Hey Developer’s, I’m back with a new topic which is Combinatorics in the series of statistics foundations.

### Quick Refresher

Given a set of elements, all distinct arrangements of these elements are called the permutation of a set.

The number of ways you can arrange N elements out of N possibilities is…

P(n,n) = N X (N-1) X (N-2) X …. X 2 X 1 = N! permutations

So, let’s get started …

### Combinatorics

Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures.

Combinatorial Methods

Given a bag of balls B = {R,G,B,W,O}

If we pull out 2 balls after replacement, the number of arrangements is : P(5,2) = 5!/3! = 20

What if we don’t care about the order, i.e,. we consider {R,B} and {B,R} to be the same pick ?

How can we calculate the number of unique picks now ?

Problem: Find the number of subsets of a set(remember, the order doesn’t matter)

Given a bag of balls B = {R,G,B,W,O}

The total number of picks (including duplicate sets) is P(5,2) – And each of the picks has 2! possible arrangements.

So we reduce the total number of picks by the number of arrangements in each pick:

C(n,k) = P(n,k)/k! = n!/k!(n-k)!

In our case, C(5,2) = 5!/2!(5-2)!

Here they are: {RG,RB,RW,RO,GB,GW,GO,BW,BO,WO}

#### Next Post will be on Binomial Coefficients

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