Pascal's Triangle Formula

Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The triangle starts with a "1" at the top, and each subsequent row contains the coefficients of the binomial expansion.

Here is an example of Pascal's Triangle:

The formula to calculate any element in Pascal's Triangle is based on binomial coefficients, denoted as:

\dbinom{n}{k} = \dfrac{n!}{k!(n - k)!}

Pascal's triangle is a beautiful concept of probability developed by the famous mathematician Blaise Pascal, which is used to find coefficients in the expansion of any binomial expression. It is a method to know the binomial coefficients of terms of binomial expression (x + y)ⁿ, where n can be any positive integer and x, and y are real numbers.

This triangle is used in different types of probability conditions. Here each row represents the coefficient of expansion of (x + y)ⁿ.

(x + y)⁰ = 1
(x + y)¹ = 1x + 1y
(x + y)² = 1x² + 2xy + 1y²
(x + y)³ = 1x³ + 3x²y + 3xy² + 1y³
(x + y)⁴ = 1x⁴ + 4x³y + 6x²y² + 4xy³+1y⁴
(x + y)⁵ = 1x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + 1y⁵

Here the power of y in any expansion of (x + y)ⁿ represents the column of Pascal's Triangle.

• n represents the row of Pascal's triangle.
• Row and column are 0 indexed in Pascal's Triangle.

Pascal's Triangle Construction

It's quite simple to make a Pascal's Triangle.

1. Start by creating the top row (the 0th row) with just the number 1.
2. In the following rows, each new number in Pascal's Triangle is the sum of the two numbers directly above it.
3. For example, to find the number in row 4, column 2:
   • Add the numbers from row 3, column 1, and row 3, column 2.
   • So, the number in row 4, column 2 is 1 + 2 = 3.

Properties of Pascal's Triangle

1. Each number in Pascal's Triangle is the sum of two numbers above it.
2. Numbers in a row are symmetric.
3. Each number represents a binomial coefficient.
4. Numbers on the left and right sides of the triangle are always 1.
5. nth row contains (n+1) numbers in it.

Read more: Important Facts about Pascal's Triangle

Pascal's Triangle Other Formulas

The Pascal's Triangle formula to find the element in the n-th row and k-th column of the triangle is:

\binom{p}{q} = \binom{p-1}{q-1} + \binom{p-1}{q}

Here, 0 ≤ q ≤ p, p is a non-negative number

Alternatively, the formula to find the number in the nnn-th row and r-th column is given by:

\binom{p}{q} = \frac{p!}{q!(p-q)!}

And another recursive expression is:

\binom{p}{q} = \binom{p}{q-1} + \binom{p-1}{q-1}

Pascal's Triangle Binomial Expansion

As we already know tascal's triangle defines the binomial coefficients of terms of binomial expression (x + y)ⁿ, So the expansion of (x + y)ⁿ is:

(x + y)ⁿ = a₀xⁿ + a₁x^n-1 + ......a_n-1xy^n-1 + a_nyⁿ

Practice Questions on Pascal's Triangle

Question 1: Find the coefficient of the term x^2y in the expansion of (x + y)³.
Solution:

Method 1:

We look at the row 3rd row of Pascal's Triangle because n is 3 and 1st column of the Pascal's Triangle because power of y is 1 in the term x^2y. So the coefficient is 3.

Method 2:

We simply apply ⁿC_r where n = 3, r = 1.
So coefficient of x^2y in the expansion of (x + y)³ is ³C₁ = 3

Question 2: Find the coefficient of the term x²y² in the expansion of (4x + 3y)⁴.
Solution:

Method 1:

We look at the row 4th row of Pascal's Triangle because n is 4 and 2nd column of the Pascal's Triangle because power of y is 2 in the term x²y². So number in Pascal's Triangle is 6. 

But we see that coefficient of x is 4 and y is 3 now since power of x is 2 and y is 2 in the term x²y² so pascal Triangle number will be multiplied by 4² and 3² to find the coefficient.

Coefficient = 6 x 4² x 3² = 864

Method 2:

We simply apply ⁿC_r where n = 4, r = 2.

So Pascal Triangle number of term x²y² in the expansion of (4x +3y)⁴ is ⁴C₂ = 6.
But we see that coefficient of x is 4 and y is 3 now since power of x is 2 and y is 2 in the term x²y² so pascal Triangle number will be multiplied by 4² and 3² to find the coefficient.

Coefficient = 6 x 4² x 3² = 864

Question 3: Write the 6th row of the Pascal's Triangle

Solution:

6th row can be written as :  ⁶C₀ ⁶C₁ ⁶C₂ ⁶C₃ ⁶C₄ ⁶C₅ ⁶C₆
1, 6, 15, 20, 15, 6, 1

Question 4: Find the coefficient of the term x⁴ in the expansion of (2x + y)⁴.

Solution:

Method 1:

We look at the row 4th row of Pascal's Triangle because n is 4 and 0th column of the Pascal's Triangle because power of y is 0 in the term x⁴. So number in Pascal's Triangle is 1.

But we see that coefficient of x is 2 and y is 0 now since power of x is 4 and y is 0 in the term x⁴ so Pascal Triangle's number will be multiplied by 2⁴ and 1⁰ to find the coefficient.

Coefficient = 1 x 2⁴ x 1⁰= 16 

Method 2:

We simply apply ⁿC_r where n = 4, r = 0.

So Pascal Triangle number of term x⁴ in the expansion of (2x + y)⁴ is ⁴C₀ = 1.
But we see that coefficient of x is 2 and y is 0 now since power of x is 4 and y is 0 in the term x⁴ so Pascal Triangle's number will be multiplied by 2⁴ and 1⁰ to find the coefficient.

Coefficient = 1

Question 5: Find the coefficient of the term xy² in the expansion of (2x + y)³.

Solution:

Method 1:

We look at the row 3rd row of Pascal's Triangle because n is 3 and 2nd column of the Pascal's Triangle because power of y is 2 in the term xy². So number in Pascal's Triangle is 3.

But we see that coefficient of x is 2 and y is 1 now since power of x is 2 and y is 1 in the term xy² so Pascal Triangle's number will be multiplied by 2¹ and 1² to find the coefficient.

Coefficient = 3 x 2¹ x 1² = 6  

Method 2:

We simply apply ⁿC_r where n = 3, r = 2.

So Pascal Triangle number of term xy² in the expansion of (2x + y)³ is ³C₂ = 3.

But we see that coefficient of x is 2 and y is 1 now since power of x is 2 and y is 1 in the term xy^2 so Pascal Triangle's number will be multiplied by 2¹ and 1² to find the coefficient.

Coefficient = 3 x 2¹ x 1² = 6