# 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:

$(1 2 3), (1 3 2), (2 1 3), (2 3 1), (3 1 2), (3 2 1)$

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:

$(4 2 3 1), (2 4 3 1), (2 3 4 1), (2 3 1 4)$

Method 2:

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

\begin{align} 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{align}

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{align} \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{align}

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

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