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:

$ (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) $

Binomial Coefficients

The combinations of n objects taken k at a time are the possible choices of k different elements from a collection of n items.

The number of combinations is described by the binomial coefficient $ \binom{n}{k} $

$ \binom{n}{k} = \dfrac{n!}{k! (n-k)!} $

We can define the binomial coefficient for the case when n is not an integer. Define:

$ \binom{r}{k} = \dfrac{r(r-1)(\dots)(r-k+1)}{k(k-1)\dots(1)} = \prod_{1 \leq j \leq k} \left( \frac{r+1-j}{j} \right) \qquad k \in \mathbb{Z}, k \geq 0 $

$ \binom{r}{k} = 0 \qquad k < 0 $

Computing these numbers in a table leads to Pascal's Triangle.

Binomial coefficients satisfy thousands of identities; these can be classified as follows:

• Representation by factorials
• Symmetry condition
• Moving in and out of brackets
• Addition formulas
• Summation formulas
• Binomial theorem
• negating upper index
• Sums of products

Representation by Factorials

$ \binom{n}{k} = \dfrac{n!}{k! (n-k)!} $

Holds for integers n and k >= 0

Symmetry Condition

$ \binom{n}{k} = \binom{n}{n-k} $

Holds for all integers k.

Moving In and Out of Brackets

$ \binom{r}{k} = \frac{r}{k} \binom{r-1}{k-1} \qquad k \neq 0 $

This is useful for combining binomial coefficient with other parts of an expression.

By elementary transformation, we have:

$ k \binom{r}{k} = r \binom{r-1}{k-1} $

valid for all integers k, and

$ \frac{1}{r} \binom{r}{k} = \frac{1}{k} \binom{r-1}{k-1} $

valid when not dividing by zero.

Similarly:

$ \binom{r}{k} = \dfrac{r}{r-k} \binom{r-1}{k} $

for k != r.

Addition formulas

To add binomials, we have

$ \binom{r}{k} = \binom{r-1}{k} + \binom{r-1}{k-1} $

for integer k. This is equivalent to summing the two entries in Pascal's triangle above to the left and above to the right.

Alternatively, we have:

$ r \binom{r-1}{k} + r \binom{r-1}{k-1} = (r-k) \binom{r}{k} + k \binom{r}{k} = r \binom{r}{k} $

This is useful for proofs by induction on r, when r is an integer.

Summation formulas

Two important summation formulas. The first comes from applying the summation identity given above, repeatedly:

$ \sum_{0 \leq k \leq n} \binom{r+k}{k} = \binom{r}{0} + \binom{r+1}{1} + \dots + \binom{r+n}{n} = \binom{r+n+1}{n} \qquad n \geq 0 $

We also have:

$ \sum_{0 \geq k \geq m} \binom{k}{m} = \binom{0}{m} + \binom{1}{m} + \dots + \binom{n}{m} = \binom{n+1}{m+1} $

Valid for integer m and integer n, both >= 0.