Discrete Mathematics: An Open Introduction, 4th Edition

In 1653, Blaise Pascal, concerned with questions that would lay the foundation of probability theory, collected several facts about a triangular array of numbers in his Treatise on Arithmetical Triangle. This arrangement of numbers appeared as early as the 10th century in China, India, and Persia. The Chinese and Persian treatment of the triangle was in service of what we would now consider algebra: finding \(n\)th roots, essentially solving polynomial equations. The numbers in the triangle appear as solutions to counting problems in Indian texts: from six tastes, how many combinations of one, or two, or three,... can you make? European mathematicians in the 14th century presented the triangle as a table of figurate numbers (numbers that can be arranged in a geometric shape), which were themselves the centerpiece of the work of Pythagoras and his followers.

So what is this remarkable triangle that holds the secrets of so many different mathematical problems? Behold, Pascal’s triangle:

Spend some time gazing at the beauty of this triangle. What do you notice? What do you wonder? Look specifically at the 5th row (we call the 1 on the top row 0, so row 5 is 1, 5, 10, 10, 5, 1). How do the numbers in this row relate to the numbers above them? Notice that \(5 = 1+4\) and \(10 = 4+6\text{.}\) Does this occur anywhere else in the triangle?

Indeed, every number in the triangle is the sum of the two numbers above it. Let’s take this as our definition of Pascal’s triangle. We can then generate as many rows of the triangle as we like. It is this additive definition that was used in China and Persia to find \(n\)th roots, and we will briefly mention this use at the end of this section. However, we are interested in counting questions, so our main goal now is to observe how the numbers of Pascal’s triangle are answers to a variety of counting questions.

Here are some apparently different discrete objects we can count: lattice paths, bit strings, subsets, and pizzas. We will give an example of each type of counting problem (and say what these things even are). As we will see, the numbers in Pascal’s triangle are the answers to all of these questions.

Before we jump in, a little bit of notation. Let’s give each number in Pascal’s triangle a name, based on its position. Think of each number as being in a row and a column: rows are counted down, starting at 0, and columns are counted in from the left, also starting at 0. The entry in row \(n\) and column \(k\) will be denoted \(\binom{n}{k}\text{.}\) For example, the \(\binom{6}{3} = 20\text{,}\) since that is the value in row 6, column 3. For reasons that will become clear soon, we pronounce \(\binom{n}{k}\) as “\(n\) choose \(k\text{.}\)” We can rewrite the triangle with these names:

The integer lattice is the set of all points in the Cartesian plane for which both the \(x\) and \(y\) coordinates are integers. If you like to draw graphs on graph paper, the lattice is the set of all the intersections of the grid lines.

A lattice path is one of the shortest possible paths connecting two points on the lattice, moving only horizontally and vertically. For example, here are three possible lattice paths from the point \((0,0)\) to \((3,2)\text{:}\)

Notice that to ensure the path is the shortest possible, each move must be either to the right or up. Additionally, in this case, no matter what path we take, we must make three steps right and two steps up. No matter in what order we make these steps, there will always be five steps. Thus each path has length five.

Let’s take what we learned from the rook paths (which are, gasp, actually lattice paths). Consider the lattice shown below:

Any lattice path from (0,0) to (3,2) must pass through exactly one of \(A\) and \(B\text{.}\) The point \(A\) is 4 steps away from (0,0) and two of them are in the \(x\) direction. The last step is also in the \(x\) direction, so the paths from (0,0) to (3,2) that pass through \(A\) are exactly the six strings we listed above that end in an \(x\text{.}\) For the paths that pass through point \(B\text{,}\) the last step will be in the \(y\) direction, so the paths from (0,0) to (3,2) that pass through \(B\) are exactly the four strings we listed above that end in a \(y\text{.}\) So the total number of paths to (3,2) is just \(6+4\text{.}\)

The general observation here is that to find the number of paths that start at \((0,0)\) and end at \((m,n)\text{,}\) we can find the number of paths to the point directly to the left of the endpoint, \((m-1,n)\) and add the number of paths to the point directly below the endpoint, \((m,n-1)\text{.}\) This is exactly the same way that Pascal’s triangle is generated! Indeed, if we rotate the lattice appropriately, so the point \((0,0)\) is at the top of the triangle and the axes along the sides of the triangle, we see that the numbers in Pascal’s triangle give us exactly the number of paths to each lattice point.

To make this observation helpful for actually finding the number of paths from the origin to a given point, we note that it is the length of the path that determines the row of Pascal’s triangle, and the number of steps in the \(y\) direction that says how far into the triangle we are -- the column of Pascal’s triangle.

The number of bits (0’s or 1’s) in the string is the length of the string; the strings above have lengths 4, 1, 4, and 10 respectively. We also can ask how many of the bits are 1’s. The number of 1’s in a bit string is the weight of the string; the weights of the above strings are 2, 0, 4, and 5 respectively.

For example, the elements of the set \(\B^3_2\) are the bit strings 011, 101, and 110. Those are the only strings containing three bits, exactly two of which are 1’s.

The counting questions: How many 5-bit strings have weight 3? In other words, we are asking for the cardinality \(|\B^5_3|\text{.}\)

Now we have two good reasons to believe that Pascal’s triangle tells us the number of bit strings of a given weight: There is a one-to-one correspondence between lattice paths and bit strings, and the same recursive relationship holds for bit strings as it does for generating Pascal’s triangle. So we can now use the triangle to count bit strings.

A subset of a set \(A\) is any set all of whose elements are also in \(A\text{.}\) Think of starting with the set \(A\) and removing some (or none or all) of its elements: the resulting set is a subset of \(A\text{.}\) (More information about sets can be found in Section 0.2 and Section 5.1.)

Again, we see there are ten. In fact, we have listed them in the same order as we listed the ten 5-bit strings of weight 3 and the ten lattice paths from (0,0) to (3,2). Wait, does this even make sense? In what way is a subset the same as a bit-string?

Think of each bit in a bit string as representing one of the elements in a set. The set \(A\) has five elements, so we need five bits to represent a subset of \(A\text{.}\) If the bit in position \(n\) is a 0, that means we do not include \(n\) in our subset, while a 1 in that position tells us that \(n\) is in the subset. Three 1’s means we have said, “yes” to three elements.

This example illustrates that, once again, Pascal’s triangle can give us the answer to a counting question. The number of \(k\)-element subsets of a set with \(n\) elements is the same as the number of \(n\)-bit strings of weight \(k\text{,}\) and that is the number in row \(n\text{,}\) column \(k\) of the triangle: \(\binom{n}{k}\text{.}\)

At this point I’m sure we are all getting pretty hungry, so let’s get some pizza. But which pizza shall we order? Let’s not overdo it and just choose three toppings from the ten available. How many different pizzas can we order?

Aha! So that’s why we care about counting subsets!! Each pizza choice is nothing more than a 3-element subset of the set of 10 toppings. We now know how to count this: \(\binom{10}{3} = 120\) different pizzas.

What if we want an Italian soda with dinner? Let’s say we want to add two different flavored syrups from the 13 available. How many different sodas are possible? This is just the number of 2-element subsets of a set with 13 elements: \(\binom{13}{2} = 78\text{.}\)

This is why we pronounce \(\binom{n}{k}\) as “\(n\) choose \(k\)”. It is the number of ways to choose \(k\) items from a collection of \(n\) items, since choosing \(k\) elements is exactly how you build a \(k\)-element subset.

We can view counting lattice paths as choosing \(k\) out of \(n\) things: Of the \(n\) steps on the path, we choose \(k\) of them to be in the \(x\) direction. Bit strings can also be thought of in this way: Of the \(n\) bits in the string, we choose \(k\) of them to be 1’s.

We can now answer all sorts of real-world counting problems, as long as they are really nothing more than asking for the number of subsets of a set. Pascal’s triangle contains all these answers.

Earlier we said that one of the original uses for Pascal’s triangle was to solve problems in algebra. What does counting subsets (or bit strings or lattice paths) have to do with algebra?

Suppose you expand the binomial expression \((x+1)^6\) (i.e., multiply the binomial \(x+1\) by itself six times). This can be tedious to do by hand, but a computer algebra system such as SageMath can do this easily.

Say we want to find the coefficient of the \(x^2y\) therm. We collect like terms, collecting all the terms in which we have chosen \(x\) two times (and \(y\) the other one time). Alternatively, the \(x^2y\) term comes from all the strings with two \(x\) and one \(y\text{,}\) just like a bit string or lattice path. No matter how you think of it, the result is that we have \(\binom{3}{2} = 3\) terms with the form \(x^2y\text{.}\)

Hopefully it is clear that this generalizes to the expansion of \((x+y)^n\) for any positive integer \(n\text{.}\) This is known as the binomial theorem.

For this reason, the numbers in Pascal’s triangle are often called binomial coefficients.