Counting combinations

Think about a $4$ -digit ATM PIN, or a coin flipped five times in a row. How many different PINs exist? How many different sequences of heads and tails can that coin produce? You could try to list them all, but for even modest numbers the list explodes into the thousands before you have written ten lines. There is a shortcut hiding in plain sight: whenever the same number of choices repeats, independently, at every step of a task, the total count of possibilities is not found by adding or multiplying-by-the-number-of-steps — it is found by raising the number of choices to a power. This is exponential notation doing real counting work, and it turns a tedious listing problem into one line of arithmetic. This article extends the Class 8 "Power Play" idea of counting with exponents, keeping every worked number from the syllabus article and adding more practice, a proof of why it works, and the traps that catch students first.

The core insight

When a task has $n$ independent steps and every step offers the same $a$ choices, the total number of outcomes is $a^n$ , not $a \times n$ .

The core idea

Many counting problems share one hidden shape: a task broken into a fixed number of steps, where each step offers the same number of choices, and every step's choice is made independently of the others (nothing about an earlier choice restricts a later one). Recognising that shape is the whole skill — once you see it, the count is a single power.

Steps and choices. Call the number of choices available at each step $a$ , and the number of steps $n$ . A $3$ -character PIN using digits $0$ – $9$ has $a = 10$ (ten possible digits) and $n = 3$ (three positions).
Why multiplication, not addition. After step $1$ there are $a$ possible partial outcomes. Each of those $a$ outcomes can be followed by any of the $a$ choices at step $2$ , giving $a \times a$ combined outcomes so far. Continuing this for all $n$ steps multiplies $a$ by itself $n$ times.
The formula. Once the pattern is set up, the total number of outcomes is:

\text{Total outcomes} = a^n

where: $a$ is the number of choices repeated at every step, and $n$ is the number of independent steps.

Where this shape shows up. Repeated coin tosses ( $a = 2$ per toss), dial locks with the same symbol set on every dial, passwords built from a fixed character set with repetition allowed, and any "menu repeated $n$ times" situation. The one non-negotiable condition is that $a$ stays the same at every step — if the choices shrink (say, digits cannot repeat), $a^n$ no longer applies, and that harder counting method belongs to a later grade.

Figure

A tree diagram for two coin tosses showing two branches from the start each splitting into two more branches giving four final outcomes HH HT TH TT illustrating that repeated choices multiply

Derivation AWhy repeated independent choices give a power

The goal is to show, from nothing more than the definition of multiplication and the meaning of a power, that a task with $n$ steps and the same $a$ choices at every independent step must produce exactly $a^n$ total outcomes — not $a \times n$ , and not any other combination — so that the shortcut can be trusted rather than just memorised.

Step 1. Start with one step. If there is only $n = 1$ step and it offers $a$ choices, the number of outcomes is obviously $a$ . This matches $a^1 = a$ , so the formula holds for the smallest case.

Step 2. Add a second step. For each of the $a$ outcomes of step $1$ , step $2$ independently offers $a$ fresh choices (independence is what makes this true — nothing from step $1$ removes or adds options at step $2$ ). So every one of the $a$ step- $1$ outcomes branches into $a$ step- $2$ outcomes, giving $a$ groups of $a$ , which is $a \times a = a^2$ total outcomes after two steps.

Step 3. Extend the pattern one step at a time. Suppose after $k$ steps there are $a^k$ outcomes (this matches Steps 1 and 2 for $k = 1$ and $k = 2$ ). Adding one more independent step with $a$ choices multiplies every existing outcome by $a$ again, since each of the $a^k$ partial outcomes can be followed by any of the $a$ new choices. That gives $a^k \times a = a^{k+1}$ outcomes after $k+1$ steps — exactly one more power of $a$ , matching the exponent rule $a^k \times a = a^{k+1}$ .

Step 4. Conclude for $n$ steps. Repeating Step 3 from $k = 1$ up to $k = n-1$ shows the outcome count climbs $a, a^2, a^3, \ldots, a^n$ one multiplication at a time, so after all $n$ steps the total is $a^n$ . This is why the shortcut works: it is repeated multiplication by $a$ , exactly $n$ times, which is the definition of $a^n$ .

Worked example

A student is asked to create a $3$ -character login code using only the digits $0$ – $9$ , and digits are allowed to repeat. How many different codes are possible?

Step 1. Identify $a$ , the number of choices at each position: there are $10$ possible digits ( $0, 1, 2, \ldots, 9$ ) for every character.

Step 2. Identify $n$ , the number of positions: there are $3$ characters in the code, and each character's digit is chosen independently of the others (repetition is allowed).

Step 3. Apply the formula: since the same $10$ choices repeat independently across all $3$ positions, the total number of codes is $10^3 = 10 \times 10 \times 10 = \mathbf{1000}$ , running from $000$ to $999$ .

Competitive stretch

Extend the same lock to $4$ characters, but now suppose the first character must be a letter from a set of $26$ letters, while the remaining $3$ characters are still digits $0$ – $9$ . How many codes are possible?

Step 1. This task no longer has the same $a$ at every step, so the plain power rule $a^n$ cannot be applied to the whole code at once — it applies separately within each block of identical choices.

Step 2. The first position has $26$ independent choices (letters); the remaining $3$ positions each repeat the same $10$ digit choices, giving $10^3 = 1000$ combinations for that block by the same reasoning as above.

Step 3. Because the letter choice and the digit block are also chosen independently of each other, the counts multiply together (an extension of the same idea): total codes $= 26 \times 10^3 = 26 \times 1000 = \mathbf{26{,}000}$ .

Common traps

Multiplying instead of exponentiating: writing $10 \times 3 = 30$ instead of $10^3 = 1000$ for the $3$ -digit code — the number of choices must be raised to the power of the number of steps, not multiplied by it.
Using $a^n$ when the choices are not identical at every step: if a code cannot repeat digits, the choices shrink at each step ( $10$ , then $9$ , then $8$ , …), so a plain power overcounts and this needs a different method from a later grade.
Ignoring dependence between steps: the shortcut only works when every step's choice is genuinely free of the earlier ones; if an earlier choice restricts what comes later (as in the no-repeat case), independence fails and $a^n$ is wrong.
Mixing up counting with likelihood: finding that there are $32$ possible $5$ -toss sequences answers "how many outcomes exist," which is a different question from "how likely is one particular outcome" — probability is a separate topic.

Ask the tutor: Want to try a mixed counting problem — like a lock with letters in some positions and digits in others — and check your reasoning step by step?

How is a^n different from just multiplying a and n together

$a^n$ means $a$ multiplied by itself $n$ times — for the login code, $10^3 = 10 \times 10 \times 10 = 1000$ . Plain multiplication $a \times n$ would give $10 \times 3 = 30$ , which massively undercounts because it ignores that every earlier choice branches into fresh choices at the next step. The power notation is a shorthand for repeated multiplication, exactly matching how the choices actually combine.

Why does the number of choices have to stay the same at every step for this to work

The formula $a^n$ comes from multiplying the same number $a$ by itself $n$ times, so it only makes sense if every step genuinely offers that same $a$ choices. If a step offers a different number of choices — say the choices shrink because repetition is not allowed — the count is still found by multiplying the choices at each step together, but the numbers being multiplied are no longer identical, so it cannot be written as a single power $a^n$ . That situation needs the more general counting method taught in later grades.

Is this the same as permutations and combinations that older students learn

Not quite. What you are learning here is a special, simpler case: counting outcomes when the same number of choices repeats at every step with repetition allowed, expressed neatly using exponential notation. Full permutations and combinations ( $nPr$ , $nCr$ ) handle the more general situation, including cases where repetition is not allowed or choices change from step to step. That formal topic is introduced in later grades; for now, the exponent shortcut $a^n$ is the tool you need.

Counting combinations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads