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

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

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

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

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 $\hspace{0.2em} 1 \hspace{0.2em}$.

$\hspace{0.2em} n! \hspace{0.2em}$ 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.

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.

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.

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.

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