Basic Counting Principles

Learning goals

Multiply the choices at each stage of a sequence
Add group sizes when one choice comes from separate, non-overlapping groups
Decide from the wording whether to multiply or to add
Count arrangements with choices that shrink by one at each stage
Raise choices to a power when every stage repeats the same options
Draw a tree or a grid to see why the multiplication works

The multiplication principle

Start with the lunch, but shrink it so we can see everything. Say there are $2$ sandwiches, ham and veg, and $3$ drinks, tea, milk, and juice. A tree diagram lays out the choices in stages. Draw one branch for each sandwich, then from the end of every sandwich branch draw one branch for each drink. Each path from left to right makes one complete lunch, and the tips on the far right are all the lunches there are.

A tree for choosing one of 2 sandwiches and then one of 3 drinks. Each of the 2 sandwich branches splits into the same 3 drink branches, giving 2 groups of 3 tips. The 6 tips on the right are every possible lunch.

The tree makes the count easy to see. There are $2$ sandwich branches, and each one sprouts the same $3$ drink branches, so the tips come in $2$ equal groups of $3$ . That is $3 + 3 = 6$ tips, and counting equal groups is exactly what multiplication does, so $2 \times 3 = 6$ lunches. This is the multiplication principle, also called the fundamental counting principle, and the same argument works for any two independent stages.

Why $m$ choices then $n$ choices give $m \times n$ outcomes#

Suppose the first stage has $m$ possible choices, and that whichever one you pick, the second stage always offers the same $n$ choices. Draw the tree. From one starting point, draw a branch for each first-stage choice, which gives $m$ branches and finishes the first stage.

Now grow the second stage. At the tip of every one of those $m$ branches the second stage offers $n$ choices, so draw $n$ new branches there. Each first-stage branch sprouts exactly $n$ branches of its own. The tip of each of those second-stage branches is one complete outcome: a first choice paired with a second choice.

Count the tips. There are $m$ first-stage branches, and each ends in $n$ tips, so the tips fall into $m$ equal groups of $n$ . Adding $n$ a total of $m$ times is the definition of multiplication, so the number of tips is

\underbrace{n + n + \cdots + n}_{m \text{ groups}} = m \times n.

Every outcome is one tip and every tip is one outcome, with none repeated and none left out. So the number of combined outcomes is exactly the number of tips, $m \times n$ .

Trees, lists, and grids

A tree is one way to organize a count, and two others are worth keeping in your toolkit. They earn that place because they let you verify a count by hand. Reading the tree’s tips from top to bottom gives an organized list of every outcome. That list reads: ham with tea, ham with milk, ham with juice, veg with tea, veg with milk, veg with juice. Listing in a fixed order means, for instance, holding the sandwich, running through the drinks, then moving to the next sandwich. That is what keeps you from skipping an outcome or counting one twice. Six items on the list, matching $2 \times 3$ .

A grid shows the same count as a rectangle. Put the sandwiches down the side and the drinks across the top, and each cell is one lunch. A rectangle with $2$ rows of $3$ cells holds $2 \times 3 = 6$ cells. That is the multiplication principle in picture form, since the area of a rectangle is rows times columns.

The same 6 lunches as a grid: 2 sandwich rows and 3 drink columns. Each cell is one outcome, so the count is 2 rows times 3 columns, the same multiplication as the tree.

Worked example 1 Count lunches at the full diner

The diner from the opening offers $4$ sandwiches and $3$ drinks, and a lunch is one sandwich and one drink. How many different lunches are possible?

This is a sequence of two stages: first choose a sandwich, then choose a drink. The stages are independent, since every sandwich can pair with every drink. Apply the multiplication principle with $m = 4$ and $n = 3$ :

4 \times 3 = 12.

There are $12$ possible lunches. As a check, a tree would draw $4$ sandwich branches, each splitting into $3$ drinks. That gives $4$ groups of $3$ tips, which is $12$ tips in all.

Three or more stages

The principle does not stop at two stages. If you make a third independent choice, every one of the existing outcomes splits again, once for each new option. So you multiply by the third stage’s count too. In general, the number of ways to complete a sequence of independent stages is the product of the number of choices at each stage:

(\text{stage 1 choices}) \times (\text{stage 2 choices}) \times (\text{stage 3 choices}) \times \cdots

The reasoning is the same each time. Adding one more stage with $k$ choices replaces every outcome so far with $k$ versions of it, which multiplies the running total by $k$ .

Worked example 2 Build a sundae in three stages

A sundae is built in three stages: choose $1$ of $3$ ice-cream flavors, then $1$ of $2$ sauces, then $1$ of $4$ toppings. How many different sundaes are possible?

Each stage is an independent choice, so multiply the three stage counts together:

3 \times 2 \times 4 = 24.

There are $24$ sundaes. You can see the multiplying happen step by step. The flavor and sauce alone give $3 \times 2 = 6$ combinations, and then each of those $6$ splits into $4$ topping choices, $6 \times 4 = 24$ .

Check your understanding

A smoothie shop lets you pick one of $3$ sizes and one of $5$ flavors. How many different smoothies (one size and one flavor) are possible?

15

8

5

3

The addition principle

Not every problem is a sequence of choices. Sometimes you make a single choice, picking one item from one of several separate groups. When the groups do not overlap, you add their sizes instead of multiplying.

Suppose a prize shelf holds $5$ books and $3$ board games, and you take exactly one item home. Your prize is either a book or a game, never both at once. Because no item belongs to both groups, you can slide the two groups together into one pile of $5 + 3 = 8$ items. Choosing one prize is then just choosing one item from that pile. So there are $8$ possible prizes. This is the addition principle. If you choose one item from groups that share no members, the number of choices is the sum of the group sizes.

The contrast with the multiplication principle is the entire point, so hold the two side by side. A sequence of choices, “this and then that,” multiplies, because each first choice opens up a fresh set of second choices. A single choice from separate groups, “this or that,” adds, because the groups sit beside each other and you land in only one of them. The numbers can be identical while the answers differ: $5$ and $3$ give $5 \times 3 = 15$ for one book and one game. But $5 + 3 = 8$ is the count for one prize that is a book or a game.

Worked example 3 Add or multiply? Reading the problem

A club must send one representative to a meeting. The representative may be any of the $7$ seniors or any of the $6$ juniors. How many choices does the club have?

The club picks one person, and that person comes from one of two separate groups (a senior or a junior, not both). No one is counted in both groups, so this is a single choice from separate groups: add.

7 + 6 = 13.

There are $13$ choices. Now compare a different task: send one senior and one junior as a pair of representatives. That is a sequence of two choices, so it multiplies, $7 \times 6 = 42$ . The word that decides it is “or” (one representative, add) versus “and” (a senior and a junior, multiply).

Check your understanding

A lunch special lets you choose your single main course from $4$ hot dishes or $3$ cold salads, and you choose exactly one dish. How many choices do you have for your one dish?

12

7

4

1

Counting arrangements stage by stage

A special kind of counting asks in how many orders a set of items can be placed. Take $3$ different books, A, B, and C, that you want to line up on a shelf. Treat each shelf position as a stage. For the first position you may use any of the $3$ books. Once that book is placed it is used up, so only $2$ books remain for the second position. After that, just $1$ book remains for the third. The choices shrink at each stage, and by the multiplication principle you multiply them:

3 \times 2 \times 1 = 6.

Arranging 3 books in a row, one position at a time. The first position has 3 choices, the second has 2 (one book is now used), and the last has 1. Multiplying the shrinking choices gives 3 times 2 times 1 = 6 orders.

This is not a new rule. It is the multiplication principle applied to stages whose choice counts shrink by one each time. The counts shrink because each item you place is no longer available for the next position. You do not always fill every position from the whole set, either. If you only want the first few places in order, just multiply the shrinking counts for as many stages as you fill, and stop.

Worked example 4 Hand out gold, silver, and bronze

Six runners finish a race, and the top three earn gold, silver, and bronze medals. In how many ways can the three medals be awarded?

Award the medals as a sequence of three stages, watching the choices shrink. Any of the $6$ runners can win gold. Once gold is decided, $5$ runners remain for silver, and then $4$ remain for bronze:

6 \times 5 \times 4 = 120.

There are $120$ ways. The choices drop from $6$ to $5$ to $4$ because a runner who already has a medal cannot receive another. That is exactly the shrinking that arrangements always produce. Notice we stopped after three stages, since only three medals are given out.

Check your understanding

In how many different orders can $4$ different books be arranged in a row on a shelf?

24

12

256

8

Repeated choices and exponents

When the same number of choices repeats at every stage, the product becomes a power, which connects counting straight back to exponents. Flip a coin and there are $2$ outcomes, heads or tails. Flip it again and each of those splits into $2$ , giving $2 \times 2 = 4$ . Flip it a third time and you reach $2 \times 2 \times 2 = 8$ . For $n$ flips you multiply $n$ twos:

\underbrace{2 \times 2 \times \cdots \times 2}_{n \text{ flips}} = 2^n.

The exponent counts how many stages there are, and the base counts the choices at each stage. The same shape appears whenever choices repeat. A $4$ -digit code, where each digit is independently one of the $10$ digits $0$ through $9$ , has

10 \times 10 \times 10 \times 10 = 10^4 = 10{,}000

possible values, which is why a $4$ -digit PIN has exactly ten thousand settings. Watch the order of the base and the exponent: $2^n$ (two choices, $n$ times) and $n^2$ ( $n$ choices, twice) are different counts. So keep straight which number is the choices and which is the number of stages.

Worked example 5 How many 4-digit codes?

A lock uses a $4$ -digit code, and each of the four positions can be any digit from $0$ to $9$ , independently of the others. How many codes are possible?

Each position is an independent stage with $10$ choices, and the same $10$ choices repeat at all four positions. Multiply four tens, which is a power of ten:

10 \times 10 \times 10 \times 10 = 10^4 = 10{,}000.

There are $10{,}000$ codes, from $0000$ to $9999$ . The choices do not shrink here, because digits may repeat (a code like $7727$ is allowed), so every stage keeps all $10$ options. That repetition is what turns the product into the clean power $10^4$ .

Counting the equally likely possibilities

These rules matter beyond menus and medals. The reason is that they answer a question the next lesson depends on: how many possible results does an experiment have? Rolling two dice, drawing a card then a second card, spinning a spinner twice: each of these is a sequence of stages. The multiplication principle counts the experiment’s outcomes without forcing you to list them. When those outcomes are all equally likely, that count is the foundation of the next step. In the lesson on probability you will measure how likely an event is. You will do that by comparing the number of outcomes you want against this total number of possibilities. For now the job is just to find that total quickly and correctly, which is exactly what these principles do.

Common mistakes

Adding when you should multiply, or the reverse. A sequence of choices (“one of these and then one of those”) multiplies. A single choice from separate groups (“one item, either from here or from there”) adds. Read the problem for “and” versus “or” before you reach for an operation.
Forgetting that arrangement choices shrink. When you place items in order, each item used is gone, so the counts fall $4, 3, 2, 1$ . Writing $4 \times 4 \times 4 \times 4$ wrongly lets an item repeat.
Letting choices shrink when repetition is allowed. A $4$ -digit code can repeat digits, so every position keeps all $10$ choices, giving $10^4$ , not $10 \times 9 \times 8 \times 7$ . Shrink the counts only when an item cannot be reused.
Swapping the base and the exponent. With repeated choices, the base is the number of choices per stage and the exponent is the number of stages. Flipping a coin $5$ times is $2^5 = 32$ , not $5^2 = 25$ .
Multiplying stages that are not independent. The multiplication principle assumes each stage offers the same number of choices no matter what came before. If an earlier choice changes the later count, adjust the later count (as the shrinking in arrangements does) rather than blindly multiplying.
Double counting or skipping when listing. When you verify by listing, move through the choices in a fixed order so every outcome appears exactly once. A scattered list is how outcomes get repeated or lost.

Key takeaways

The multiplication principle: for a sequence of independent stages, multiply the number of choices at each stage. Two stages with $m$ and $n$ choices give $m \times n$ outcomes, because the outcomes form $m$ equal groups of $n$ .
The addition principle: to make one choice from separate, non-overlapping groups, add the group sizes.
Which to use is the key skill: “and / then” (a sequence) multiplies, “or / one item” (separate groups) adds.
Arrangements count orderings by stages whose choices shrink by one each time, as in $3 \times 2 \times 1 = 6$ , because each item placed is then used up.
Repeated choices turn into a power: the same $c$ choices at each of $n$ stages give $c^n$ outcomes. Examples are $2^n$ coin-flip results, or $10^4$ four-digit codes.
These counts give the total number of equally likely possibilities, the foundation for measuring probability in the next lesson.

Basic Counting Principles

Learning goals

The multiplication principle

Trees, lists, and grids

Three or more stages

The addition principle

Counting arrangements stage by stage

Repeated choices and exponents

Counting the equally likely possibilities

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)