The Art of Counting

David M. Bressoud

July 24, 2006

1 Introduction

Counting is the most basic mathematical activity we do, yet it can also be one of the most complex.It lies at the heart of any calculation of probabilities with discrete objects. Finding the number ofconfigurations that satisfy a given set of rules is a recurring problem that one finds in chemistry,in physics, and in almost every discipline of mathematics. Following a short refresher on binomialcoefficients, this paper will focus on two of my favorite counting problems, problems that illustratethe power of recursion:

1. Slicing cheese: Given a large block of cheese, how many pieces can we get with six straightslices?

2. Square ice: How many ways can we place 25 water molecules (H2O) into a square lattice sothat each oxygen atom along the top or bottom is adjacent to three hydrogen atoms, and allother oxygen atoms are surrounded by exactly four hydrogen atoms?

The title of this article is borrowed from a book of selected writings by Paul Erdos [9], one of thegreat counters of all time.

2 Binomial Coefficients

Recursion is a basic technique of counting. Any formula that connects previous calculations tothe next is called “recursive.” A good example of this is given by binomial coefficients such as

(52

)which represents the number of ways of choosing two objects from a selection of five. If our objectsare the five letters {A,B, C, D, E}, then the possible choices can be listed as

AB, AC, AD, AE,BC, BD, BE, CD, CE, DE.

I have intentionally separated these. The first line consists of those choices that include the letterA, the second line those that do not include A. If we include A, then there are four ways of choosing

1

0

0

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$%&= 1

1

0

!"#

$%&= 1

1

1

!"#$%&= 1

2

0

!"#

$%&= 1

2

1

!"#

$%&= 2

2

2

!"#

$%&= 1

3

0

!"#

$%&= 1

3

1

!"#

$%&= 3

3

2

!"#

$%&= 3

3

3

!"#

$%&= 1

4

0

!"#

$%&= 1

4

1

!"#

$%&= 4

4

2

!"#

$%&= 6

4

3

!"#

$%&= 4

4

4

!"#

$%&= 1

5

0

!"#

$%&= 1

5

1

!"#

$%&= 5

5

2

!"#

$%&= 10

5

3

!"#

$%&= 10

5

4

!"#

$%&= 5

5

5

!"#

$%&= 1

+ + + +

+ + +

+ +

+

Figure 1: The recursion that generates Pascal’s triangle.

because there are four other letters, and we must select one of them,(

41

)= 4. If our choice does

not include A, then we must pick two letters from the remaining four:(

42

)= 6.

This idea works in general. Given n letters from which we must choose k, either we do choose thefirst letter, leaving k − 1 choices from n − 1 letters, or we do not choose the first letter, leaving kchoices from n− 1 letters: (n

k

)=

(n− 1k − 1

)+

(n− 1

k

).

When we combine this recursive formula with the observation that there is always one way ofchoosing nothing and one way of choosing everything, we can build Pascal’s triangle, obtainingeach row from the row above by adding pairs of adjacent entries (see Figure 1).

This triangle carries Pascal’s name because of the book he wrote in 1654, Treatise on the Arith-metical Triangle, that both explained and popularized this triangular arrangement of binomialcoefficients, but it is much older. Figure 2 shows a page from Siyuan yujian by the Chinese math-ematician Zhu Shijie, written in 1303. He refers to it as an ancient method. The earliest reliablereference is to a work of Jia Xian from around 1100. There are hints that it might have been knownas much as a hundred years earlier in China, in India, and/or in the Islamic Baghdad-Cairo axis.

2

Figure 2: “Pascal’s” triangle from Siyuan yujian, by Zhu Shijie, written in 1303. Reprinted from[5, p. 135].

3 Slicing Cheese

How many regions in space can we get if we cut space by six planes? This, our first countingproblem, was popularized by George Polya in the short film he created for the MathematicalAssociation of America in 1965, Let Us Teach Guessing [7]. The first published solution was byJakob Steiner [10] in 1826. See also articles by Alexanderson and Wetzel [1, 2]. This problem isrelated to the one studied by Jean Pederson in an earlier Mathematical Adventure [6], but in hercase, many of the planes were parallel. Here we want no two planes to be parallel and no morethan three planes to meet in any single point. In other words, we want to get as many regions aswe can.

The situation starts out simply (see Figure 3):

0 planes: 1 region1 plane: 2 regions2 planes: 4 regions3 planes: 8 regions.

But the fourth plane does not leave us with 16 regions, but rather only 15.

To understand what happens, it is useful to back up and look at a simpler situation. Simpler thanspace cut by planes would be a plane cut by lines, but we are going to back up even further, tothe number of pieces of a line cut by points. An uncut line consists of one piece. Each new cutpoint increases the number of pieces by one. So a line cut by n points has n + 1 pieces. We now

3

Figure 3: Space sliced by three planes.

consider a plane cut by lines. The first line gives us one additional piece. The second line givesus two additional pieces, and the third line gives us three additional pieces. A fourth line gives usfour new pieces, and so on. When we add a new line, say the nth line, it cuts each of the previousn− 1 lines. The new line is cut into n pieces. Each of those pieces cuts one region in the plane intotwo regions (see Figure 4). Since the nth line adds n new regions, the total number of regions ona plane cut by n lines is

1 + 1 + 2 + 3 + · · ·+ n.

What happens when our fourth plane cuts through space? It meets each of the previous threeplanes. Each of the previous planes cuts the new plane along a line. Our new fourth plane is cutby three lines. It will be cut into seven pieces, each of which creates one new region in space. Wehad eight regions in space. The fourth plane adds seven new regions. We now have fifteen regions.You should be able to see the pattern that enables us to construct the following table:

n line by point plane by line space by plane0 1 1 11 2 2 22 3 4 43 4 7 84 5 11 155 6 16 266 7 22 ?

I leave it to you to determine the number of regions in space cut by six planes.

Something else is going on here. You should be very suspicious that each entry in this table is thesum of the entry directly above and the entry above and to the left. That is exactly the recursion

4

Figure 4: A fifth line cuts through a plane already cut by four lines.

of Pascal’s triangle. If we list of elements of Pascal’s triangle so that they line up according to thisrecursion, we can see what is happening:

n line/point plane/line space/plane k = 0 = 1 = 2 = 3 = 4 = 5 = 60 1 1 1 11 2 2 2 1 12 3 4 4 1 2 13 4 7 8 1 3 3 14 5 11 15 1 4 6 4 15 6 16 26 1 5 10 10 5 16 7 22 ? 1 6 15 20 15 6 1

The number of pieces on a line cut by points is the sum of the first two columns of Pascal’s triangle.For a plane cut by lines, it is the sum of the first three columns. The formula for the number ofpieces in space cut by n planes is (n

0

)+

(n

1

)+

(n

2

)+

(n

3

).

This leaves several questions that I invite you to explore:

1. Can you make sense of this formula? Given n planes in space, why should the number ofregions equal the number of ways choosing none of the planes plus the number of ways ofchoosing one of the planes plus the number of ways of choosing two of the planes plus thenumber of ways of choosing three of the planes?

2. What is the formula for the number of finite regions formed by n planes?

3. What happens in higher dimensions, and what does that mean?

5

H

O

H

O

H

O

H

O

H

O

H

H

H H

H

H

H

O

H

O

H

O

H

O

H

O

H

H

H H

H

H

H

O

H

O

H

O

H

O

H

O

H

H

H H

H

H

H

O

H

O

H

O

H

O

H

O

H

H

H H

H

H

H

O

H

O

H

O

H

O

H

O

H

Figure 5: A 5× 5 sheet of square ice.

4 Square Ice

A sheet of square ice (see Figure 5) is a grid of oxygen atoms (O) and hydrogen atoms (H). Exceptalong the top and bottom rows, each oxygen atom is surrounded by four hydrogen atoms. Alongthe top and bottom, there are three hydrogen atoms next to each oxygen atom. If we have a squarearrangement of oxygen atoms, then there will be exactly twice as many hydrogen atoms as oxygenatoms. Each oxygen atom must connect to exactly two of the neighboring hydrogen atoms. Howmany ways can this be done?

As stated, this problem is artificial. It is extremely difficult to get water to freeze into such a squarelattice. But other materials do form square lattices, and this model is a useful description forother physical phenomena. David Robbins, a mathematician at the Institute for Defense Analysis,stumbled upon this question in 1980 when studying an algorithm for evaluating determinants, analgorithm that had been devised by Charles Dodgson [4], better known as Lewis Carroll, the authorof Through the Looking Glass and Alice’s Adventures in Wonderland . Robbins own account of hisdiscovery can be found in [8].

We shall follow Robbins’s approach and start by changing this into another, equivalent problem.In each row, we record the molecules for which the oxygen atom is not connected to the hydrogenatom immediately below. We have one such molecule in the first row, two in the second, and soon. For the sheet of square ice in Figure 5, this is molecule 2 in the first row, molecules 1 and 4 inthe second row, and molecules 1, 2, and 5 in the third row. This sheet of square ice corresponds tothe triangle:

6

21 4

1 2 51 2 4 5

1 2 3 4 5

Which triangles correspond to sheets of square ice? If a number appears in one row and not inthe row below, then the oxygen atom in the row from which it disappeared must be connected tothe hydrogen atoms directly above and below. We call this a vertical molecule. If a numberdoes not appear in one row but does appear in the row below, the the oxygen atom in the row inwhich it appears must be connected to the hydrogen atoms to the left and the right. We call this ahorizontal molecule. Knowing where the horizontal and vertical molecules are located uniquelydetermines our triangle, and vice-versa.

Notice that in any row or column, any two vertical molecules must be separated by a horizontalmolecule. And any two horizontal molecules must be separated by a vertical molecule. Furthermore,once we have placed our horizontal and vertical molecules, all of the remaining molecular bondsare uniquely determined. If a number in one of the rows of our triangle does not appear in thenext row, then there must be two new numbers in that next row, one of which is smaller than thenumber that disappeared, one of which is larger. In general, if k numbers in a given row do notappear in the next row, then k + 1 new numbers must appear, and for the jth number from theleft that disappears, 1 ≤ j ≤ k, there will be j new smaller numbers that appear, k − j + 1 newlarger numbers that appear.

This is the same as saying that for any three adjacent numbers in the triangle,

ab c

we always have b ≤ a ≤ c and b < c. These inequalities guarantee that there will be a correspondingsheet of square ice. We call a triangle that meets these conditions a monotone triangle. Ourexample is a monotone triangle of size 5.

Now, how many monotone triangles of size 5 are there?

We start by listing all of the possible rows of each length (see Figure 6). Each monotone trianglecorresponds to a choice of rows. In the figure, the choices that correspond to our example aremarked as a path that starts at 12345 and ends at 2. How many such paths are there? Each pathmust go through one row of each length, but we are not free to take any row we want. The rows oflength k that are allowed will depend on the the row of length k+1 from which the path is coming.

For the row of length 4: 1245, there is only one path from 12345. But for the row of length 3: 125,we could go through 1235 or 1245. To get to 124, we could go through 1234, 1235, or 1245. Thereare four paths to 134, and five paths to 234.

Consider 14, a row of length 2. Any path to 14 must pass through 124, 125, 134, 135, or 145.

7

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

Figure 6: A choice of rows for the monotone triangles of size 5.

We can calculate the number of paths to each of these: 3, 2, 4, 3, and 2, respectively. There area total of 3 + 2 + 4 + 3 + 2 = 14 different paths from 12345 to 14. Figure 7 shows the numberof paths to each possible row. When we add up the number of paths to 1, 2, 3, 4, or 5, we get42 + 105 + 135 + 105 + 42 = 429. There are 429 possible monotone triangles of size 5.

This recursion can be programmed fairly easily (Mathematica code can be found in [3, pp. 62–63]or downloaded from www.macalester.edu/∼bressoud/books/PnC/prf-n-con.nb, section 2.3). IfAn is the number of monotone triangles of size n, then the first ten values are

A1 = 1,A2 = 2,A3 = 7,A4 = 42, (2 · 3 · 7),A5 = 429, (3 · 11 · 13),A6 = 7436, (22 · 11 · 132),A7 = 218348, (22 · 132 · 17 · 19),A8 = 10850216, (23 · 13 · 1722 · 1922),A9 = 911835460, (22 · 5 · 1722 · 193 · 23),

A10 = 129534272700, (22 · 3 · 52 · 7 · 17 · 193 · 232).

The factorizations of these numbers are included because they are quite remarkable and suggestthat something very interesting is happening. Most numbers as large as 129534272700 have largeprime divisors. The numbers we have found all have small prime divisors. There is hope that theremight be a nice formula for An.

David Robbins next step was to work not with the numbers An, the number of monotone trianglesof size n, but to use the values he had found for An,k, 1 ≤ k ≤ n, the number of monotone triangles

8

123 124 125 134 135 145 234 235 345 2

12345

1234 1235 1245 1345 2345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14 7

42 105 135 105 42

23 14 26 23 7 14 14 7

2 2 4 3 2 5 4

Figure 7: The number of configurations that can lie below each row of a monotone triangles of size5.

of size n with a k at the apex. As we have seen,

A5,1 = 42, A5,2 = 105, A5,3 = 135, A5,4 = 105, and A5,5 = 42.

We build a triangle of the values An,k:

11 1

2 3 27 14 14 7

42 105 135 105 42429 1287 2002 2002 1287 429

7436 26026 47320 56784 47320 26026 7436

Notice that the first entry in each row is the sum of the entries in the row above. This is easyto explain. A monotone triangle with a 1 at the apex must have all 1s down the left-hand edge.The number of ways of filling in the rest of the triangle is just the number of monotone trianglesof the next smaller size. What about the second entry in each row? If we look at the ratio of thesecond entry to the first, we see that the ratios increase by 1/2: 1, 3/2, 2, 5/2, 3, 7/2, . . .. There aremany representations for each of the other ratios, but Robbins realized that they could be writtenas shown below:

9

11 2/2 1

2 3/2 3 2/3 27 4/2 14 5/5 14 2/4 7

42 5/2 105 9/7 135 7/9 105 2/5 42429 6/2 1287 14/9 2002 16/16 2002 9/14 1287 2/6 429

7436 7/2 26026 20/11 47320 30/25 56784 25/30 47320 11/20 26026 2/7 7436

Take a moment to look at those ratios and see if you can pick out the pattern.

Once we know the values of An−1,k for 1 ≤ k ≤ n− 1, we know that

An,1 =n−1∑k=1

An−1,k, (1)

so we know the first entry in the next row. If we know the values of the ratios, then the first entryuniquely determines the remaining entries in that row. We can compute any value of An,k.

Isolating just the numerators of our ratios, we get yet another triangle,

23 2

4 5 25 9 7 2

6 14 16 9 27 20 30 25 11 2

The denominators are simply the vertical reflection of these numerators. Each numerator is the sumof the two numerators lying directly above it, which suggests that Pascal’s triangle has appearedonce again. In fact, the triangle of numerators can be represented as

1 + 11 + 2 1 + 1

1 + 3 2 + 3 1 + 11 + 4 3 + 6 3 + 4 1 + 1

1 + 5 4 + 10 6 + 10 4 + 5 1 + 11 + 6 5 + 15 10 + 20 10 + 15 5 + 6 1 + 1

Each numerator and each denominator is a sum of two binomial coefficients,

An,k+1

An,k=

(n−2

n−k−1

)+

(n−1

n−k−1

)(

n−2k−1

)+

(n−1k−1

) . (2)

We have not proven equation 2. In fact, its proof would turn out to be very difficult, not completeduntil 1996 [11]. But it should be clear that it must be true. Equipped with equations (1) and (2),we can find an explicit formula for An, the number of monotone triangles of size n.

10

We first express An in terms of An−1 and the ratios:

An = An,1 + An,2 + An,3 + · · ·+ An,n

= An,1

(1 +

An,2

An,1+

An,2

An,1· An,3

An,2+ · · ·+ An,2

An,1· · · An,n

An,n−1

)= An−1

(1 +

An,2

An,1+

An,2

An,1· An,3

An,2+ · · ·+ An,2

An,1· · · An,n

An,n−1

). (3)

Our next step is to simplify the ratio An,k+1/An,k,

An,k+1

An,k=

(n−2

n−k−1

)+

(n−1

n−k−1

)(

n−2k−1

)+

(n−1k−1

)=

(n−2)!(n−k−1)!(k−1)! + (n−1)!

(n−k−1)!k!

(n−2)!(n−k−1)!(k−1)! + (n−1)!

(k−1)!(n−k)!

=(n−2)!

(n−k−1)!k!(k + n− 1)(n−2)!

(n−k)!(k−1)!(n− k + n− 1)

=(n− k)(n + k − 1)

k(2n− k − 1). (4)

Now we simplify the product of ratios,

t∏k=1

An,k+1

An,k=

(n− 1)(n− 2) · · · (n− t) · n(n + 1) · · · (n + t− 1)1 · 2 · · · t · (2n− 2)(2n− 3) · · · (2n− t− 1)

=(n + t− 1)!(2n− t− 2)!t! (n− t− 1)! (2n− 2)!

=(n− 1)!2

(2n− 2)!· (n + t− 1)!

t! (n− 1)!· (2n− t− 2)!(n− t− 1)!(n− 1)!

=(n− 1)!2

(2n− 2)!·(

n + t− 1n− 1

)·(

2n− t− 2n− 1

)(5)

Combining this with equation (3), we have that

An = An−1

n−1∑t=0

(n− 1)!2

(2n− 2)!·(

n + t− 1n− 1

)·(

2n− t− 2n− 1

)

= An−1(n− 1)!2

(2n− 2)!

n−1∑t=0

(n + t− 1

n− 1

)·(

2n− t− 2n− 1

)(6)

11

We next prove that (3n− 22n− 1

)=

n−1∑t=0

(n + t− 1

n− 1

)·(

2n− t− 2n− 1

). (7)

The left side of this equation counts the number of ways of choosing 2n − 1 objects from a set of3n − 2. Let the set from which we are choosing consist of the integers {1, 2, 3, . . . , 3n − 2}. Whatis the nth integer that we choose? It cannot be smaller than n or larger than 2n − 1 (becausewe still have to choose n − 1 more integers). Let the nth integer that we choose be n + t where0 ≤ t ≤ n − 1. For this choice of t, we still need to choose n − 1 integers from first n + t − 1 andn− 1 integers from the last 2n− t− 2. That is precisely what the tth summand counts.

We now use equation (7) to finish the formula for An:

An = An−1(n− 1)!2

(2n− 2)!

(3n− 22n− 1

)= An−1

(n− 1)!2 (3n− 2)!(2n− 2)! (2n− 1)! (n− 1)!

= An−1(n− 1)! (3n− 2)!(2n− 2)! (2n− 1)!

. (8)

We know that A1 = 1, and so

An =1! · 4!2! · 3!

· 2! · 7!4! · 5!

· 3! · 10!6! · 7!

· · · (n− 1)! (3n− 2)!(2n− 2)! (2n− 1)!

=1! · 2! · · · (n− 1)! · 4! · 7! · · · (3n− 2)!

1! · 2! · 3! · · · (2n− 1)!

=4! · 7! · · · (3n− 2)!

n! · (n + 1)! · · · (2n− 1)!

=n−1∏j=0

(3j + 1)!(n + j)!

. (9)

And there it is! We have found a simple formula for the number of monotone triangles of size n.For example,

A5 =1!5!· 4!6!· 7!7!· 10!

8!· 13!

9!=

9 · 10 · 10 · 11 · 12 · 132 · 3 · 4 · 5 · 5 · 6

= 3 · 11 · 13 = 429.

For more on this problem and an explanation of the proof of equation (2), see [3]. Again, I leaveyou with several questions to explore:

1. In Figure 7, count the number of paths by starting at the top. For each possible row, findthe numbers of paths from 1, 2, 3, 4, or 5 to that row.

2. Every time we have a vertical molecule, its position is marked by an integer that appearsin one row of a monotone triangle but not in the row below. When we trace the path thatcorresponds to this monotone triangle, we will have a path segment for which the row at the

12

upper end has at least one integer that does not appear in the row at the bottom end. Weassign a weight to each path segment, equal to the number of integers in the row above thatdo not appear in the row below. The number of vertical molecules in the sheet of square iceis the sum of the weights along the corresponding path. In Figure 7, find all paths of weight4.

3. In Figure 7, instead of finding the total number of paths from 12345 to each row, break itdown so that at each row you record the number of paths of each possible weight. How manyof the monotone triangles of size 5 correspond to paths of weight 0, 1, 2, 3?

4. For any size monotone triangle, how many paths of weight 0 are there?

5. Let Mn denote the set of monotone triangles of size n. For each M in Mn, let w(M) be theweight of the corresponding path. Calculate the sum∑

M∈Mn

2w(M)

for n = 1, 2, 3, 4, and 5. Guess the general formula for this sum.

References

[1] Gerald L. Alexanderson and John E. Wetzel. Simple partitions of space. Mathematics Maga-zine, 51:220–225, 1978.

[2] Gerald L. Alexanderson and John E. Wetzel. Arrangements of planes in space. DiscreteMathematics, 34:219–240, 1981.

[3] David M. Bressoud. Proofs and Confirmations: The Story of the Alternating Sign MatrixConjecture. Mathematical Association of America, Washington, DC, 1999.

[4] Charles L. Dodgson. Condensation of determinants. Proceedings of the Royal Society, London,15:150–155, 1866.

[5] Joseph Needham. Science and Civilisation in China, volume 3. Cambridge University Press,1959.

[6] Jean Pederson. Platonic divisions of space. In David F. Hayes and Tatiana Shubin, editors,Mathematical adventures for students and amateurs, pages 117–133. Mathematical Associationof America, Washington, DC, 2004.

[7] George Polya. Let Us Teach Guessing. video. The Mathematical Association of America,Washington, DC, 1965.

[8] David P. Robbins. The story of 1, 2, 7, 42, 429, 7436, . . .. The Mathematical Intelligencer,13:12–19, 1991.

[9] Joel Spencer, editor. Paul Erdos: The art of counting, selected writings. MIT Press, Cambridge,MA, 1973.

13

[10] J. Steiner. Einige gesetze uber die theilung der ebene und des raumes. J. reine angew. Math.,1:349–364, 1826.

[11] Doron Zeilberger. Proof of the refined alternating sign matrix conjecture. New York Journalof Mathematics, 2:59–68, 1996.

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