Ⅵ Counting, Probability & Statistics · Stage 32 — Counting & the Binomial Theorem · 32.1 The Two Counting PrinciplesAll lessons →
Stage 32 · Counting, Permutations, Combinations & the Binomial Theorem

The Two Counting Principles

Add for kinds, multiply for steps — find the count without writing the list.

Ages 14 to 18 · Reasoning, one step at a time
Three shirts, then two pants: the tree splits into 3 branches, each into 2 more, so it ends in 6 leaves — and 6 is exactly 3 × 2. Counting is reading the leaves without naming them one by one.

Suppose you wanted to know how many different outfits you can make from 3 shirts and 2 pairs of pants. You could write them all out — shirt 1 with pants A, shirt 1 with pants B, shirt 2 with pants A — and then count the lines. That works for a tiny closet. But ask how many 4-digit PINs a phone allows, or how many license plates a state can print, and the list runs to thousands or millions; nobody writes that out. Counting is the art of finding the total N without ever writing the list. Almost everything comes from two plain habits of mind: add when a job splits into separate kinds (an "or"), and multiply when a job is a relay of steps (an "and"). Get those two straight and the rest of this whole stage falls into place.

32.1.1 Why count without listing

Here is the honest motivation. To count the outfits above, the list has just six lines, so listing is fine. But the number of choices explodes the moment a job has several steps. A 4-digit PIN draws each digit from 10 symbols (0 through 9), and there are four digits, so there are 10 × 10 × 10 × 10 = 10,000 possible PINs. You are not going to write ten thousand lines. A car plate with two letters then four digits has 26 × 26 × 10 × 10 × 10 × 10 = 6,760,000 possibilities — over six and a half million.

So the goal is a count, the single number N, found by reasoning about the choices rather than by enumeration. Two principles do the heavy lifting, and the entire trick is telling which one a problem needs.

The one question to ask

Is the job a choice between separate kinds (do this or that — and only one happens) — or a relay of steps (do this and then that — both happen)? "Or" means add. "And" means multiply.

32.1.2 The Addition Principle — counting by cases

Imagine you need to get out of town for the day. There are 3 trains you could take or 2 buses. You will take exactly one of them — a single trip is either a train trip or a bus trip, never both at once. The cases don't overlap. To count all the ways to leave, you just pool them:

N = trains + buses = 3 + 2 = 5 ways.

That is the Addition Principle: if a task can be done in one of several disjoint cases — case 1 in m1 ways, case 2 in m2 ways, and so on, with no way counted in two cases — then the number of ways to do the task is

N = m1 + m2 + … + mk.

Two disjoint cases — take a train or take a bus. Pool them: 3 + 2 = 5 ways to leave town. Nothing is in both buckets, so there is no double-counting.
Watch — disjoint or you'll double-count

Adding is only legal when the cases can't happen together. If some option counted as a "train" also counted as a "bus," you'd add it twice. When in doubt, make sure every outcome lands in exactly one bucket before you add.

32.1.3 The Multiplication Principle — counting by steps

Now a different shape of problem. An outfit is a shirt and a pair of pants — both have to be chosen; finishing the shirt does not finish the outfit. This is a relay of steps. With 3 shirts, every single shirt can be paired with each of the 2 pants, so each shirt opens 2 outfits and there are three shirts:

3 shirts × 2 pants = 6 outfits.

That is the Multiplication Principle: if a task is completed by a sequence of independent steps — step 1 in m1 ways, step 2 in m2 ways, and so on, where the count at each step does not depend on the earlier choices — then the number of ways to do the whole task is

N = m1 × m2 × … × mk.

The tree at the top of this lesson is the picture of exactly this. The root splits into 3 shirt-branches; each of those splits into 2 pants-branches; the tree ends in 3 × 2 = 6 leaves, and every leaf is one complete outfit. Dial the numbers yourself.

Try it Outfit tree — multiply the steps, count the leaves
Set the number of shirts and pants. Watch the leaf count and the product N = m × n stay locked together.
Shirts m 3
Pants n 2

32.1.4 "Cases" or "steps"? — the or/and test

Everything hinges on one decision. Read the problem and ask: does a single choice finish the job, or does it take a relay of choices to finish? If picking a train already gets you out of town, one choice finished it — those are cases, joined by "or", so you add. If picking a shirt leaves you still needing pants, the job isn't done — those are steps, joined by "and", so you multiply.

Try saying the problem out loud with "or" and with "and" and hear which one is true: "I take a train or a bus" (true → add) versus "I pick a shirt and I pick pants" (true → multiply). The little word is the whole answer.

Try it Trip planner — flip between "or" and "and"
Toggle the scenario. "OR" pools disjoint routes (add); "AND" chains two legs of one trip (multiply). The green total updates with the matching arithmetic.
Example — both words in one problem

"Travel from home to the lake: you may go by land (3 roads) or by water (2 ferries)." One trip is land or water → add: 3 + 2 = 5 ways. But "drive to the dock (3 roads) and then take a ferry (2 ferries)" is one trip with two legs → multiply: 3 × 2 = 6 ways. Same numbers, opposite operation — because the first is "or" and the second is "and."

32.1.5 Tree diagrams & listing — count the leaves

A tree diagram is the honest bookkeeping behind the Multiplication Principle. Each step is a level of branching; following one path from root to leaf builds one complete outcome; and the leaves are every outcome there is. Because each of the m1 first-branches grows the same m2 second-branches, and so on, the number of leaves is exactly the product m1 × m2 × … — counting the leaves and multiplying the steps must give the same number, always.

So a tree is both a way to list (read the paths) and a way to count (the leaf total is the product). On a small problem you can do both and check they agree; on a large one you trust the product and skip the drawing.

From A to C there are 3 ways A→B and 2 ways B→C. The tree has 3 first-branches, each with 2 second-branches, ending in 3 × 2 = 6 leaves — six routes A→C. Counting the leaves equals multiplying the steps.

32.1.6 Both principles together — split, then step

Real "how many" questions usually want both tools, in this order: first split into broad cases and add, then inside each case go step by step and multiply. A PIN or a license plate is the cleanest version of the multiply-by-steps idea — a fixed number of slots, each filled independently.

If a slot can hold any of k symbols and there are s slots, every slot multiplies the count by k, so

N = k × k × … × k (s factors) = ks.

A 4-digit PIN is 104 = 10,000; a 6-character code from the 26 letters is 266, more than three hundred million. Dial the slots and symbols below and watch the count grow like a power.

Try it PIN & plate counter — equal steps make a power
Each slot holds any of k symbols, independently. With s slots the count is k multiplied by itself s times — that is ks.
Slots s 4
Symbols k 10

And the "handle the awkward slot first" step: how many license plates have the form letter–letter–digit–digit–digit–digit, but the very first letter may not be the banned letter O (to avoid confusion with zero)? The first letter has 25 choices, the second letter 26, and each of the four digits 10, giving 25 × 26 × 10 × 10 × 10 × 10 = 6,500,000. That is still pure multiplication — every plate fits one mould — so one relay of steps does it.

Now a question that genuinely needs both tools. How many 4-character codes start with a letter or a digit in slot 1, followed by 3 free digits? The first slot's two kinds can't overlap — a symbol is a letter or a digit, never both — so split into two disjoint cases and add; inside each case the four slots are a relay, so step and multiply:

The two cases are disjoint, so add them: 26,000 + 10,000 = 36,000, which is just (26 + 10) × 1000 = 36 × 1000 = 36,000. Split into cases (add), then step within each case (multiply) — that is both principles in one problem, in the order you will use them all chapter long.

What to carry forward

Two habits, one question. Ask "or or and?" — separate kinds, or a relay of steps — and the operation is decided.

Addition PrincipleMultiplication Principle
The little word"or""and"
Shape of the jobseparate kinds / casesa relay of steps
How many happenexactly one caseevery step happens
Operationaddmultiply
FormulaN = m1 + m2 + …N = m1 × m2 × …
Must holdcases disjointsteps independent
Example3 trains + 2 buses = 53 shirts × 2 pants = 6

Up next: when those multiplied steps are an ordered pick, you get a permutation — and the slots offer shrinking choices.

Exercises

  1. A café offers 4 kinds of tea or 3 kinds of coffee. You will order exactly one drink. How many choices do you have?
    Answer
    One drink is a tea or a coffee — disjoint cases, so add: 4 + 3 = 7 choices.
  2. A diner combo is one sandwich and one side and one drink. There are 5 sandwiches, 3 sides, and 4 drinks. How many different combos?
    Answer
    A combo is a relay of three steps — sandwich and side and drink — so multiply: 5 × 3 × 4 = 60 combos.
  3. A 3-character code uses one letter (26 options) followed by two digits (10 options each). How many codes are possible?
    Answer
    Three independent slots: 26 × 10 × 10 = 2,600 codes. (Same as 26 × 102.)
  4. How many 4-digit PINs are possible if each digit is 0–9? How many if no digit may repeat?
    Answer
    Free: each slot has 10 choices, so 10 × 10 × 10 × 10 = 10,000. No repeats: the choices shrink as digits are used, 10 × 9 × 8 × 7 = 5,040 (a preview of permutations).
  5. To travel A → B → C there are 3 roads from A to B and 4 roads from B to C. How many routes from A to C? Now suppose there is also 1 direct flight A → C. How many ways to get from A to C in total?
    Answer
    A route through B is a relay of two legs → multiply: 3 × 4 = 12 routes. Going through B or flying direct are disjoint cases → add: 12 + 1 = 13 ways. (Multiply within a case, add across cases.)
  6. A plate is letter–letter–digit–digit (26 letters, 10 digits), but the first character may not be the letter O. How many plates?
    Answer
    Handle the restricted slot first: 25 choices for the first letter, then 26, 10, 10 for the rest. Multiply: 25 × 26 × 10 × 10 = 65,000 plates.

🎯 Quick check

Six questions to lock it in. Tap the answer you think is right.

§ For teachers and parents

This lesson builds the two fundamental counting principles — addition for disjoint cases ("or"), multiplication for independent steps ("and") — and the tree-diagram intuition behind them, the groundwork for CCSS HSS-CP.B.9 (use permutations and combinations to compute probabilities of compound events). The single decision students must master is whether a task is an "or" (separate kinds, add) or an "and" (a relay of steps, multiply); the trip-planner and tree widgets let them feel the difference, and the PIN counter shows how equal independent steps produce a power ks. Permutations (32.2) and combinations (32.3) are just the multiplication principle applied to ordered and unordered picks.

eastmath.com · Stage 32 · 32.1 The Two Counting Principles · Reasoning, one step at a time