Pascal's Triangle

Pascal’s Triangle is a kind of number pattern. The numbers are so arranged that they reflect as a triangle. Firstly, 1 is placed at the top, and then we start putting the numbers in a triangular pattern. The numbers which we get in each step are the addition of the above two numbers. It is similar to the concept of triangular numbers.

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History of Pascal’s Triangle

Blaise Pascal was born at Clermont-Ferrand, in the Auvergne region of France on June 19, 1623. In 1653 he wrote the Treatise on the Arithmetical Triangle which today is known as the Pascal’s Triangle. Although other mathematicians in Persia and China had independently discovered the triangle in the eleventh century, most of the properties and applications of the triangle were discovered by Pascal.

This triangle was among many of Pascal’s contributions to mathematics. He also came up with significant theorems in geometry, discovered the foundations of probability and calculus and also invented the Pascaline-calculator. Still, he is best known for his contributions to the Pascal triangle.

Pascal’s Triangle Definition

Most people are introduced to Pascal’s triangle through an arbitrary-seeming set of rules. Begin with 1 on the top and with 1’s running down the two sides of a triangle. Each new number lies between two numbers and below them, and its value is the sum of the two numbers above it. The theoretical triangle is infinite and continues downward forever, but only the first 6 lines appear in figure 1. More rows of Pascal’s triangle are listed in the last figure of this article. A different way to describe the triangle is to view the first line is an infinite sequence of zeros except for a single 1. To obtain successive lines, add every adjacent pair of numbers and write the sum between and below them. The non-zero part is Pascal’s triangle.

Construction of Pascal’s Triangle

The easiest way to construct the triangle is to start at row zero and write only the number one. From there, to obtain the numbers in the following rows, add the number directly above and to the left of the number with the number above and to the right of it. If there are no numbers on the left or right side, replace a zero for that missing number and proceed with the addition. Here is an illustration of rows zero to five.

Pascals triangle

From the above figure, if we see diagonally, the first diagonal line is the list of one’s, the second line is the list of counting numbers, the third diagonal is the list of triangular numbers and so on.

How to use Pascal’s Triangle

Pascal’s triangle can be used in various probability conditions. Suppose if we are tossing the coin one time, then there are only two possibilities of getting outcomes, either Head (H) or Tail (T).

If we toss it two times, then there are one possibility of getting both heads HH and both as tails TT, but there are two possibilities of getting at least a Head or a Tail, i.e. HT or TH.

Now you may consider how Pascal’s triangle will help here. So let’s see the table given here based on the number of tosses and outcomes.

Number of Tosses Number of Outcomes Pascal’s Triangle
1 H


2     HH



1, 2, 1





We can also extend it by increasing the number of tosses.

Pascal’s Triangle Patterns

1) Addition of the Rows: One of the interesting properties of the triangle is that the sum of numbers in a row is equal to 2n

where n corresponds to the number of the row:

1 = 1 = 20

1 + 1 = 2 = 21

1 + 2 + 1 = 4 = 22

1 + 3 + 3 + 1 = 8 = 23

1 + 4 + 6 + 4 + 1 = 16 = 24

2) Prime Numbers in the Triangle: Another pattern visible in the triangle deals with prime numbers. If a row starts with a prime number or is a prime numbered row, all the numbers that are in that row (not counting the 1’s) are divisible by that prime. If we look at row 5 (1 5 10 10 51), we can see that 5 and 10 are divisible by 5. However, for a composite numbered row, such as row 8 (1 8 28 56 70 56 28 8 1), 28 and 70 are not divisible by 8.

3) Fibonacci Sequence in the Triangle: By adding the numbers in the diagonals of the Pascal triangle the Fibonacci sequence can be obtained as seen in the figure given below.

Pascal's Number Application- Fibonacci series

There are various ways to show the Fibonacci numbers on the Pascal triangle. R. Knott was able to find the Fibonacci appearing as sums of “rows” in the Pascal triangle. He moved all the rows over by one place and here the sums of the columns would represent the Fibonacci numbers.

Properties of Pascal’s Triangle

    • Each number is the sum of the two numbers above it.
    • The outside numbers are all 1.
    • The triangle is symmetric.
    • The first diagonal shows the counting numbers.
    • The sums of the rows give the powers of 2.
    • Each row gives the digits of the powers of 11.
    • Each entry is an appropriate “choose number.”
    • And those are the “binomial coefficients.”
  • The Fibonacci numbers are there along diagonals.

Here is an 18 lined version of the pascal’s triangle;

Pascal's Triangle Properties


The formula to find the entry of an element in the nth row and kth column of a pascal’s triangle is given by:

\({n \choose k}\). The elements of the following rows and columns can be found using the formula given below.

\({n \choose k}= {n-1 \choose k-1}+ {n-1 \choose k}\)

Here, n is any non-negative integer and 0 ≤ k ≤ n.

The above notation can be written as:

\({n \choose k}\) (i.e., n choose k) = C(n, k) = nCk = n!/[k!(n – k)!]

This pattern of getting binomial coefficients is called Pascal’s rule.

Pascal’s Triangle Binomial Expansion

Pascal’s triangle defines the coefficients which appear in binomial expansions. That means the nth row of Pascal’s triangle comprises the coefficients of the expanded expression of the polynomial (x + y)n.

The expansion of (x + y)n is:

(x + y)n = a0xn + a1xn-1y + a2xn-2y2 + … + an-1xyn-1 + anyn

where the coefficients of the form ak are precisely the numbers in the nth row of Pascal’s triangle. This can be expressed as:

\(a_{k}= {n \choose k}\)

For example, let us expand the expression (x + y)n for n = 3.

(x + y)3 = 3C0x3 + 3C1 x2y + 3C2 xy2 + 3C3 x0y3

= (1)x3 + (3)x2y + (3)xy2 + (1)y3

Here, the coefficients 1, 3, 3, 1 represent elements in the 3rd row of the pascal’s triangle.

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