Pascal’s Triangle is one of the most fascinating and intriguing objects in math, I was first introduced to it in the sterile context of high school algebra as an aside. Only decades later did I discover its true grandeur when I rediscovered it in the context of combinatorics and number theory. I will attempt to provide some, hopefully, enlightening images but I feel that mathematical notation is both important and beautifully concise and allows one to have a transcendent experience with math so we will start with a formula, one of my “Eleven Equations”. The more I learn the more I feel it is more than justified to be in that list. I also feel that its centrality in combinatorics and probability theory makes it an equation that every math or computer science geek or enthusiast should be able to both recite off of the top their heads and be able to apply, if you don’t agree maybe this post will inspire you to think differently. The equation is that of the binomial coefficient formula and it describes each element of Pascal’s Triangle:
The above formula evaluates under the rules that c ≤ r, r ≤ 0 and 0! = 1. I prefer the use the letters r and c over the traditional n and k as in n chose k, one can as easily say r choose c. The reason I prefer r and c is that r means row and c means column and these values can be viewed as a way to address positions within Pascal’s Triangle, for example using the binomial coefficient notation from the above formula Pascal’s Triangle can be represented as:
As you can see in the above representation each position is the address (r,c) given by the coefficient notation. Evaluating the expressions above with the Binomial Coefficient formula yields the more common representation of Pascal’s Triangle:
For our purposes we are interested in the one variable version which can be written as:
This can be expanded to:
This can be written in "Pascal’s Triangle Form" as:
These types of introductions often focus on the additive properties as well. Each element is the sum of the two elements above it, which can be illustrated as:
The logic of this representation assumes that there is a zero on each side to add to propagate the ones on both side, we can rewrite this using the binomial coefficient notation as:
I admit this is where it starts to get more interesting to me. Not to denigrate its importance but I never really got that excited by the whole binomial theorem thing, although the series e stuff is pretty cool. If we pick 5 choose 3 in the above representation, it can be written as the following sum:
If you look at the above representation you can see that this can be done for all entries that have two parents, this can be generalized as:
Powers of Two
Another interesting and extremely important pattern in Pascal’s Triangle is the fact that the values of each row sum to the power of two of the row number, which the following representation depicts:
This can be written as the following formula, I think this also qualifies as an identity:
As I mentioned at the beginning the binomial coefficient formula is read as r choose c. I was trying to get back to that but I took the long way, so looking at our examples above our 5 choose 3 means if I have five things how many ways can I choose three of them which is 10 ways . This is useful for counting, in combinatorics and probability theory I could take a deck of 52 cards and count how many 5 card hands I could choose and calculate probabilities of getting various hands, these are pretty standard applications of these types of counting problems. However, if we step back and look at this counting, where each row represents a set of items, then each column represents the number of subsets of size c, this another reason I prefer c here it means both column and cardinality of the subset. We can create a representation of Pascal’s Triangle to show this, but it does become unwieldy pretty quickly due to the exponential growth of each row, we would get:
In this representation each number represents a set element, these can map to any sets with these cardinalities. Each column is wrapped by parentheses. The ones on the left are created by the empty set and the ones on the right are created by the set itself. So each row in Pascal’s triangle gives the structure of the powerset of a set with the cardinality of the row number. Also it is not a coincidence that each row sums to a power of two since the cardinality of the powerset is the cardinality of the set raised to the power of 2.
Since I’ve had lattice theory on my mind lately, I couldn’t help but calling on our old friend the powerset lattice, as you can see in the context of the above each powerset lattice is going to have a row from Pascal’s triangle imprinted on it, each antichain consists of sets of the same cardinality and the number sets in the antichain maps back to a column:
Principle of Inclusion and Exclusion (P.I.E)
The principle of inclusion and exclusion is used to count the size (cardinality) of a union of sets based on the cardinality of the sets and their intersections, the most common form, which appears in many probability texts, is:
|A ∪ B| = |A| + |B| - |A ∩ B|
The reason for this is that if you take the size of the union two sets, if they have any common elements then they counted twice, so subtracting the intersection removes the duplicate count. This type of problem, counting things more than once and then needing to remove the duplicates comes up in other counting problems as well. The venn diagram below shows the intersection of two sets, these are the elements that get counted twice in |A| + |B|:
Now as a disclaimer this Pascal’s triangle correlation is strictly my own observation so it might be wrong, but it seems to hold. If you take the formula from above and add a term for the empty set, which is not needed as it evaluates to zero you get |∅| + |A| + |B| - |A ∩ B|. Now you can see the number of elements for each term with number of sets (0, 1, 2) are 1 2 1, row 2 of Pascal’s triangle, for three you have 1 3 3 1:
|A ∪ B ∪ C| =
|A| + |B| + |C|
- |A ∩ B| - |A ∩ C| - |B ∩ C|
+ |A ∩ B ∩ C|
And for four we have (1 4 6 4 1)
|A ∪ B ∪ C ∪ D | =
|A| + |B| + |C|+ |D|
- |A ∩ B| - |A ∩ C| - |A ∩ D| - |B ∩ C| - |B ∩ D| - |C ∩ D|
+ |A ∩ B ∩ C| + |A ∩ B ∩ D| + |A ∩ C ∩ D| + |B ∩ C ∩ D|
- |A ∩ B ∩ C ∩ D |
This does not prove it but I would bet the pattern continues. Like I said this is my own observation, so it might be wrong, if you know either way please let me know.
The binomial distribution is given by the following formula:
I converted it to my r,c nomenclature, I may be going too far using it here but I wanted to keep consistent, remember r=n, and c=k. It is a discrete probability distribution for the number of successes of r independent yes/no experiments each of which yields success with probability p. If you set p=1/2, the symmetric case, then the row number r divided by 2r gives you the binomial distribution. By the central limit theorem this symmetric case approaches the normal distribution as r increases.
Central Binomial Coefficient and Symmetry
The above representation shows the symmetry in Pascal’s Triangle, each row is colored in a way that highlights the numbers that are the same, the first and last columns consisting of ones match each other obviously, but in each row the second column matches the second to last column, and the same for the third column and so one. This gives rise to, you guessed it, another combinatorial identity:
The columns highlighted in the gray box are special case called the central binomial coefficient. If you notice each even numbered row has an odd number of columns with the central one only occurring only once, it still follows the above identity but the identity evaluates to a value equaling itself. The central binomial coefficient is given by the following form, I use n here since we don’t care about row/column in this case:
There is an area of math that deals with figurate numbers, these are numbers formed by arranging dots to form various polygons which form sequences as you expand size of each polygon, I looked at some interesting properties of triangular numbers in a previous post, for example here are the first 5 triangular numbers:
Triangular numbers are created by adding up the natural numbers, but there is a binomial formula which evaluates to a closed formula:
The idea of figurate numbers can be extended into higher dimensions, so not only do you have numbers that create polygons you can create polyhedrons as well. When you add up triangular numbers, you get the tetrahedral number which can be described by an analogous formula:
This can be visually depicted as stacking up the triangular number to form tetrahedrons:
Pascal’s Triangle has a number of relations with figurate numbers, but the most notable are the triangular and tetrahedral numbers, shown in red and blue respectively:
Also if you look at the binomial expression for these you will notice that the column is fixed, for triangular numbers it is c=2 and for tetrahedral numbers it is c=3, this maps our addressable form to the second and third columns as shown.
The above tabular form is based on a table from Concrete Mathematics, 2nd Edition by Ronald L. Graham, Donald E. Knuth and Oren Patashnik. Again you can see column one, where c=1, is the natural numbers, column two is the trianglular numbers, the summation of natural numbers and three is the tetrahedral numbers, the summation of the triangular numbers. This would imply that this pattern continues to progress, and it does. The next column are the pentatope numbers, which describe a series of four dimensional polytopes, a polytope is an extension of a polyhedron into higher dimensions, called pentachorons which are formed by stacking three dimensional tetrahedrons in four dimensions. These numbers are given by the formula:
And these numbers progress into higher dimensions, each summing the numbers immediately below them. These are generalized as the pyramidal numbers and are described by the following table:
|Number||Summation Form||Binomial Form||Closed Form|
Powers of Eleven
There is another interesting pattern in Pascal’s Triangle each successive row is equivalent to a power eleven of as you can see in the following table:
Oops, I fibbed a little here, 115 does not equal the row (1 5 10 10 5 1), this breaks down due to the need to carry 1 to make ten. However, the pattern will hold if you push up the number base, for example 11 hexadecimal, which I will write as 1116, to the 5th power is (1116)5 = 15AA51, remember A equals 10 so now the pattern holds, and it will continue to hold as long as you keep upping the number base, but remember (1116)5≠ 115.
There is another interesting relation to another mathematically interesting triangle, the Sierpinski Triangle aka Sierpinski gasket is a fractal that can be constructed by taking an equilateral triangle and removing the center which is an equilateral triangle of ¼ area and then repeating this on each of the three remaining triangles, and so on. It can also be created by applying modulo 2 to each element in Pascal’s Triangle as shown below. The zeros are grayed out to highlight the pattern:
And the List Goes On...
As we saw we ventured into several areas of mathematics, combinatorics, number theory, probability theory, fractal geometry and geometry (topology), and lattice theory. There are many other patterns such as relations to polytopes, cellular automata, a pattern with prime numbers and more. The Wikipedia page has many interesting patterns, some of which I used here also two other interesting sites are Christopher Olah’s post and this.