the integers d1 and d2 in the word σ(1) · · · σ(n).

For example, if σ = 3 5 1 4 2 7 6 then �(σ) = 3 5 2 4 1 7 6.

2. If σ(d1) = d2 and σ(dn−m) �= d1, then the permutation σ ′ = �(σ) is obtained by mul-tiplying the permutation σ on the right by the transposition (d1, dn−m). In other terms,we obtain the permutation σ ′ = �(σ) by exchanging the integers d1 and dn−m in theword σ(1) · · · σ(n). For example, if σ = 2 3 1 4 7 5 6 then �(σ) = 2 3 7 4 1 5 6

3. If σ(d1) = d2 and σ(dn−m) = d1 or if σ(d2) = d1 and σ(d1) �= d2, then let di be thelargest anti-exceedant non fixed point of σ , with d2 < di < dn−m and σ(di+1) �= di

(if such an integer di exists). The permutation σ ′ = �(σ) is obtained by multiplying thepermutation σ on the right by the transposition (di di+1). In other terms, σ ′ is obtainedfrom σ by exchanging the integers di and di+1 in the word σ(1) · · · σ(n). For example,if σ = 2 6 7 4 3 5 1 then the integer di is equal to 6 and the permutation σ ′ is equal to2 7 6 4 3 5 1.

Proposition 26 The map � preserves the set of fixed points, that is, an integer � ∈ [n] is afixed point for the permutation σ if and only if it is a fixed point for the permutation �(σ).

Proof Notice that the two integers i and j that need to be exchanged in σ to compute �(σ)

are both non fixed points and we have i �= j . After the exchange, we have σ ′(σ−1(i)) =j �= σ−1(i) and σ ′(σ−1( j)) = i �= σ−1( j). Proposition 27 The map � changes the parity of a given permutation, that is, if σ is an evenpermutation then �(σ) is an odd permutation and vice-versa.

Proof The action of � consists in multiplying a permutation by a transposition. This opera-tion changes the parity. Proposition 28 The map preserves the set of anti-excedances of all permutations, that is,the permutation σ has an anti-excedance in an integer � of the set [n] if and only if �(σ)

has an anti-excedance in �.

Proof If σ is a permutation such that σ(d1) = d2 and σ(dn−m) = d1, then the exchange ofthe integers d1 and dn−m does not change the fact that we have anti-excedances in σ−1(d1)

and an excedance in σ−1(dn−m). In all other cases, we exchange two consecutive non fixedpoints di and di+1, as images of the two integers σ−1(di ) and σ−1(di+1). We have

σ−1(di ) < di = σ(σ−1(di )) ⇔ σ−1(di ) < di+1 = σ ′(σ−1(di+1))

and

σ−1(di+1) < di+1 = σ(σ−1(di+1)) ⇔ σ−1(di+1) < di = σ ′(σ−1(di ))

so, σ has an excedance in σ−1(di ) (respectively, in σ−1(di+1)) if and only if σ ′ has an ex-cedance in σ−1(di ) (respectively, in σ−1(di+1)). In either case, the set of excedances of σ ′is the same as the set of excedances of σ . Corollary 29 The map � preserves the number of fixed points of a given permutation, aswell as the number of anti-excedances.

Theorem 30 The map � is a bijection on the set Fn,F\Kn,F onto itself.

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Proof Notice that if � is defined on a permutation σ , then � is also defined on σ ′ = �(σ),because it is impossible to obtain a critical permutation as image of another permutation.Notice that the two integers that need to be exchanged in σ ′ to compute �(σ ′) are the same asthe two integers that need to be exchanged in σ to compute σ ′ = �(σ) from σ . Therefore, ifτ is the transposition such that �(σ) = στ then �(�(σ)) = �(στ) = σττ = σ . Therefore,� is an involution and hence is a bijection. Theorem 31 For all integers n, k and m such that 0 ≤ m ≤ k ≤ n and for all subset F ⊂ [n]of size m, one has

Dn,k,F − Pn,k,F = (−1)n−m . (37)

Proof For all integer k = m + 1, . . . , n and for all subset F of [n] of size m, there exists aunique permutation �n,F,k−m of the set Sn,k which is an element of the set Kn,F . Further-more, this permutation �n,F,k−m is always an even permutation if and only if the integern − m is odd and vice-versa. Corollary 32 The map � is a bijection on the set Fn,m\Kn,m onto itself.

Corollary 33 For all integers n, k and m such that 0 ≤ m ≤ k ≤ n, one has

Dn,k,m − Pn,k,m = (−1)n−m(

n

m

). (38)

Acknowledgments The author is indebted to Roberto Mantaci for an enlightening discussion on the subject.

References

1. Désarménien, J.: Une autre interprétation du nombre de dérangements. Actes 8e Sém. Lothar. Comb.,IRMA Strasbourg, 11–16 (1984)

2. Foata, D., Schützenberger, M.P.: Théorie Géometrique des Polynômes Eulériens. Lect. Notes in Math.,vol. 138. Springer, Berlin (1970)

3. Knuth, D.: The Art of Computer Programming, vol. 3. Sorting and Searching, 2nd edn. Addison-Wesley,Reading (1981)

4. Mantaci, R.: Sur la distribution des anti-excédances dans le groupe symétrique et dans ses sous-groupes. Theoret. Comput. Sci. A 117, 243–253 (1993)

5. Mantaci, R.: Binomial coefficients and anti-exceedances of even permutations: a combinatorial proof.J. Combin. Theory A 63(2), 330–337 (1993)

6. Mantaci, R., Rakotondrajao, F.: A permutation representation that knows what “Eulerian” means. DiscreteMath. Theor. Comput. Sci. 4, 101–108 (2001)

7. Mantaci, R., Rakotondrajao, F.: Exceedingly deranging! Adv. Appl. Math. 30(1/2), 177–188 (2003)8. Rakotondrajao, F.: k-fixed-points-permutations. Integers: Electron. J. Combin. Number Theory 7, A36

(2007)9. Rakotondrajao, F.: Thèse de Doctorat, Université d’Antananarivo, Madagascar (November 1999)

10. Remmel, J.: A note on a recursion for the number of derangements. Eur. J. Combin. 4(4), 371–373 (1983)11. Riordan, J.: An Introduction to Combinatorial Analysis. Wiley, New York (1958)12. Wilf, H.S.: A bijection in the theory of derangements. Math. Mag. 57(1), 37–40 (1984)

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Permutations by number of fixed points and anti-excedances