Combinatorics of permutations
part 3
Marilena Barnabei
Università di Bologna
Inversions of a permutation
Let σ = x1x2…xn be a permutation in Sn. The pair of values
(xi, xj) is said to be an inversion of σ if i < j and xi > xj.
The number of inversions of σ is denoted by the symbol inv(σ).
Example σ = 3 1 5 2 4 has 4 inversions, namely,
(3, 1), (3, 2), (5, 2), (5, 4).
Obviously,
⎛ ⎞
0 ≤ inv(σ ) ≤ ⎜n ⎟,
⎝ 2 ⎠
and the minimum value is attained only by the identity
permutation
€ ε, while the maximum is attained only by the
reverse permutation ψ.
Denote by Invn(x) the generating function of inversions of the
permutations in Sn:
Inv n (x) =
∑ x inv(σ ).
σ ∈Sn
€
Theorem For every integer n ≥ 2 we have:
(*)
Inv n (x) = (1+ x)(1+ x + x 2 )L(1+ x + x 2 +K + x n−1).
Proof We want to prove that each one of the n! monomials we
obtain by performing the products in the right-hand side of (*)
€
corresponds to a permutation in Sn. We proceed by induction
on n.
For n = 2, the permutation 1 2 has no inversions (hence, it
corresponds to the term 1), and 2 1 has one inversion (and it
corresponds to the term x). Hence the assertion is true.
Suppose now the assertion true for Sn–1. If σ is a permutation
in Sn–1, we insert (in n different ways) the symbol n in σ to get
a permutation σ in Sn. Inserting n in the i-th position gives rise
to a number of new inversions (with respect to the inversions
of σ) equal to the number of elements to the right of n, namely,
n–i. Hence, the monomial corresponding to σ is now multiplied
by xn–i, and we get
Invn(x) = (1 + x + x2 + ... + xn–1) Invn–1(x).
This gives the assertion.
Denote by B(n,k) the number of permutations in Sn with k
inversions, namely, the coefficient of xk in the polynomial
Invn(x).
The sequence B(n,k) has a symmetry similar to the symmetry
of Eulerian numbers, namely,
B(n,k) = B(n,
( ) − k).
n
2
In fact, if a permutation σ has k inversions, its reverse σr
⎛n ⎞
has ⎜ ⎟ − k inversions.
⎝ 2 ⎠€
Moreover:
€
Theorem For every n, the sequence
B(n,0), B(n,1),K, B(n,
is log-concave.
( ))
n
2
Proof It is easily seen that the convolution product of log
€ sequences is log-concave. Since, for every k, the
-concave
sequence of coefficients of the polynomial
1 + x + ... + xk
is obviously log-concave, the assertion follows from the
preceding result.
The inversion table
k
n
0
1
2
3
4
5
6
5
3
1
7
8
9
10
4
1
1
1
2
1
1
3
1
2
2
1
4
1
3
5
6
5
1
4
9 15 20 22 20 15 9
Theorem For every k ≤ n we have:
B(n+1,k) = B(n+1,k–1) + B(n,k).
Proof Let σ = x1x2…xn+1 be a permutation in Sn+1 with k ≤ n
inversions. If xn+1 = n+1, we can delete the last element of σ
leaving the number of inversions unchanged, hence obtaining
a permutation in Sn with k inversions. Every such permutation
can be obtained in this way.
If xi = n+1 for i ≤ n, interchange xi and xi+1. The resulting
permutation σ has now k–1 inversions. Note that in σ the
element n+1 is not at the first position. On the other hand, in
every permutation in Sn+1 whose number of inversions is
k–1 < k ≤ n
the greatest symbol occupies a position different from the first
one, since, when the symbol n+1 is at the first position, the
number of inversions of the permutation is at least n. Hence, in
this way we get all permutations in Sn+1 with k–1 inversions.
Major index
Let σ be a permutation in Sn. The major index of σ is defined to
be the sum of its descents :
maj(σ ) =
∑i.
i∈Des(σ )
Example Let σ = 3 5 2 4 6 1. Then Des(σ) = {2,5}, hence
maj(σ) = 7.
€
Theorem The two statistics “number of inversions” and “major
index” are equidistributed in Sn, namely,
∑ x inv(σ ) = ∑ x maj(σ ).
σ ∈Sn
σ ∈Sn
Proof We define recursively a bijection φ of Sn into itself, such
€
that for every
σ we have maj(σ) = inv(φ(σ)).
If n ≤ 2, set φ(σ) = σ.
If n > 2, we add numbers to φ(σ) one at a time: suppose σ =
σ1σ2…σn and begin by setting φ(1) = σ1, φ(2) = σ1σ2. To find φ(3),
start with σ1σ2σ3. Then if σ3 > σ2, draw a bar after each element
of σ1σ2σ3 which is less than σ3, while if σ3 < σ2, draw a bar after
each element of σ1σ2σ3 which is greater than σ3. Also add a bar
before σ1.
For example, if σ = 4137562, we now have |41|3.
Now regard the numbers between two consecutive bars as
“blocks”, and in each block, move the last element to the
beginning, and finally remove all of the bars.
We end up with φ(3) = 143.
Proceeding inductively, we begin by adding σi to the end of
φ(i−1). Then if σi > σi−1,draw a bar after each element of φ(i−1)
which is less than σi, while if σi < σi−1, draw a bar after each
element of φ(i−1) which is greater than σi . Also draw a bar
before the first element of φ(i−1). Then in each block, move the
last element to the beginning, and finally remove all of the
bars.
If σ = 4137562, the successive stages of the algorithm yield
143 = φ(3)
|1|4|3|7 → 1437 = φ(4)
|1437|5 → 71435 = φ(5)
|71|4|3|5|6 → 174356 = φ(6)
|17|4|3|5|6|2 → 7143562 = φ(7),
so φ(4137562) = 7143562.
Note that maj(4137562) = 11 = inv(7143562).
It is easily seen by induction that the map φ is a bijection.
Bruhat order
The Bruhat order relation over Sn is defined as follows:
σ  < τ if σ can be obtained from τ multiplying it by a finite
number of transpositions, each one of which involves the
elements of an inversion of τ.
Example The Hasse diagrams of S2 and S3 are the following:
321
21
12
231
312
132
213
123
S2
S3
Theorem The set Sn endowed with the Bruhat order is ranked.
Proof We show that, if the permutation τ = x1x2…xn covers σ,
then τ has exactly one more inversion than σ.
In fact, if τ covers σ, σ can be obviously obtained from τ by
interchanging the elements xi, xj (with xj < xi and i < j) of a single
inversion. Hence, σ = τ (xi xj).
Suppose that τ has at least two inversions more than σ. This
implies that there exists an index k such that i < k < j and
xj < xk < xi. In this case, setting π = τ (xi xk), we have σ < π < τ,
and τ does not cover σ.
Hence, all saturated chains between the minimum of the poset
(namely, the identity permutation) and a given permutation σ
have length inv(σ), and this is the rank of σ.
Let σ = x1x2…xn be a permutation. For every pair (a, b) with
1 ≤ a, b ≤ n, set
σ(a, b) = |{i ≤ a; xi ≥ b}|.
In the graphical representation of σ, σ(a, b) is the number of
points of σ lying in the rectangle with basis a and height
n–b+1, attached to the north-west corner of the grid:
σ = 3 7 1 4 2 5 6;
σ(4, 3) = 3.
Theorem σ ≤ τ in the Bruhat order if and only if
σ(a, b) ≤ τ(a, b) for every a, b.
Proof Suppose σ ≤ τ. Note that, interchanging in τ the
elements of any inversion, the numbers τ(a, b) can not
increase. Hence σ(a, b) ≤ τ(a, b) for every a, b.
Conversely, suppose σ(a, b) ≤ τ(a, b) for every a, b. Let
σ = x1x2…xn, τ = y1y2…yn, and set a = n–1.
The condition σ(n–1, b) ≤ τ(n–1, b) for every b implies xn ≥ yn.
If xn = yn, we can delete the last element from σ and τ, and
show that σ ≤ τ by induction.
If xn > yn, then in τ there is an element yi such that yi = xn and
i < n. This implies that (yi, yn) is an inversion of τ , the
permutation π = τ (yi, yn) is such that π < τ, and π has the
same last element than σ. It is easy to verify that
σ(a, b) ≤ π(a, b) for every a, b, hence we can proceed again by
induction.
€
τ = 3 7 1 4 2 5 6;
τ(4, 3) = 3.
σ = 1 2 3 4 7 5 6;
σ(4, 3) = 2.
Theorem σ ≤ τ in the Bruhat order whenever σ‒1 ≤ τ‒1.
Proof By the preceding result it suffices to show that, if σ(a, b)
≤ τ(a, b) for every a, b, then σ‒1(a, b) ≤ τ‒1(a, b) for every a, b.
Recall that the graphical representation of σ‒1 is obtained from
that of σ by flipping it around the main diagonal.
We must prove that, if we choose any south-east rectangle, it
contains at most as many points of σ as of τ.
Choose two values a and b, and suppose that the north-west
rectangle with basis a and height b contains A points of σ and
B points of τ; then the other rectangles contain the number of
points in the figure:
A
a–A
n–A–b+1
A+b–a–1
B
a–B
σ
Since, by hypothesis, A ≤ B, we have also
A+b–a–1 ≤ B+b–a–1, and the assertion is true.
n–B–b+1
B+b–a–1
τ
The cycle decomposition of a permutation
The type of a permutation σ is defined to be the n-tuple (a1,
a2, ..., an), where ai denotes the number of cycles of length i in
the cycle decomposition of σ.
N.B. Sometimes, the type of σ is defined also to be the
sequence of its cycle lengths, written in non-increasing order.
In this case, of course, the type is a partition of the integer n.
Example Let σ = 2 8 3 7 6 5 1 4 9 = (9) (56) (3) (12847).
The type of σ is either the 9-tuple (2, 1, 0, 0, 1, 0, 0, 0, 0), or
the partition of 9 λ = (5, 2, 1, 1).
Recall that two permutations σ, τ are conjugate (namely, there
exists a permutation γ such that τ = γ-1 σ γ) if and only if they
have the same type.
Theorem Let (a1, a2, ..., an) be an n-tuple of non-negative
integers such that
n
∑ i ai = n.
i=1
The number of permutations in Sn of type (a1, a2, ..., an) equals
n!
€
a1
a2
a1! a2!Lan ! 1 2 Ln
an
.
Proof Write the integers 1, 2, ..., n in any order (n! ways).
Then €
put some parentheses, denoting the cycles, in such a
way that the first a1 elements are the 1-cycles of the
permutation, the following 2a2 form the 2-cycles, and so on.
In this way we get all permutations of type (a1, a2, ..., an).
Every such permutation appears many times. In fact, for every
i, the ai cycles of length i can be rearranged (in i! ways), and
within every cycle the elements can be cyclically permuted (i
cyclic permutations for each one of the ai cycles).
Example Let n = 4, and consider the type (2, 1, 0, 0). The
permutation (1) (3) (24) comes from 4 = 2!. 21 permutations,
namely,
1324

(1) (3) (24)
1342

(1) (3) (42)
3124

(3) (1) (24)
3142

(3) (1) (42)
The signless Stirling numbers of the
first kind
Denote by c(n,k) the number of permutations in Sn which can
be decomposed into k disjoint cycles. The numbers c(n,k) are
called the signless Stirling numbers of the first kind.
We have:
•  c(0,0) = 1 (convention);
•  c(n,0) = 0 for n > 0;
•  c(n,1) = (n–1)!;
•  c(n,n) = 1;
⎛n ⎞
•  c(n,n–1) = ⎜ ⎟.
⎝ 2 ⎠
€
Theorem For every n, k ≥ 0 we have:
c(n +1,k) = n c(n,k) + c(n,k − 1).
Proof Consider a permutation in Sn+1 with k cycles; the element
n+1 can
€ be a fixed point, and hence it is a cycle of length 1;
then, if we remove it, we get a permutation in Sn with k–1 cycles.
Otherwise, the element n+1 belongs to a cycle of length at least
2. If we remove this element we get a permutation in Sn with k
cycles. In this last case, the same permutation in Sn is obtained
n times.
Example
The permutation in S8 whose cycle decomposition is
(5 6 8) (2 4 3) (1 7)
can be obtained by deleting the symbol 9 from the following 8
permutations in S9:
(5 9 6 8) (2 4 3) (1 7)
(5 6 8 9) (2 4 3) (1 7)
(5 6 9 8) (2 4 3) (1 7)
(5 6 8) (2 9 4 3) (1 7)
(5 6 8) (2 4 9 3) (1 7)
(5 6 8) (2 4 3 9) (1 7)
(5 6 8) (2 4 3) (1 9 7)
(5 6 8) (2 4 3) (1 7 9)
The table of the signless Stirling
numbers of the first kind
k
n
0
1
2
3
4
5
1
1
2
0
1
3
0
0
1
4
0
2
3
1
5
0
6
11
6
1
6
0
24
50
35
10
7
0
720 1764 1624 735 175
6
7
21
1
1
Theorem For every positive integer n we have:
n
x(x +1)L(x + n − 1) =
∑ c(n,k) x k .
k =0
Proof Denote by b(n,k) the coefficient of xk in the polynomial
pn(x)
€ = x(x+1)...(x+n–1). We have of course
b(n,0) = 0 for every n > 0, and b(0,0) = 1. Moreover, since
pn(x) = (x+n–1) pn–1(x),
the coefficients b(n,k) satisfy the same recurrence as the
numbers c(n,k), namely
b(n+1,k) = b(n,k–1) + n b(n,k).
This implies that the two sequences coincide.
Corollary For every n > 0 the polynomial
c(n,1) x + c(n,2) x2 + ... + c(n,n) xn
has only real zeros, hence, the sequence of its coefficients is
log-concave.
The Stirling numbers of the first kind
The Stirling numbers of the first kind s(n,k) are defined to be
the coefficients in the expansion of the polynomials (x)n with
respect to the basis xk:
(x)n =
n
∑ s(n,k) x k .
k =0
Since, of course,
(x)n+1
€ = (x–n) (x)n = x (x)n – n (x)n,
the sequence s(n,k) satisfies the recurrence
s(n +1,k) = s(n,k − 1) − n s(n,k).
Comparing this with the recurrence satisfied by the sequence
c(n,k), and considering the initial values, we get
€
s(n,k) = (−1)n−k c(n,k).
€
Theorem Consider the generating function
n
F(x,u) =
∑
un
∑ s(n,k)x n! .
k
n≥0 k =0
We have:
€
F(x,u) = (1+u)x.
Proof The coefficient of un in F(x,u) is
n
1
(x)n ⎛ x ⎞
k
s(n,k) x =
= ⎜ ⎟,
∑
n!
n! ⎝ n ⎠
k =0
and this is precisely the coefficient of un in (1+u)x.
€
Recall that, for every n, we have
n
x =
n
∑ S(n,k) (x)k ,
k =0
where S(n,k) are the Stirling numbers of the second kind.
Hence, we
€ get :
⎧1 if n = m
∑ s(n,k)S(k,m) = ⎨
⎩0 if n ≠ m
k
€
Permutations as product of transpositions
Recall that every permutation can be written as product of
transpostions (= permutations consisting of one cycle of
length 2 and n–2 cycles of length 1).
There are many ways of writing a permutation as a product of
(not necessarily disjoint) transpositions. We want to determine
the number of such minimal representations.
We can restrict our attention to cyclic permutations, recalling
that a cycle of length k is the product of (at least) k–1
transpositions.
First of all, we can determine a criterion that allows us to
decide whether the product of n–1 transpositions is a cycle of
length n.
Let t1, t2, ..., tq be transpositions of Sn.
We associate to the q-tuple (t1, t2, ..., tq) a graph G(t1, t2, ..., tq)
with vertex set {1, 2, ..., n} and q edges, as follows: two vertices
a, b are adjacent in the graph if there exists an index such that
ti = (a, b).
Conversely, given a graph with n vertices with labels from 1 to n
and q edges, we can associate to it a set of q transpositions in
Sn, and hence q! (not necessarily distinct) permutations
obtained by multiplying the q transpositions, in every possible
order.
Examples
1. Consider the 5 transpositions in S6:
t1 = (1, 2), t2 = (1, 4), t3 = (1, 5), t4 = (2, 3),
t5 = (3, 4).
The graph G(t1, t2, t3, t4, t5) is:
2
1
3
6
4
5
2.  Consider now
t1 = (1, 3), t2 = (1, 4),
t3 = (2, 5),
t4 = (2, 6),
The graph G(t1, t2, t3, t4, t5) is:
1
2
3
6
4
5
t5 = (3, 4),
Recall that a tree is a connected graph without circuits. As a
consequence, a tree with n vertices has n–1 edges.
Theorem Let t1, t2, ..., tn–1 be distinct transpositions in Sn. The
product σ = t1 t2 ... tn–1 is a cyclic permutation if and only if the
graph G(t1, t2, ..., tn–1) is a tree.
Proof Suppose σ is a cycle. Then for every a = 2, ..., n there
exists an integer k such that σ k(1) = a. This implies that in the
graph there is a path (= a sequence of adjacent edges) from 1
to a, hence, the graph is connected. Since G has n–1 edges, it
is a tree.
Conversely, suppose that the graph G(t1, t2, ..., tn–1) is a tree.
We prove that the permutation σ is a cycle by induction on n.
For n = 2 the assertion is trivially verified. Suppose now the
assertion true for n–1. Consider a leaf a (= a vertex incident
with a single edge) of the graph G(t1, t2, ..., tn–1). Deleting from
G the vertex a and the unique edge incident with it
(corresponding to some transposition tk = (a, b)) we obtain a
tree G’ with n–1 vertices and n–2 edges. By induction
hypothesis, this tree corresponds to a cyclic permutation τ in
Sn–1. Since a is a leaf of G, there is a unique transposition
among t1, t2, ..., tn–1 containing a, namely, tk. This implies that
the product τ tk is a cycle of length n.
Example Consider the 5 transpositions in S6:
t1 = (1, 2),
t2 = (1, 4),
t3 = (1, 6),
t4 = (2, 3),
The graph G(t2, t3, t4, t5, t6) is a tree:
1
4
6
2
3
5
and the product
t1 t2 t3 t4 t5 = 3 4 5 6 2 1 = (135246)
is a cycle.
t5 = (2, 5).
Corollary Let t1, t2, ..., tn–1 distinct transpositions in Sn. If the
product t1 t2 ... tn–1 is a cyclic permutation, then every other
product tτ(1) tτ(2) ... tτ(n–1) is a cyclic permutation, for every
index permutation τ.
Example Consider the transpositions of the preceding
example:
t1 = (1, 2), t2 = (1, 4), t3 = (1, 6), t4 = (2, 3),
t5 = (2, 5).
The associated graph is the tree
1
4
6
2
3
5
Multiplying the same transpositions in a different order, we get
for example the cycle
t3 t2 t4 t5 t1 = 6 3 5 2 1 4 = (164235).
We need to know the number of distinct trees with n vertices
labelled 1, 2, ..., n.
Theorem (Cayley) Foe every n ≥ 2 the number of trees with
vertex set [n] = {1, 2, ..., n} is nn–2.
Proof Denote by Tn the number of trees with vertex set [n]. We
count in two different ways the number of sequences of edges
that can be added to the empty graph to obtain a rooted tree (=
a tree with a selected vertex, called root). We can costruct one
of such sequences by choosing one of the Tn trees with n
vertices, choosing the root (n choices), and choosing one of
the (n–1)! possibile sequences of adding the edges of the
chosen graph. Hence, n! Tn possible choices.
Another way of counting the rooted trees is adding step by
step an edge to the empty graph, counting the number of
choices available at each step. Suppose we have chosen n–k
edges. At this point, the graph we obtained is certainly without
circuits, but not necessarily connected: it will be the union of k
rooted trees. At this step we have n(k–1) possible choices for
the next edge: its first vertex v can be chosen among the n
available vertices, and the second vertex r can be chosen
among the roots of the trees not containing the vertex v (k–1
choices). After adding the edge (v, r), we obtain the union of
k–1 trees: the tree containing v and the tree containing r are
now the same. We choose the root of the tree containing v to
be the root of the “new” tree.
Multiplying the number of choices we have at the first, second
step,. etc., we obtain the total number of choices, namely,
n
∏ n(k − 1) = nn−1(n − 1)! = nn−2 n!.
k =2
This gives the assertion.
€
Example The Tn trees for the first values of n:
n=2
1
n=3
2
1
2
1
2
3
3
3
1
2
n=4
1
4
1
4
1
3
1
2
2
3
3
2
2
4
3
4
2
4
2
3
2
4
2
1
1
3
1
4
3
1
3
4
3
4
3
2
3
1
3
4
1
2
1
4
2
4
2
1
1
4
2
4
4
2
4
3
1
3
1
3
1
2
2
3
Theorem There are nn–2 ways of writing a given cyclic
permutation in Sn as the product of n–1 transpositions.
Proof We can consider without loss of generality the cyclic
permutation (12...n). We will define a bijection between the set
of all trees with vertex set [n] and the set of (n–1)-tuples of
transpositions (t1, t2, ..., tn–1) such that
t1 t2 ... tn–1 = (12...n).
Let T be a tree with vertex set [n]. For every vertex i = 2,3,...,n
there exists in T a path from 1 to i, whose last edge is (a, i).
Denote by si the transposition (a i), and set τ = s2 s3 ... sn. By a
previous result τ is a cyclic permutation.
Define now a new permutation π by setting for every k =
1,2,...,n,
π(k) = τk–1(1).
It is easily checked that π τ π–1 = (12...n).
For every i set
ti = π si π–1.
We have:
•  t2, t3, ..., tn are transpositions;
•  t2 t3 ... tn = π τ π–1 = (12...n).
We have defined a function Φ that maps a tree T to an (n–1)
-tuple (t2, t3, ..., tn) of transpositions such that
t2 t3 ... tn = (12...n).
We must now prove that Φ is a bijection. To do this, we observe
that, given a tree T with vertex set [n], the map Φ associates
with T an (n–1)-tuple of transpositions (t2, t3, ..., tn) such that t1
t2 ... tn–1 = (12...n) and a permutation π previously defined. Note
that T is exactly the tree G(s2, s3, ..., sn) associated with the set
of transpositions {s2, s3, ..., sn}, with si = π–1ti π.
This implies that, if we modify the labels of the vertices of T
according to the permutation π–1, we get the tree associated
with {t2, t3, ..., tn}. Hence, the map Φ is injective.
Given now an (n–1)-tuple of transpositions (t2, t3, ..., tn) with
t1 t2 ... tn–1 = (12...n), consider the tree G(t2, t3, ..., tn). For every i
= 2, 3, ..., n, let xi be the further vertex of the edge ti in the path
from the vertex 1 to the vertex i in G. We define a permutation
ξ in Sn by setting
ξ(1) = 1,
ξ(xi) = i for i = 2, 3, ..., n.
Denote by T the tree obtained from G by relabelling its vertices
according to the permutation ξ. We have Φ(T) = (t2, t3, ..., tn).
In fact, the permutation ξ coincides with the permutation π
associated with T, since the edges of T correspond to the
transpositions ξ–1ti ξ.
Example Consider the tree T:
3
4
6
2
5
1
7
8
The map associates with T the transpositions
t2 = (14), t3 = (24), t4 = (34), t5 = (16), t6 = (56), t7 = (17), t8 = (1 8),
and the permutation
π = 1 3 4 2 6 5 7 8, with π–1 = 1 4 2 3 6 5 7 8.
The tree G(t2, t3, ..., t8) is
2
3
5
4
6
1
7
8
and it is obtained from T by relabelling the vertices according to
π–1.
Multisets
Let U be a non-empty set. A multiset in U of length n is a map
m :U → N
such that
∑m(x) = n.
x∈U
€
For every x in U, the value m(x) is called the multiplicity of x in
m.
€ of m is defined to be the set
The support
Supp(m) = {x ∈ U; m(x) > 0}.
If m is a multiset in U of length n, with
Supp(m) = {x1, x2, ..., xk},
the following notation is often used:
h
h
m = {x1 1 ,x2 2 ,K,xk
hk
},
where hi = m(xi), or also,
€
m = {x1,K,x1, x2 ,K,x2 ,K,xk ,K,xk }.
1
424
3 1424
3
1424
3
h1 times
h2 times
k
We have of course
∑ hi = n.
i=1
€
€
hk times
Theorem Let U be a finite set of cardinality s. The number of
multisets in U of length n is
⎛s + n − 1⎞
⎜
⎟.
⎝ n ⎠
Proof Without loss of generality, suppose U = {1, 2, ...., s}.
Given a multiset in U of length n, m = {x1, x2, ..., xn}, with
€
x1 ≤ x2 ≤ ... ≤ xn, for every i set
yi = xi + i – 1.
The elements y1, y2, ..., yn are distinct, hence they are a subset
of {1, 2, ...., s+n–1}.
So, we have defined a bijection between the family of multisets
in U of length n and the family of subsets of {1, 2, ...., s+n–1}
of cardinality n. This gives the assertion.
Permutations of multisets
and multinomial coefficients
Let
h
h
h
m = {x1 1 ,x2 2 ,K,xk k }
be a multiset in U = {1, 2, ..., s} of length n.
A permutation of m is a list of length n in the set U where the
symbol
€ x1 appears h1 times, the symbol x2 appears h2 times,
and so on.
It is immediately seen that the number of permutations of m is
n!
(*)
.
h1!h2!Lhk !
The integer (*) is called multinomial coefficient, and is denoted
as
⎛
⎞
n
n!
€
= ⎜
⎟.
h1!h2!Lhk ! ⎝ h1,h2 ,K,hk ⎠
We have
€
⎛ n ⎞ ⎛ n ⎞
⎜
⎟ = ⎜ ⎟,
h
,h
⎝ 1 2 ⎠ ⎝ h1⎠
⎛
⎞ ⎛ n ⎞ ⎛n − h1⎞ ⎛n − h1 − h2 −K − hk −1⎞
n
⎜
⎟ = ⎜ ⎟ ⎜
⎟K ⎜
⎟.
h
,h
,K,h
h
⎝ 1 2
hk
⎠
k ⎠ ⎝ 1⎠ ⎝ h2 ⎠ ⎝
€
Given a permutation p = y1 y2 ... yn of the multiset m, we define
an inversion of p to be a pair of integers (i, j) with i < j and
yi > yj.
For example, p = 3 2 2 1 4 has the 5 inversions (1, 2), (1, 3),
(1, 4), (2, 4), (3, 4).
We want to determine an expression for the generating
function
∑ q inv(p),
p∈Sm
where Sm denotes the set of permutations of the multiset m.
To this €
aim, we need the notion of Gaussian coefficient.
Gaussian coefficients
Let q be a variable. For every positive integer n, set
2
[n]q = 1+ q + q +K + q
n−1
1− q n
=
,
1− q
([n]q is called a Gaussian integer), and
€
[n]q != [n]q [n − 1]q L[1]q =
(1− q n )(1− q n−1)L(1− q)
(1− q)
n
,
(Gaussian factorial).
€ Note that, for q = 1, the polynomial [n]q takes the value n, and
the Gaussian factorial [n]q! takes the value n!.
We now define the (n, k)-th Gaussian coefficient (or q-binomial
coefficient) to be
⎧(1− q n )L(1− q n−k +1)
⎪⎪
if k ≤ n
[n]q !
⎡n⎤
k
.
⎢⎣k ⎥⎦ = [k] ![n − k] ! = ⎨ (1− q )L(1− q)
q
q
q
⎪
⎪⎩
0
if k > n
Example
€
⎡4⎤
(1− q 4 )(1− q 3 )
= (1+ q 2 )(1+ q + q 2 ) = 1+ q + 2q 2 + q 3 + q 4 .
⎢ ⎥ =
⎣2⎦q (1− q 2 )(1− q)
Properties:
⎡n⎤ ⎡ n ⎤
⎢⎣k ⎥⎦ = ⎢⎣n − k ⎥⎦ ;
q
q
⎡n⎤
⎡n⎤
=
1=
⎢⎣0⎥⎦
⎢⎣n⎥⎦ ;
q
q
⎡n⎤
⎡ n ⎤
=
[n]
=
q ⎢
⎢⎣1⎥⎦
⎥⎦ ;
n
−
1
⎣
q
q
⎡n⎤
⎡n − 1⎤ ⎡n − 1⎤
k ⎡n − 1⎤
n−k ⎡n − 1⎤
=
q
+
=
+
q
⎢⎣k ⎥⎦
⎢⎣ k ⎥⎦ ⎢⎣k − 1⎥⎦ ⎢⎣ k ⎥⎦
⎢⎣k − 1⎥⎦ .
q
q
q
q
q
€
The Gaussian coefficients are apparently rational functions in q,
bt they are indeed polynomials with integer coefficients.
In fact, we have:
Theorem Let n, k be non negative integers, with k ≤ n. Denote
by an,k,i the number of subsets of cardinality k of the set
{1, 2, ..., n} such that the sum of their elements equals
⎛k +1⎞
i + ⎜
⎟.
⎝ 2 ⎠
Then
€
k(n−k)
⎡n⎤
i
⎢ ⎥ = ∑ an,k,i q .
⎣k ⎦q
i=0
Proof Let S be a subset of cardinality k of {1, 2, ..., n} such
that the sum of its elements is i + ½ k(k+1).
€ not contain the symbol n, it corresponds to a subset of
If S does
{1, 2, ..., n–1} of cardinality k such that the sum of its elements
is i + ½ k(k+1).
Otherwise, S corresponds to a subset of {1, 2, ..., n–1} of
cardinality k–1 such that the sum of its elements is
i + ½ k(k+1) – n = k–n+i + ½ k(k–1).
This implies that
an,k,i = an–1,k,i + an–1,k–1,k–n+i .
Set now
k(n−k)
An,k (q) =
∑
an,k,i q i .
i=0
The preceding considerations yield the following recurrence for
the polynomials An,k(q):
€
An,k(q) = An–1,k(q) + qn–k An–1,k–1(q),
and this is the recurrence satisfied by the Gaussian coefficients.
Since the inital values of the two sequences are the same, the
assertion is true.
Example Let n = 4, k = 2. Consider all the 2-element subsets
of the set {1,2,3,4}. Two of these sum up to 5, namely,{1,4} and
{2,3}, while all the other 4 sum up to different numbers. Hence,
A4,2(q) = 1 + q + 2q2 + q3 + q4,
and this is precisely
⎡4⎤
⎢ ⎥ .
⎣2⎦q
€
Gaussian coefficients have an interesting linear algebra
interpretation:
Theorem Let q be a prime power, and let V be a vector space
of dimension over the Galois field GF(q). The number of vector
subspaces of V of dimension k is
⎡n⎤
⎢ ⎥ .
⎣k ⎦q
Proof First of all, we count the number of k-tuples of linearly
independent vectors in V. The first vector can be any non-zero
€ n–1 choices). The second vector can be any
vector of V (q
vector, except for a mutiple of the first one (qn–q choices), and
so on. In conclusion, the number of k-tuples of linearly
independent vectors in V is
(qn–1) (qn–q) ... (qn–qk–1).
Any one of these k-tuples is a basis of a subspace of V of
dimension k, and this subspace has (qk–1) (qk–q) ... (qk–qk–1)
different bases, by the same argument.
Hence, the number of such subspaces is
(q n − 1)L(q n − q k −1)
(q k − 1)L(q k − q k −1)
€
=
(1− q n )L(1− q n−k +1)
(1− q k )L(1− q)
⎡n⎤
= ⎢ ⎥ .
⎣k ⎦q
It is also possible to define the multinomial Gaussian
coefficients: if h1, h2, ..., hk are integers such that h1 + h2 + ...,
hk = n, we set
⎡
⎤
[n]q !
n
=
.
⎢
⎥
⎣h1,h2 ,K,hk ⎦q [h1]q ![h2 ]q !L[hk ]q !
h
h
h
Theorem Let m = {1 1 ,2 2 ,K,k k } be a multiset of length n,
€
and let Sm be the set of its permutations.
We have
⎡
⎤
n
€ ∑ q inv(p) = ⎢
⎥ .
⎣h1,h2 ,K,hk ⎦q
p∈S
m
Proof Consider first the case k = 2. In this case, m contains
h€
1 copies of the symbol 1 and h2 = n – h1 copies of the symbol
2. An inversion will be simply an occurrence of the symbol 2
that appears before an occurrence of the symbol 1.
We proceed by induction on n: for n = 1 the assertion is trivially
verified. Suppose the assertion true for n–1, and consider a
multiset m of length n in {1, 2}. A permutation of m can either
end by 2, and in this case its last element is involved in no
inversions, or by 1, and its last element is involved in
h2 = n – h1 inversions.
Then, by the induction hypothesis, we have
∑q
p∈Sm
inv(p)
⎡n − 1⎤
⎡
⎤
n−h1 n − 1
= ⎢
⎥ + q
⎢
⎥ .
⎣ h1 ⎦q
⎣h1 − 1⎦q
∑ q inv(p)
This implies that the polynomials
€
p∈Sm
satisfy the same recurrence
€ as
⎡ n ⎤ ⎡ n ⎤
⎢ ⎥ = ⎢
⎥ ,
⎣h1⎦q ⎣h1,h2 ⎦q
with the same initial conditions. Hence, the assertion is true.
We now prove the general case by induction on k. Suppose
€ the multiset m = {1h1 ,2h2 ,K,k hk }
that the assertion is true for
and consider the multiset m+ = {1h1 ,2h2 ,K,k hk ,(k +1)hk+1 }.
Note that every permutation p of m+ is completely determined
by the pair (p’,p”), where p’ is the permutation obtained from p
€ different from k+1, and p” is
by replacing by 1 all the elements
€
obtained from p by deleting all the occurrences of k+1.
For example, if m+ = {12, 23, 3, 42}, and p = 21243142, we have
p’ = 11141141, p” = 212312.
It is evident that the permutations p of m+ correspond bijectively
to the pairs (p’, p”), and that
inv(p) = inv(p’) + inv(p”).
Hence, denoting by 2 the multiset consisting of h1 + h2 + ..., hk
copies of 1 and hk+1 copies of 2, we get
∑ q inv(p) = ∑ q inv(pʹ′) ∑ q inv(pʹ′ʹ′).
p∈S
m+
pʹ′∈S2
pʹ′ʹ′∈Sm
⎡
⎤
€The first summand is ⎢ n ⎥ , as we have shown.
⎣hk +1⎦q
By the induction hypothesis, the second summand is
€
⎡ n − hk +1 ⎤
⎢
⎥ .
⎣h1,h2 ,K,hk ⎦q
The product of these two quantities is precisely
€
⎡
⎤
n
⎢
⎥ .
⎣h1,h2 ,K,hk ,hk +1⎦q
€