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{{ printedBook.courseTrack.name }} {{ printedBook.name }} For any integer $n$ which is positive, the product of all integers from $1$ to $n$ is called $n$ factorial,

denoted as $n!$

$n!=1⋅2⋅3⋅…⋅(n−2)⋅(n−1)⋅n$

The factorial of $n$ can alternatively be defined as the product of all the natural numbers less than $n.$ Therefore, it is also very common to see the factors listed in descending order.

$n!=n⋅(n−1)⋅(n−2)⋅…⋅3⋅2⋅1$

The table below shows the factorial of some numbers. By rewriting the expressions, it can be seen that a special pattern emerges.

$n$ | $n!$ | Simplify | Rewrite |

$1$ | $1$ | $1$ | $1$ |

$2$ | $2⋅1$ | $2$ | $2⋅1!$ |

$3$ | $3⋅2⋅1$ | $6$ | $3⋅2!$ |

$4$ | $4⋅3⋅2⋅1$ | $24$ | $4⋅3!$ |

$5$ | $5⋅4⋅3⋅2⋅1$ | $120$ | $5⋅4!$ |

$⋮$ | $⋮$ | $⋮$ | $⋮$ |

From the table above, it can be concluded that the factorial of a number follows a recursive rule.

$1!=1,n!=n⋅(n−1)!$

The definitions discussed above apply to positive integers, but it may be helpful to know the value of $0$ factorial. By convention, $0!=1.$

$0!=1$

This may seem strange since, by the definition of factorial, this means that the product of multiplying no numbers is equal to $1.$ Nevertheless, defining $0!$ as $1$ is necessary for the usage of the factorial of a number. For example, if the recursive property is used with $n=1,$ the following result is obtained.
$n!1!1!1 =n⋅(n−1)!=1⋅(1−1)!=1⋅0!=1⋅0! $
The last equality, $1=1⋅0!,$ only holds true if $0!=1.$ In a similar way, there are many applications of the factorial in different areas of mathematics that only make sense if $0!=1.$ For example, the number of ways to choose groups of $r$ elements from a set of $n$ elements — combinations of $n$ in $r$ — is given by the formula shown below.
$_{n}C_{r}=(n−r)!n!n! $
In particular, there is only **one** way to make groups of $n$ elements from a set having $n$ elements. This is by making a group using all the elements. This is what happens when substituting $n=n$ and $r=n.$
$_{n}C_{n}11 =(n−n)!n!n! =0!n!n! =0!1 $
Once more, for everything to be consistent, the only value that can be assigned to $0!$ is $1.$