Balls Into Boxes

7/30/2019 Balls Into Boxes

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Counting Permutations

by Putting Balls into Boxes

Ira M. Gessel

Brandeis University

C&O@40 Conference

June 19, 2007
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I will tell you shamelessly what my bottom line is: It is placing

balls into boxes.

Gian-Carlo Rota, Indiscrete Thoughts
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Eulerian polynomials

If is a permutation of [n] = {1, 2, . . . , n}, a descent of is ani, with 1 i n 1, such that (i) > (i + 1).

Example: 1 3 2 6 4 5 has two descents.

Let Sn be the group of permutations of [n]. How manypermutations in Sn have i descents? Let us define the Eulerian

polynomials

En(t) =

Sn

tdes().

Then E1(t) = 1, E2(t) = 1 + t, E3(t) = 1 + 4t + t2.
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Theorem.

k=1

kntk1 =

Sn

tdes()

(1 t)n+1

Proof. kn

is the number of placements of n balls, labeled 1, 2,. . . , n, into k boxes. We will associate to each placement a

permutation Sn so that the total contribution from istdes()/(1 t)n+1. We represent a placement

2 5 1 3 4

more compactly as

2 5

|1

| |3

|4

|The balls in each box are in increasing order.
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We remove the bars from the placement 2 5 | 1 | | 3 | 4 | to getthe permutation 2 5 1 3 4.

So we can think of a placement as a permutation with bars in it:a barred permutation.

Which barred permutations correspond to 2 5 1 3 4?
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Which barred permutations correspond to 2 5 1 3 4? We need

at least one bar in every descent, so we start with

2 5 | 1 3 4
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Which barred permutations correspond to 2 5 1 3 4? We need

at least one bar in every descent, so we start with

2 5 | 1 3 4

Then we put any number of additional bars in each of the 6

spaces to get 2 5 | 1 | | 3 | 4 | . We assign the weight t to eachbar. Then the contribution from this permutation is

t(1 + t + t2 + )6 = t(1 t)6
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In general, the contribution from a permutation of [n] is

tdes()

(1 t)n+1

so

k=1

kntk1 =

Sn

tdes()

(1 t)n+1 .

Note that if there are k boxes then there are k

1 bars.
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The Method of Barred Permutations

A barred permutation is a permutation of balls numbered 1 to n

with bars in it. Between (and before and after) the bars are

boxes and between (and before and after) the balls are spaces.

spaces

boxes

0

21

0

52

|1

1 |2

3

|4

3 |

4

3

4 |

5

5|6We can count barred permutations in two ways:

1) Start with bars, and put balls into boxes.

2) Start with a permutation, and put bars into spaces.

Note: The method of barred permutations is closely related to

the method of P-partitions (MacMahon, Knuth, Stanley).
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2-descents

We consider barred permutations in which consecutive balls

cannot be in the same box.

3 | 5 7 || 1 | 2 6 | 4

How many placements of n balls in k boxes are there?
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2-descents

We consider barred permutations in which consecutive balls

cannot be in the same box.

3 | 5 7 || 1 | 2 6 | 4

How many placements of n balls in k boxes are there?

Ball 1: k boxes

Ball 2: k 1 boxesBall 3: k 1 boxes

. . .

Ball n: k 1 boxes

So there are k(k 1)n1 placements.
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If we start with a permutation (for example, 4 1 2 5 3 6) we must

put a bar in each descent, but also in any space where m is

followed by m+ 1:
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followed by m+ 1:4 | 1 | 2 5 | 3 6
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followed by m+ 1:4 | 1 | 2 5 | 3 6

Then we put an arbitrary number of additional bars in each

space:

|| 4 | 1 || 2 5 | 3 || 6 |
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followed by m+ 1:4 | 1 | 2 5 | 3 6

Then we put an arbitrary number of additional bars in each

space:

|| 4 | 1 || 2 5 | 3 || 6 |We call i a 2-descent of if (i) + 2 > (i + 1). Then by thesame reasoning as before,

k=1

k(k 1)n1tk1 = S

n

t2-des()

(1 t)n+1 .
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We could define r-descents similarly: (i) + r > (i + 1). The

same reasoning would give

k=1

k(k1)(k2) (kr+2)(kr+1)nr+1tk1 =

Sntr-des()

(1 t)n+1 .

(Foata-Schtzenberger)
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We could define r-descents similarly: (i) + r > (i + 1). The

same reasoning would give

k=1

k(k1)(k2) (kr+2)(kr+1)nr+1tk1 =

Sntr-des()

(1 t)n+1 .

(Foata-Schtzenberger)

Note that k(k 1)(k 2) (k r + 2)(k r + 1)nr+1 is thechromatic polynomial of the graph with vertex set [n] in whichtwo vertices are adjacent if and only if they differ by at most

r 1; i.e., they are not allowed in the same box. A similar resultholds for the chromatic polynomial of any chordal graph.

Si d P t ti
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Signed Permutations

A signed permutation of [n] is permutation of [n] in which the

entries may have minus signs:

4 2 1 5 3

Its convenient to write i for i so well write this signedpermutation as

4 2 1 5 3

Sometimes its useful to think of a signed permutation as a

permutation of the set {n, n+ 1, . . . , n 1, n} (with orwithout 0) with the property that (

i) =

(i). We denote by

Bn the set (or group) of signed permutations of [n]. (This is thehyperoctahedral group, the Coxeter group of type Bn.)

Descents of signed permutations are defined as usual except

that if (1) < 0 then 0 is a descent of . (Think of (0) = 0.)
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Theorem. (Steingrmsson)

k=0

(2k + 1)n

tk

=

Bn

tdes()

(1 t)n+1
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Theorem. (Steingrmsson)

k=0

(2k + 1)n

tk

=

Bn

tdes()

(1 t)n+1

To prove this formula, we count barred signed permutations.

2 4|

3 1 5| |

6

In the first box (box 0) only positive numbers can appear but in

the other boxes, positive and negative numbers can appear.

How many barred permutations have k bars? For each of the n

balls, we can make it positive and put it in any of k + 1 boxes ormake it negative and put it in any of k boxes. So there are

(k + 1) + k = 2k + 1 possibilities for each ball, so (2k + 1)n inall.

Flag descents
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Flag descents

Adin, Brenti, and Roichman (2001) defined the flag-descent

number of a signed permutation by

fdes() = 2des ((1) < 0).In other words all descents are counted twice, except that a

descent in position 0 is counted only once. They proved

k=1

kntk1 =B

n

tfdes()

(1 t)(1 t2)n.

Flag descents
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Flag descents

Adin, Brenti, and Roichman (2001) defined the flag-descent

number of a signed permutation by

fdes() = 2des ((1) < 0).In other words all descents are counted twice, except that a

descent in position 0 is counted only once. They proved

k=1

kntk1 =B

n

tfdes()

(1 t)(1 t2)n.

Comparing this with our formula for Eulerian polynomials,

k=1

k

n

t

k1

=

Sn

tdes()

(1 t)n+1 ,we see that their result is equivalent to

Bn

tfdes() = (1 + t)n

Sn

tdes().

Their proof was by induction, but we can give a direct proof
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p y , g p

using barred permutations. Lets first go back to the

enumeration of signed permutations by descents. Instead of

looking at the sequence (1) (2) (n), lets look at

(n) (1) (0) (1) (n),

where (i) = (i) and in particular, (0) = 0. We consideronly symmetric" barred permutations, for example

| 1 || 3 2 | 0 | 2 3 ||1|

or

| 1 || 3 | 2 0 2 | 3 ||1|

These symmetric barred permutations have 2(n+ 1) spacesand 2k bars for some k. (So if we want to weight all the bars

equally, we should weight each one by

t rather than t).

If we know the right half of a such a barred permutation (the

part to the right of the 0) then the left half is determined. So

what have we gained?
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We can give a slightly different argument from before that the

number of barred permutations with 2k bars (corresponding to

k bars before) is (2k + 1)n. With 2k bars there are now 2k + 1spaces, so we put the numbers 1, 2,. . . , n arbitrarily into these

2k + 1 spaces, and then the locations of 1, 2, . . . , n (and0) are determined

| | | | | | | |
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| | | | | | | || | | 3 | 2 | | |1|
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| | | | | | | || | | 3 | 2 | | |1|

| 1 | | 3 | 2 0 2 | 3 | |1|

For flag descents we do the same thing but without 0 For
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For flag descents, we do the same thing, but without 0. For

example, if we start with the permutation 3 2 1, adding in its

negative half gives

1 2 3 3 2 1

There are now 7 spaces, rather than 8, and they are paired,

except for the central space. Then if we count the barred

permutations corresponding to this permutation (weighting

each bar with t), the bars in the n noncentral spaces come in

pairs, but the bars in the center space do not.

|| 2 | 1 3 ||| 3 1 | 2 ||

For flag descents we do the same thing but without 0 For
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For flag descents, we do the same thing, but without 0. For

example, if we start with the permutation 3 2 1, adding in its

negative half gives

1 2 3 3 2 1

There are now 7 spaces, rather than 8, and they are paired,

except for the central space. Then if we count the barred

permutations corresponding to this permutation (weighting

each bar with t), the bars in the n noncentral spaces come in

pairs, but the bars in the center space do not.

|| 2 | 1 3 ||| 3 1 | 2 ||

So the sum of the weights of the barred permutation

corresponding to a given permutation is

tfdes()

(1 t)(1 t2)n
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To get the other side of the equation, we note that the number

of bars need not be even; it can be any number. If there arek 1 bars, there are k boxes, and thus kn ways to put 1, 2, . . . ,n into the boxes, and as before the locations of 1, 2, . . . , n are

determined.
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To get the other side of the equation, we note that the number

of bars need not be even; it can be any number. If there arek 1 bars, there are k boxes, and thus kn ways to put 1, 2, . . . ,n into the boxes, and as before the locations of 1, 2, . . . , n are

determined.

So we have Adin, Brenti, and Roichmans identity

k=1

kntk1 =

Bn

tfdes()

(1 t)(1 t2)n.

We do not have a bijective proof of
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Bn

tfdes() = (1 + t)n

Sn

tdes().

However, our approach gives a refinement of this formula, by

telling us which permutations in Bn correspond to each

permutation in Sn.

For Sn, let B() be the set of 2n signed permutations in Bnobtained from by the following procedure:

First we cut into two parts:1425637 142 5637

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Bn

tfdes() = (1 + t)n

Sn

tdes().


telling us which permutations in Bn correspond to eachpermutation in Sn.



Next we reverse and negate the first part:

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Bn

tfdes() = (1 + t)n

Sn

tdes().


telling us which permutations in Bn correspond to eachpermutation in Sn.



Next we reverse and negate the first part:

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Finally we shuffle the two parts:256

4

137
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Then

B()

tfdes() = (1 + t)ntdes()
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Then

B()

tfdes() = (1 + t)ntdes()

Proof sketch. The set of barred permutations of (in Sn) is the

same as the set of barred permutations of elements of B() (inBn). Therefore

tdes()

(1 t)n+1 =

B() tfdes(

)

(1 t)(1 t2)n.

Example.

3 | 1 ||| 2 2 3 | 1 || 1 | 3 23 1 2 1 3 2
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Balls Into Boxes