In this explainer, we will learn how to solve problems on Pascal’s triangle.
Pascal’s triangle is one of the most fascinating structures we can build from a simple number pattern. It is fascinating to see the connections between such a simple construction and many other areas of mathematics.
Pascal’s triangle can be formed by starting with a one at the top and then placing two ones below. Then, each element of a row is equal to the sum of the two elements above. Hence, in the figure below, we can see that the two is the sum of the two ones above.
To complete the next row, we can consider the pairwise sum of the elements of this row. The first entry will be 1. We can think of this as the sum of 0 and 1 as shown.
The next element is the sum of 1 and 2 as shown below.
Similarly, the following element is the sum of 2 and 1 as shown.
The final element, like the first, can be thought of as the sum of 1 and 0 as follows.
Continuing this pattern, we arrive at what is known as Pascal’s triangle.
Pascal’s triangle is a triangular array of the numbers which satisfy the property that each element is equal to the sum of the two elements above. The rows are enumerated from the top such that the first row is numbered .
Similarly, the elements of each row are enumerated from up to . The first eight rows of Pascal’s triangle are shown below.
Although, in much of the Western world, the triangle is named after the French mathematician Blaise Pascal, it was, in fact, well known to mathematicians centuries before him in places such as China, Persia, and India. To this day, it is know by different names in these places.
Pascal’s triangle has many interesting properties. We will begin by looking as some of the simple patterns which exist in the triangle.
Some of the most obvious patterns are related to the diagonals: for example, the first diagonal only contains ones, whereas the second contains consecutive integers.
More interestingly, the third diagonal contains the triangle numbers, and the fourth contains the tetrahedral numbers.
Furthermore, we can see there is reflectional symmetry about the center.
Example 1: Elements in Pascal’s Triangle
What is the second element in the 500th row of Pascal’s triangle?
Recall that the second elements of each row of Pascal’s triangle are consecutive integers. At this point, we might be tempted to immediately jump to the conclusion that it will therefore be 500. However, we need to be a little more careful than this. Recall that the first row only contains 1. Hence, there is no second element. The first row with a second element is the second row, which consists of two ones. Therefore, the second element in this row is 1 and not 2. Hence, the second element of the 500th row of Pascal’s triangle will be 499.
Example 2: Patterns in Pascal’s Triangle
A partially filled-in picture of Pascal’s triangle is shown. By noticing the patterns, or otherwise, find the values of , , , and .
We begin by considering the elements of the third diagonal. There is a clear pattern to go from one element to the other: to go from the first to the second, we add two; then to go from the second to the third, we add 3.
We can extend this pattern as follows.
Both and are the elements in this row. Therefore, and .
We now consider element . This element is actually also in the third diagonal—the one that is in the other direction—and it is the sixth element. Hence, .
Finally, we see that is in the second diagonal. This diagonal contains consecutive positive integers. Hence, since it is the eleventh element, its value will simply be 11.
Therefore, our final answer is , , , and .
Example 3: Sums along Diagonals in Pascal’s Triangle
The figure shows a section of Pascal’s triangle. Without using a calculator, find the sum of the highlighted elements.
For this question, we could simply sum the individual elements. However, we can actually use the properties of Pascal’s triangle to quickly evaluate the sum of these elements. We will start from the smallest element in the row: the 1. Clearly the sum of this element is simply 1, which we can see is the element below to the right as shown in the figure.
We now consider the first two elements and notice that their sum is the element below the second element to the right.
Similarly, the sum of the first three elements is the sum of the first two elements and the third element. From the defining property of Pascal’s triangle, we see that this is the element directly below these two.
By continuing this pattern, we see that the sum of the highlighted elements will be the element below the last element to the right as shown.
Hence, the sum of the highlighted elements is 5 005.
We will now turn our attention to the relationships between adjacent elements. Clearly, any element is the sum of the two elements above. However, there are other relationships between these three terms. When discussing this, we will make use of the enumeration of the rows which starts from . Similarly, within a given row, we will enumerate the elements by , where is the first element in the row.
Let us consider the multipliers that relate the terms on the diagonals between the rows enumerated by and . The figure below represents the multiplier that takes us from the row to the row going diagonally left.
It appears that there is a general form relating these terms which we can represent as multiplication by , where is the enumeration of the elements in the row. This is in fact the correct relationship that holds for any two elements which are related on a left diagonal. Notice that both and are related to the lower elements. Similarly, we can look at the relationship between elements on the right diagonal. The figure below represents the multiplier that takes us from the row to the row going diagonally right.
Once again, a clear pattern appears which we can represent as multiplication by . This again generalizes to any two elements connected on a right diagonal. Once again, both and are related to the lower elements.
We will now consider the relationship between two consecutive elements on the same row. The figure below represents the multiplier that takes us from the th element to the th element of the row.
Once again there is a clear pattern here: to move from the th element to the th, we multiply by .
We summarize these relationships in the following figure.
Example 4: Relationship between Adjacent Elements in Pascal’s Triangle
The figure shows 7 adjacent elements in Pascal’s triangle.
Given that 5 985 is the 18th element on the 22nd row, find , , , , , and .
Firstly, we would like to write down the values of and for the given element. Recall that we begin enumeration in and from zero. Hence, the 18th element is the element for which and the 22nd row is the row for which .
We will use the relationships between consecutive elements in Pascal’s triangle. We have represented the multipliers which connect adjacent elements in the figure below.
We now turn our attention to finding . Recall that, within a given row, to move from the th element to the th, we multiply by . Hence, to move from the th to th, element we multiply by the reciprocal: . Hence,
We can now use the relationship between the left diagonals to find . Recall that, to move down the left diagonal (from the th element on the th row to the th element on the th row), we multiply by . Hence, to move in the opposite direction, we multiply by the reciprocal: . Hence,
Similarly, by using the relationship between elements on the right diagonal, we can find . Recall that, to move down the right diagonal (from the th element on the th row to the th element on the th row), we multiply by . Hence, to move in the opposite direction, we multiply by the reciprocal: . Hence,
It is worth checking that , since, if this were not true, we would have made a mistake. Checking this, we find we that our three elements meet this condition. We now consider . We can use the relationship between elements in the same row and we can find by multiplying by . Hence,
Finally, we can find using the defining property of Pascal’s triangle: . Hence,
Similarly, we can find by evaluating . Hence,
Hence, the final answer is
Using this relationship between the elements, we can find a general formula for an element. We consider the element enumerated on the row enumerated . We consider how to get to this element from the element at the beginning of the row which is always equal to 1. To get to the second element (), we multiply by . Then, to get to the next element, we multiply by . Hence, to get from the first to the third, we multiply by . Multiplying this by the multiplier to get to the next element, we have the following formula for the fourth element (at position ):
We can continue this pattern until we get to the element at position , which gives us the formula
Using factorial notation, we can write this more succinctly. Recall that the factorial of a positive integer is the product of all the positive integers less than or equal to . We denote the factorial as . Hence,
Using factorial notation, we can rewrite this as
Multiplying by , we have
This formula might be familiar because it is actually the formula for the combination which is also sometimes referred to as the binomial coefficient . Hence, in finding a general formula, we have also demonstrated that there is a strong connection between Pascal’s triangle and combinatorics. In addition, we have also shown that Pascal’s triangle is the triangle made of the binomial coefficients (which is, in fact, an alternative definition of the triangle).
In the rest of this explainer, we will spend a little time exploring these two connections, starting with the connection between Pascal’s triangle and the binomial coefficients.
Pascal’s triangle can be used to find the coefficients of the terms in the expansion of . The figure demonstrates this.
This relationship is often captured in the binomial theorem.
For an integer , where
Sometimes the following notation is used in place of : , , , , and .
Example 5: Sum of a Row of Pascal’s Triangle
What is the sum of the terms in the 30th row of Pascal’s triangle?
Recall that the 30th row of Pascal’s triangle is the row we enumerate . Given the connection between the elements of Pascal’s triangle and the binomial coefficients, we can restate this as the problem of evaluating
We could of course simply evaluate this with a calculator. However, by using the binomial theorem, we can seriously simplify the process of evaluating this expression. Notice that if we set and , we have
Hence, the sum of the terms in the 30th row of Pascal’s triangle is .
The property that the last example explored is not unique to the 30th row. In fact, the sum of consecutive rows of Pascal’s triangle is consecutive powers of two. In particular, the row enumerated has a sum of .
Example 6: Sum of Every Second Element in a Row of Pascal’s Triangle
What is sum of every other element in the 1 000th row of Pascal’s triangle?
Recall that the 1 000th row of Pascal’s triangle is the row we enumerate . For a question like this, the numbers are clearly going to be too large for us to evaluate in a long-handed way. Therefore, we need to find an alternative way. Firstly, using the connection between Pascal’s triangle and the binomial coefficients, we can restate this as the problem of evaluating
We will denote this sum . Furthermore, we will denote the sum of the other terms . Hence,
We know that the sum of all the terms of this row will be equal to . Hence, . We now would like to find another expression in terms of and so that we can solve for . Using the binomial theorem, we can find such an expression by setting , , and as follows:
We can rewrite this as
Since , we can add these two equations to get
Hence, . Therefore, the sum of every other element in the 1 000th row of Pascal’s triangle is .
Finally, we will consider the connection between Pascal’s triangle and combinatorics. We can consider Pascal’s triangle as a graph where each entry represents a node. At each node, we have a choice of going left or right. Given this idea, the entries in Pascal’s triangles can be interpreted as the number of distinct paths that lead through the graph from the top element to the given element. For example, if we consider the third element of the fourth row, we can see there are three distinct paths to get from the top of the triangle to the given element.
We can use this property to solve combinatorics and probabilistic problems as the next example will demonstrate.
Example 7: Pascal’s Triangle and Combinatorics
Ramy is playing a game where a ball is dropped onto an array of pegs, from vertically above the top peg, and it bounces down to numbered buckets at the bottom.
He only gets a prize if it falls into buckets 3 or 7. Find the probability that he gets a prize given that there is an even probability it will fall to the left or the right of any given peg.
The probability that a ball ends in a particular bucket will be equal to the number of possible paths to that bucket divided by the total number of paths through the array. Recall that the entries of Pascal’s triangle can be interpreted as the number of possible paths through the triangle. Hence, the number of paths to each bucket can be found using pascals triangle. In the given figure, we have 9 buckets. Therefore, we would like to consider the ninth row of Pascal’s triangle, which is the row for which . We can write this row out by either using the general formula for the entries or simply reproducing the first nine rows of the triangle. Here we will use the general formula.
The ninth row of Pascal’s triangle is represented below.
Evaluating each of these can express the elements of this row as follows.
We can write the number of paths to each bucket in the diagram as follows.
To find the total number of paths, we could add these numbers up or use the fact that the sum of the elements of the th row is . Hence, the total number of paths through the array is . Hence, the probability that the ball ends up in the third bucket is . Similarly, the probability that the ball ends up in the seventh bucket is also . Therefore, the probability that Ramy wins a prize is given by the sum of these: .
There are many other interesting patterns and properties of Pascal’s triangle. For example, it is interesting to explore the patterns we get by considering the location of all the odd numbers in Pascal’s triangle, or the connection between Pascal’s triangles and powers of 11.
- Pascal’s triangle can be defined in two ways:
- the triangle which satisfies the property that each element is equal to the sum of the two elements above,
- the triangle of the binomial coefficients.
- Pascal’s triangle has many symmetries and interesting properties such as the following:
- The entries in the first diagonal are all one; in the second, they are the consecutive positive integers; in the third, they are the triangle numbers; and in the fourth, they are the tetrahedral numbers.
- Each row is symmetric about its middle.
- The sum of the elements in the th row is .
- The alternating sum of any row is zero.
- The general formula for the th element in the th row is .
- Pascal’s triangle can be used to help expand binomials.
- Pascal’s triangle has useful applications in combinatorics and probability.