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The factorial is an important function in combinatorics and analysis, used to determine the number of ways to arrange objects.


The factorial is defined for positive integers as $n!=n \cdot (n-1) \cdots 2 \cdot 1$ Alternatively, a recursive definition for the factorial is: $n!=n \cdot (n-1)!$.

Additional Information

By convention, $0!$ is given the value $1$.

The gamma function is a generalization of the factorial to values other than nonnegative integers.

Prime factorization

Since $n!$ is the product of all positive integers not exceeding $n$, it is clear that it is divisible by all primes $p\le n$ and not divisible by any prime $p>n$. But what is the power of a prime $p\le n$ in the prime factorization of $n!$? We can find it as the sum of powers of $p$ in all the factors $1,2,\dots, n$ but rather than counting the power of $p$ in each factor, we shall count the number of factors divisible by a given power of $p$. Among the numbers $1,2,\dots,n$ exactly $\left\lfloor\frac n{p^k}\right\rfloor$ are divisible by $p^k$ (here $\lfloor\cdot\rfloor$ is the floor function). Now we will use the Lebesgue counting principle that says that the sum of several non-negative integers $a_1,\dots, a_n$ (the powers of $p$ in numbers $1,2,\dots,n$ in our case) can be found as $\sum_{k\ge 1}\#\{j: a_j\ge k\}$. This immediately gives the formula

$\left\lfloor\frac n{p}\right\rfloor+ \left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\dots$

for the power of $p$ in the prime factorization of $n!$. The series is formally infinite, but the terms become $0$ pretty fast. For example, the power of $7$ in $100!$ is just $\left\lfloor\frac {100}{7}\right\rfloor+ \left\lfloor\frac {100}{49}\right\rfloor=14+2=16$ ($7^3=343$ is already greater than $100$).


The factorial is used in the definitions of combinations and permutations, as $n!$ is the number of ways to order $n$ distinct objects.


See also

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