Math · Probability & Statistics · Combinatorial Analysis· Part 1 of 7

Combinatorial Analysis — Series Overview

May 2026 · 8 min read

Probability theory is, at its heart, the mathematics of counting: counting how many ways an event can happen, divided by how many ways anything can happen. Sheldon Ross opens A First Course in Probability with an entire chapter devoted to combinatorial analysis for exactly that reason. If you cannot count outcomes carefully, every formula you write afterwards will quietly leak.

This series rebuilds Chapter 1 of Ross from scratch. We will derive every formula instead of memorising it, animate the constructions whenever they help, and translate each idea into the kind of mental model you can use under exam pressure.

What This Series Covers

Each post stands alone but they form a deliberate progression: the basic principle of counting feeds permutations, permutations feed combinations, combinations feed the binomial theorem, and the binomial theorem generalises into multinomial coefficients and stars-and-bars. By the end you will be able to look at almost any counting problem and reduce it to a known shape.

PART 2

The Basic Principle of Counting

Multiplication rule, generalised k-stage rule, addition rule for disjoint cases, complement counting. Tree-of-choices animation.

PART 3

Permutations

$n!$, $P(n,r)$, permutations with identical objects, circular permutations. Slot-filling animation showing where every factor comes from.

PART 4

Combinations & Pascal's Triangle

$\binom{n}{r}$, Pascal's identity, hockey-stick, Vandermonde. Animation that grows Pascal's triangle row by row.

PART 5

The Binomial Theorem

Why $(x+y)^n = \sum \binom{n}{k}x^{n-k}y^k$, combinatorial proofs, identities you'll meet again in probability and CP.

PART 6

Multinomial Coefficients

Distributing into more than two groups, the multinomial theorem, and how it generalises everything before it.

PART 7

Integer Solutions & Stars and Bars

Counting non-negative and positive solutions of $x_1 + x_2 + \dots + x_r = n$. Star-and-bar placement animation.

Prerequisites

Almost none. You should be comfortable with:

Basic algebra and exponents.
The factorial function $n! = n \cdot (n-1) \cdots 2 \cdot 1$ with the convention $0! = 1$.
Sigma notation $\sum_{k=0}^{n} a_k$.

Everything else — including the binomial coefficient $\binom{n}{k}$ — is built up step by step inside the series.

Notation Cheat Sheet

Symbol	Reads as	Means
$n!$	"n factorial"	$n \cdot (n-1) \cdot (n-2) \cdots 1$, with $0! = 1$.
$P(n, r)$	"permutations of n taken r at a time"	$\dfrac{n!}{(n-r)!}$ — ordered arrangements of $r$ items chosen from $n$.
$\binom{n}{r}$	"n choose r"	$\dfrac{n!}{r!\,(n-r)!}$ — unordered subsets of size $r$.
$\binom{n}{n_1, n_2, \dots, n_k}$	"multinomial coefficient"	$\dfrac{n!}{n_1!\,n_2! \cdots n_k!}$ when $n_1 + \dots + n_k = n$.
$\|S\|$	"size of S"	Number of elements in the set $S$.

Why Counting Matters for Probability

For an experiment with finitely many equally likely outcomes, the classical definition of probability is

$\Pr(E) = \dfrac{|E|}{|S|}$

where $S$ is the sample space and $E \subseteq S$ is the event. Both $|E|$ and $|S|$ are counts. Almost every probability problem in Ross's first three chapters reduces to:

Decide whether order matters.
Decide whether the items are distinguishable.
Decide whether repetition is allowed.
Pick the right tool from this series and crank the formula.

The four counting "moods". Order vs. no order, with vs. without repetition. Those four combinations cover the bulk of Ross's exercises and almost every introductory probability problem you will ever face.

Visual Preview

Here is the kind of step-by-step animation you'll see throughout the series. We pick an outfit by choosing one shirt, then one pair of pants. The total number of outfits is the product of the two stage counts — the very statement of the multiplication rule we will prove in Part 2.

▶ Visual: The Multiplication Rule on a Tree

Step through choosing one of 3 shirts and then one of 2 pants. Every leaf is a distinct outfit; counting the leaves yields $3 \times 2 = 6$.

How to Use This Series

Read the posts in order if you are new to combinatorics. If you've seen the material before, skim the “Why It Works” sections — that is where the proof intuition lives. Whenever a formula tempts you to memorise it, run the animation, count by hand, and only then trust the formula.

Each post ends with practice problems drawn from Ross plus a few extras. Solve them with pen and paper before reaching for code; combinatorics is a language and you have to speak it before you optimise it.