Factorial Calculator

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About the Factorial Calculator

The factorial calculator lets you calculate the factorial of any whole number upto $\hspace{0.2em} 1000 \hspace{0.2em}$ .

Usage Guide

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i. Valid Inputs

Your input must be a natural number (a positive integer) upto $\hspace{0.2em} 1000 \hspace{0.2em}$ .

ii. Example

If you would like to see an example of the calculator's working, just click the "example" button.

iii. Share

We would love to see you share our calculators with your family, friends, or anyone else who might find it useful.

By checking the "include calculation" checkbox, you can share your calculation as well.

Here's a quick overview of what a factorial is.

What Is a Factorial?

Factorial of a natural number $\hspace{0.2em} n \hspace{0.2em}$ is defined as the product of all natural numbers upto $\hspace{0.2em} n \hspace{0.2em}$ .

It is denoted with a $\hspace{0.2em} ! \hspace{0.2em}$ symbol. So, $\hspace{0.2em} n! \hspace{0.2em}$ would be read as $\hspace{0.2em} n \hspace{0.2em}$ factorial.

n! \hspace{0.25em} = \hspace{0.25em} n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1

Here are a couple of examples of a factorial.

$\hspace{0.2em} 3! \hspace{0.25em} = \hspace{0.25em} 3 \times 2 \times 1 \hspace{0.25em} = \hspace{0.25em} 6 \hspace{0.2em}$

$\hspace{0.2em} 5! \hspace{0.25em} = \hspace{0.25em} 5 \times 4 \times 3 \times 2 \times 1 \hspace{0.25em} = \hspace{0.25em} 120 \hspace{0.2em}$

Also,

\hspace{0.2em} 0! \hspace{0.2em}

is defined to be

\hspace{0.2em} 1 \hspace{0.2em}

Factorials as Permutations

$\hspace{0.2em} n! \hspace{0.2em}$ can also be defined as the number of ways $\hspace{0.2em} n \hspace{0.2em}$ distinct objects can be arranged among themselves. In other words, $\hspace{0.2em} n! \hspace{0.2em}$ is the number of permutations of $\hspace{0.2em} n \hspace{0.2em}$ distinct objects taken all at a time.

Let me explain.

Say there are $\hspace{0.2em} 3 \hspace{0.2em}$ books — one each of math, physics, chemistry, and english. In how many different ways can we stack them one over the other.

Think about it. To begin with we can place any of the $\hspace{0.2em} 3 \hspace{0.2em}$ books. So, we have $\hspace{0.2em} 3 \hspace{0.2em}$ options for the bottom position.

Three stacks with each with one book — The first book in the stack (possibilities)

For the next position, we have $\hspace{0.2em} 2 \hspace{0.2em}$ books remaining and we can place any of them. So we have $\hspace{0.2em} 2 \hspace{0.2em}$ options for the middle position.

Three stacks with each with two books — Two books in the stack (possibilities)

For the last position (the top), we don't really have a choice. There's one book left and that's what would go there.

Three stacks with each with three books — All three books in the stack (possibilities)

Putting it all together, we have $\hspace{0.2em} 3 \hspace{0.2em}$ options to begin with. And then for each of those three options, we have $\hspace{0.2em} 2 \hspace{0.2em}$ options. And then for each of those two options, we have just $\hspace{0.2em} 1 \hspace{0.2em}$ option (meaning no choice).

So the total number of arrangements possible would be

3 \times 2 \times 1 \hspace{0.25em} = \hspace{0.25em} 3!

See how factorial came into picture? And of course, this would be true for any number of distinct objects.