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1274 LearnersLast updated on December 2, 2025

Imagine you are taking a number and multiplying it by every whole number before it until you reach 1; that result is the factorial. Factorials help you discover how many ways things can be arranged. In this article, we will explore the concept in detail.

Factorial is the way of multiplying a number by every whole number below it, all the way down to 1. We show that a factorial is denoted by the symbol ‘!’, which is the one above the number 1 on a computer keyboard. When we say n! (read as “n factorial”), which means we multiply all the positive integers from 1 up to n.
So, the factorial definition can be written as:
\(\text{n! = 1 × 2 × 3 × ... × n or, }\\[1em] \text{n! = n × (n – 1) × (n – 2) × ... × 3 × 2 × 1}\)
Example:
Find the value of 5! (5 factorial).
Answer
\(5! = 5 × 4 × 3 × 2 × 1\)
5! = 120
So, the value of 5! is 120.
The factorial of a number n is found by multiplying all the whole numbers from n down to 1. The formula is:
For any whole number n ≥ 1, the factorial can also be written using the product notation:
From these formulas, we can get a useful pattern called the recurrence relation, which shows how each factorial is built using the previous one:
n! = n × (n − 1)!
This means you can find the larger factorial by multiplying the number n by the factorial of the number just before it.
The factorial table showcases the numbers and their factorial values. As seen below, we determine the factorial of a given number by multiplying it by the factorial of the preceding number.
i.e., \(n! = n \times (n-1) \times (n-2) \times \dots \times 1 \)
For example, to find the factorial of 6, multiply 6 by the factorial of 5:
\(6! = 6 \times 120 = 720 \quad (5! = 120) \)
Similarly, the factorial of 7:
\(7! = 7 \times 720 = 5040 \quad (6! = 720) \)
| n Factorial | \(\mathbf{n (n-1) (n-2) \dots 1 }\) | \(\mathbf{n! = n \times (n-1)! }\) | Result |
|
1 Factorial |
1 | 1 | 1 |
| 2 Factorial | \(2 × 1\) | \(= 2 × 1!\) | \(= 2\) |
|
3 Factorial |
\(3 × 2 × 1\) | \(= 3 × 2!\) | \(= 6\) |
| 4 Factorial | \(4 × 3 × 2 × 1\) | \(= 4 ×3 !\) | \(= 24\) |
|
5 Factorial |
\(5 × 4 × 3 × 2 × 1\) | \(= 5 × 4!\) | \(= 120\) |


A subfactorial of a number is written as !n. It tells us how many ways we can rearrange the n objects so that none of them stay in their original position. In other words, it counts the number of derangements, shuffled arrangements where every item is moves to a different place.
The formula for finding the subfactorial of n is:
\(!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}\)
It is easy to assume that the factorial of 0 is 0, but this is incorrect. The factorial of 0 is 1 which can be written as:
\(0! = 1\).
Factorials often follow a pattern:
\(1! = 1\)
\(2! = 2 × 1 = 2\)
\(3! = 3 × 2 × 1 = 3 × 2! = 6 \)
\(4! = 4 × 3 × 2 × 1 = 4 × 3! = 24\)
To understand the zero factorial better, let’s look at the following method:
We find \(3!\) by dividing the factorial of the succeeding number by that number:
\(3! = \frac{4!}{4} \)
\(2! = \frac{3!}{3} \)
\(1! = \frac{2!}{2} \)
\(0! = \frac{1!}{1} \)
The factorial of a number n is the product of the first n natural integers can be expressed as:
\(n! = n × (n –1) × (n – 2) ×… × 3 × 2 × 1\).
The n factorial can be mathematically represented as the product of the given number by the factorial of the preceding number:
\(n! = n × (n – 1)!\)
There is a common misconception that factorials include negative numbers. We will now learn why factorials are undefined for negative numbers. Here, we start with the factorial of 3.
\(3! = 3 \times 2 \times 1 = 6 \)
\(2! = \frac{3!}{3} = \frac{6}{3} = 2 \)
\(1! = \frac{2!}{2} = \frac{2}{2} = 1 \)
\(0! = \frac{1!}{1} = \frac{1}{1} = 1 \)
\((–1)! = 0! ÷ 0 = 1 ÷ 0\) (undefined, since division by zero is impossible).
Thus, factorials are undefined for negative numbers.
As we have learned, the factorials of n are represented as n! And is determined using the formula \(n! = n × (n – 1)!\)
If \(7! = 5,040\), then \(8! = 8 × 7! = 8 × 5,040 = 40,320.\)
The table below shows the factorials of the first 15 numbers:
| n Factorial | Value |
| 1 Factorial | 1 |
| 2 Factorial | 2 |
| 3 Factorial | 6 |
| 4 Factorial | 24 |
| 5 Factorial | 120 |
| 6 Factorial | 720 |
| 7 Factorial | 5040 |
| 8 Factorial | 40320 |
| 9 Factorial | 362880 |
| 10 Factorial | 3628800 |
| 11 Factorial | 39916800 |
| 12 Factorial | 479001600 |
| 13 Factorial | 6227020800 |
| 14 Factorial | 87178291200 |
| 15 Factorial | 1307674368000 |
For developing a deeper understanding of factorials, follow these tips and tricks given below:
The concept of factorial is an important theory and has several applications. However, students might make mistakes when solving problems related to it. Here are a few common mistakes and the easy ways to avoid them:
Factorials are of immense significance in various real-life situations. Let’s now look at a few such examples:
Determine the value of (5! ÷ 4! × 3!).
\((5! ÷ 4! × 3!) = 30\)
Let’s first calculate the factorials of 5, 4, and 3 separately:
\(5! = 5 × 4 × 3 × 2 × 1 = 120\)
\(4! = 4 × 3 × 2 × 1 = 24\)
\(3! = 3 × 2 × 1= 6\)
We now substitute these values:
\((5! ÷ 4! × 3!) = (120 ÷ 24) × 6\)
\(= 5 × 6\)
\(= 30\)
Alex has 8 different books and wants to organize them on a shelf. In how many possible ways he can organize these books?
Alex can organize the books in 40,320 different ways.
To find the possible ways in which Alex can organize the books, we use the formula:
\(n! = n (n –1) (n – 2)... 3 × 2 × 1\)
Given that, there are 8 books:
So the total number of arrangements is:
\(8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1\)
\(= 40,320\)
Therefore, Alex can organize the books in 40,320 different ways.
How many different ways can the letters in the word "EDUCATION" be arranged if all letters are used?
The word“ EDUCATION” can be arranged in 362,880 different ways.
All the letters in the given word are unique, so the total number of ways to arrange them is:
\(n! = n (n –1) (n – 2)... 3 × 2 × 1\)
\(9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880\)
So, there are 362,880 different arrangements.
Determine the value of 4!10!
\(4! × 10! = 87,091,200\)
Let’s first calculate the factorials of 4 and 10 separately:
\(4! = 4 × 3 × 2 × 1 = 24\)
\(10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800\)
We’ll now multiply the factorials:
\( 4!10! = 24 × 3,628,800 = 87,091,200\)
Find the value of (12! – 8!)
\(12! – 8! = 479,001,600 – 40,320 = 478,961,280.\)
We will first find the factorials of 12 and 8 separately:
\(12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1\)
\(= 479,001,600\)
\(8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320\)
Now, substitute these values:
\(12! – 8! = 479,001,600 – 40,320\)
\(= 478,961,280\)
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.






