Factorial

Revision as of 19:51, 29 March 2011 by Williamhu888 (talk | contribs) (Examples)

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)

See Also

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