Pascal’s Triangle

Pascal Triangle is a triangular set of numbers that sums every number as the two numbers vertically above it. It takes its name from French mathematician Blaise Pascal, born in Clermont-Ferrand on June 19, 1623. Pascal's Triangle formula goes directly to the heart of algebra and combinatorics. Pascal’s triangle can be visualised by the following Pascal’s Triangle Diagram: 

1.0Pascal Triangle Explained

Pascal’s Triangle was first written in 1653 then known as a treatise on the arithmetic triangles, by French mathematician Blaise Pascal, but The history of Pascal’s triangle goes beyond 1653. Other mathematicians of ancient times from parts of the world, including Chinese, Indian, Persian, and Italian, also discovered this unique property of Arithmetics. 

Pascal Triangle Definition

Pascal's triangle is an infinite array of numbers arranged in rows, where every row is equivalent to the coefficients of a binomial expansion. The formula used to calculate the entry in a triangle is referred to as the Pascal triangle formula. It begins with 1 at the top, and then each subsequent number is the sum of the two numbers directly above it.

2.0Pascal Triangle Construction

To construct Pascal’s Triangle, you must follow these steps:

Step 1: Start with 1. The first row consists of a single 1.

Step 2: Generate New Rows, Each new row begins and ends with 1. For numbers in between, sum the two numbers directly above them. Such as (1 + 1 = 2).

Step 3: Repeat this process to generate more rows as needed. As shown below: 

This method of constructing a Pascal’s Triangle is only valuable for small entries, but in practical usage or for large numbers of entries and variables, Pascal’s Triangle formula is used. 

3.0Binomial Expansion Pascal's Triangle Formula

The Pascal Triangle formula is used to calculate the individual entries in Pascal's Triangle, which are binomial coefficients. These coefficients are important in both binomial expansions and combinatorics. The formula calculates the values for any given row and column of the triangle.

Each entry in Pascal’s Triangle is a binomial coefficient, often written as:

$C(n,k)=\frac{n!}{k!(n-k)!}$

Here:

• n is the row number (starting from 0)
• k is the position in the row (starting from 0)
• C(n,k) represents the binomial coefficient.

Binomial Theorem and Pascal's Triangle: Interrelation

The connection between the Binomial Theorem and Pascal's Triangle is that Pascal's Triangle gives directly the binomial coefficients that are needed for the binomial expansion. Each row of Pascal's Triangle contains the coefficients for the corresponding power of the binomial expansion. For Example: The expansion of (a + b)ⁿ follows:

$(a+b)={\sum }_{k-0}^{n}C(n,k)a^{n-k}{b}^k$

4.0Applications of Pascal’s Triangle

Pascal’s triangle finds its applications in mathematics for solving various types of problems. Here are some key areas where Pascal's Triangle is used: 

• Binomial Expansion: Pascal's Triangle gives the coefficients for the expansion of binomial expressions with the help of the binomial theorem.
• Combinatorics: It can be used to calculate combinations or binomial coefficients, which is a very fundamental concept in counting problems.
• Probability Theory: Pascal's Triangle gives the binomial probabilities in experiments with two possible outcomes, like success or failure.
• Number Theory: It exposes patterns like triangular numbers and aids in the derivation of other sequences, such as Fibonacci numbers and Catalan numbers.
• Recursive Algorithms: The recursive nature of Pascal's Triangle makes it useful in designing algorithms for computing binomial coefficients efficiently.

5.0Pascal's Triangle Examples

Problem: Expand (x + y)⁴ using Pascal's Triangle.

Solution: For expansion, first, we write Pascal’s triangle up to the fourth row: 

Here, the 4th row of Pascal’s triangle is [1 4 6 4 1], now using binomial expansion: 

(x + y)⁴ = 1x⁴ + 4x³y + 6x²y² + 4xy³ + 1y⁴

Thus, the expansion is:

(x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Problem: How many ways can you choose 3 objects from a set of 5 objects?

Solution: Using the general formula of Pascal’s triangle:

$C(n,k)=\frac{n!}{k!(n-k)!}$

$C(5,3)=\frac{5!}{3!(5-3)!}=\frac{5\times 4\times 3!}{3!\times 2!}$

$C(5,3)=\frac{20}{2}$

C(5, 3)=10

In Pascal’s Triangle’s 5th row, that is, [1, 5, 10, 10, 5, 1], 10 is the 4th element in the 5th row (counting starts from 0). Therefore, there are 10 ways to choose 3 objects from a set of 5.

Problem 3: Find the 4th element in the 5th row of Pascal's Triangle.

Solution: Using the Pascal’s Triangle Formula, which is: 

$C(n,k)=\frac{n!}{k!(n-k)!}$

Here, n = 4 and k = 3

• The 5th row corresponds to (x + y)⁴. So, n = 4 because the 5th row is the one corresponding to (x + y)⁴
• In Pascal's Triangle, we start counting from 0, so the 4th element corresponds to k = 3.

Now, put values of n and k in the equation: 

$C(n,k)=\frac{4!}{3!(4-3)!}=\frac{4\times 3!}{3!\times 1}$

C(n, k)=4

The 4th element in the 5th row of Pascal's Triangle is 4.