AOCP/Permutations

Volume 1

Chapter 1: Basic Concepts

Permutations and Factorials

If we have n distinct objects to arrange in a row, and order of placement matters, we can arrange these things in n! different ways.

For the first object, we have n choices of places to put it. For the second object, we have n-1 choices of places to put it. And so on.

In general, if we have to choose k objects out of n total, and arrange them in a row, we have the number of possibilities as:

$p_{n,k}=n(n-1)(\dots )(n-k+1)$

The total number of permutations is

$p_{n,n}=n(n-1)\dots (2)(1)$

The process of constructing a permutation of n objects, given all permutations of n-1 objects, is important. If we consider the case of three objects $\{1,2,3\}$,

Here are permutations of order 3:

$(123),(132),(213),(231),(312),(321)$

Now, how to get to permutations of 4 objects?

Method 1:

For each permutation of n-1 elements, form n additional permutations by inserting the nth element in every possible open slot

Adding 4 to our set of objects and using method 1 gives:

$(4231),(2431),(2341),(2314)$

Method 2:

For each permutation of the n-1 elements, form n new permutations by first constructing the array:

${\begin{aligned}a_{1}a_{2}\dots a_{n-1}&\quad &{\frac {1}{2}}\\a_{1}a_{2}\dots a_{n-1}&\quad &{\frac {3}{2}}\\&\dots &\\a_{1}a_{2}\dots a_{n-1}&\quad &(n-{\frac {1}{2}})\end{aligned}}$

Aaaaaand... yeah. No idea.

Factorial Identities

Factorial definition:

$n!=1\cdot 2\dots n=\prod _{1\leq k\leq n}k$

$0!=1$

and with this convention,

$n!=(n-1)!n$

for all positive integers n.

10! is a useful benchmark - it is around 3.5 million.

$10!=3,628,800$

10! represents an upper ceiling on computable tasks.

To tell how large a very big factorial is going to be, use Stirling's formula:

$n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$

Thus, for 8! = 40320:

$8!\approx 4{\sqrt {\pi }}\left({\frac {8}{e}}\right)^{8}\approx 39902$

Relative error of Stirling's formula is approximately ${\dfrac {1}{12n}}$

To obtain the exact value of n! factored into primes, we can use some useful identities. First, the prime p is a divisor of n! with multiplicity:

${\begin{aligned}\mu &=&{\mbox{floor}}({\frac {n}{p}})+{\mbox{floor}}({\frac {n}{p^{2}}})+{\mbox{floor}}({\frac {n}{p^{3}}})+\dots \\&=&\sum _{k>0}{\mbox{floor}}({\frac {n}{p^{k}}})\end{aligned}}$

As an example, for n = 1000, p = 3,

$\mu ={\mbox{floor}}({\frac {1000}{3}})+{\mbox{floor}}({\frac {1000}{9}})+{\mbox{floor}}({\frac {1000}{27}})+{\mbox{floor}}({\frac {1000}{81}})+{\mbox{floor}}({\frac {1000}{243}})$

which gives

$\mu =333+111+37+12+4+1=498$

Therefore, $1000!$ is divisible by $3^{498}$, but not by $3^{499}$.

Furthermore, to speed up the calculation of the above, we can use the identity

${\mbox{floor}}\left({\frac {n}{p^{k+1}}}\right)={\mbox{floor}}\left({\frac {{\mbox{floor}}({\frac {n}{p^{k}}})}{p}}\right)$

Binomial Coefficients

See AOCP/Binomial Coefficients

Multinomial Coefficients

See AOCP/Multinomial Coefficients

Generating Functions

See AOCP/Generating Functions