Factorial

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

Definition

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

Examples (Factorials from $0!$ to $30!$ )

$0! = 1$
$1! = 1$
$2! = 2$
$3! = 6$
$4! = 24$
$5! = 120$
$6! = 720$
$7! = 5040$
$8! = 40320$
$9! = 362880$
$10! = 3628800$
$11! = 39916800$
$12! = 479001600$
$13! = 6227020800$
$14! = 87178291200$
$15! = 1307674368000$
$16! = 20922789888000$
$17! = 355687428096000$
$18! = 6402373705728000$
$19! = 121645100408832000$
$20! = 2432902008176640000$
$21! = 51090942171709440000$
$22! = 1124000727777607680000$
$23! = 25852016738884976640000$
$24! = 620448401733239439360000$
$25! = 15511210043330985984000000$
$26! = 403291461126605635584000000$
$27! = 10888869450418352160768000000$
$28! = 304888344611713860501504000000$
$29! = 8841761993739701954543616000000$
$30! = 265252859812191058636308480000000$

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

Main article: 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). The ones divisible by $p$ give one power of $p$ . The ones divisible by $p^2$ give another power of $p$ . Those divisible by $p^3$ give yet another power of $p$ . Continuing in this manner gives

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

for the power of $p$ in the prime factorization of $n!$ . The series is formally infinite, but the terms converge to $0$ rapidly, as it is the reciprocal of an exponential function. 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 $100$ ).

Uses

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

Problems

Introductory

Find the units digit of the sum

$\sum_{i=1}^{100}(i!)^{2}$

$\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,3\quad\mathrm{(D)}\,5\quad\mathrm{(E)}\,7\quad\mathrm{(F)}\,9$ (Source)

Intermediate

Let $P$ be the product of the first $100$ positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k .$

(Source)

Olympiad

Let $p_n (k)$ be the number of permutations of the set $\{ 1, \ldots , n \} , \; n \ge 1$ , which have exactly $k$ fixed points. Prove that $\sum_{k=0}^{n} k \cdot p_n (k) = n!$ .

(Source)

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