Factorial notation can often be used in questions in combinatorics. It is when you need to multiply consecutive figures. When you view n! it would be the factorial of n that is being asked regarding. n factorial is defined as the product of all of the integers from 1 to n.

In factorials your order of multiplication will not matter. Factorials utilized in many regions of mathematics, but specifically in Combinations and also Permutations. In math concepts, the factorial of a non-negative integer and, denoted by and!, is the product of positive integers lower than or equal to n.

## Factorial Function

This factorial function (symbol:! ) means to multiply a number of descending natural figures.
n! = n (n - 1) (n -- 2).......... (3) (2) (1)

The factorial as a function can be defined as.

n! = $\pi^n_{k=1}k$.

‘$\pi$’ notation is used here to indicate the continued product as k takes the values 1, 2, 3, ......…., n in succession.
Examples:

4! = 4 * 3 * 2 * 1 = 24

5! = 5 * 4 * 3 * 2 * 1 = 120

0! = 1
We can easily calculate a factorial from the previous one.

## How to Solve?

With regard to solving the factorial problems, the given factorials inside the problem should be expanded with the formula of the factorial.

For Illustration: if we are asked to search for the value of n!, then we should write the quantity n itself. Put a multiplication sign next and write several 1 less as compared to n, i. e. (n - 1). Again put a multiplication sign and write several 1 less as compared to (n - 1), when i. e. (n - 2).

Continue the procedure until reaching in 1. At the finish, the product of all numbers is to be calculated.

There could possibly be problems where numerator in addition to denominator both comprise factorial. In this kind of problems, the whole expansions on the factorials is not necessary to be multiplied. In reality, many of this terms do get cancelled out. We have to just multiply simply remaining terms.

Down below, there is a listing of few factorials.  Have a look at them!

0! = 1
1! = 1
2! = 2. 1 = 2
3! = 3 . 2. 1 = 6
4! = 4 . 3 . 2 . 1 = 24
5! = 5 . 4 . 3 . 2 . 1 = 120
6! = 6 . 5 . 4 . 3 . 2 . 1 = 720
7! = 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5040
and so on.

## Examples

Given below are some examples based on factorials.

Example 1: Find the value of 9!

Solution: 9! = 9 (9 -1) (9 - 2) (9 - 3) (9 - 4) (9 - 5) (9 - 6) (9 - 7) (9 - 8)
= 9. 8. 7. 6. 5. 4. 3. 2. 1
= 362880

Example 2: Solve $\frac{8!}{4! 5!}$

Solution:

Given: $\frac{8!}{4! 5!}$

First let's find 8!

8! = 8 * 7 * 6* 5* 4 * 3* 2 * 1

=40320

Now consider 4! 5!

First 4! = 4 * 3 * 2 * 1

= 24

5! = 5 * 4 * 3 * 2 * 1

= 120

4! 5 ! = 2880

Therefore $\frac{8!}{4! 5!}$ = $\frac{40320}{2880}$

= 14