Use the multiplication principle to count ordered selections across stages, count ordered selections with repetition as n^r and without repetition as nPr = n!/(n-r)!, and apply restriction techniques: deal with the difficulty first, fix a position, keep items together as a block, and count 'at least one' by the complement

What this dot point is asking

This is the dot point where counting really begins. The factorial notation page gave you the arithmetic engine, $n!$ and the unrolling rule; here you learn the principle that turns a real counting question into a product of factors. NESA wants three connected skills. First, the multiplication principle: when a selection is made in independent stages, you multiply the number of choices at each stage. Second, you must tell apart the two kinds of ordered selection, with repetition, which gives a power $n^r$, and without repetition, which gives a permutation $^{n}P_{r} = \dfrac{n!}{(n-r)!}$. Third, and where most of the marks and most of the mistakes live, you must handle restrictions: deal with the hardest restriction first, fix a forced position, keep items together by treating them as a block, and count an "at least one" condition through its complement.

The single tool that makes all of this visible is the filled-box slot diagram: draw one box for each stage, write the number of choices inside it, and multiply across. Almost every counting question on this dot point can be set out as boxes, and a marker can follow your reasoning at a glance.

The answer

The multiplication principle: multiply the choices across stages

Suppose a task is carried out in a sequence of stages, and the choice made at each stage does not change how many choices are available at the others. Then the number of ways to complete the whole task is the product of the numbers of choices at the separate stages. This is the multiplication principle, and it is the backbone of the entire topic.

Here is the intuition, not just the rule. Imagine choosing an outfit: a shirt from $5$, then trousers from $4$, then shoes from $3$, then a cap from $2$. For each of the $5$ shirts there are $4$ choices of trousers, giving $5 \times 4 = 20$ shirt-and-trousers combinations. Each of those $20$ can be paired with any of the $3$ pairs of shoes, giving $20 \times 3 = 60$, and each of those with any of the $2$ caps, giving $60 \times 2 = 120$. Adding would be wrong, because the stages are not alternatives; they happen together, one after the other, and "and" between independent stages means multiply.

Ordered selections WITH repetition: the power n^r

Now narrow to the most common kind of staged choice: you build an ordered arrangement of $r$ positions, and at every position you may pick any of the same $n$ objects, repeats allowed. Because choosing an object never removes it from the pool, every one of the $r$ boxes keeps all $n$ choices, and the multiplication principle gives

The clearest example is digits. How many three-digit numbers can be made from the digits $1, 2, 3, 4, 5$ if a digit may be used more than once? Each of the three positions is free to be any of the $5$ digits, so the count is $5 \times 5 \times 5 = 5^3 = 125$.

The same shape covers a four-digit PIN ($10^4 = 10\,000$ possibilities, since each of the four positions is any of the ten digits), and a string of $8$ binary digits ($2^8 = 256$). Two warnings. The base $n$ is the number of objects you choose from; the exponent $r$ is the number of positions you fill. It is easy to write $r^n$ by reflex; always ask "how many choices per box ($n$) and how many boxes ($r$)?" and the answer is choices-to-the-power-of-boxes, $n^r$.

Ordered selections WITHOUT repetition: the permutation nPr

Change one word: now repeats are not allowed. As soon as an object is used it leaves the pool, so each box has one fewer choice than the box before it. Filling $r$ positions from $n$ distinct objects gives

That descending product of exactly $r$ factors, starting at $n$, is the permutation $^{n}P_{r}$. Using the unrolling rule from the factorial page, multiply and divide by $(n-r)!$ to write it compactly:

A permutation is an ordered selection without repetition, which is why "arrangement", "in order", "first, second, third" and "no repeats" are its signal words. For instance, the number of three-letter words with no repeated letter taken from the $8$ letters $A, B, C, D, E, F, G, H$ is

Two special cases tie this back to factorials. Arranging all $n$ objects means $r = n$, and

which is exactly the "arrange $n$ distinct objects in a row" count, and the reason the convention $0! = 1$ is forced. At the other end, $^{n}P_{1} = n$ and $^{n}P_{0} = 1$ (one way to choose nothing). On a slot diagram, $n^r$ has the same number in every box; $^{n}P_{r}$ has a number that steps down by one each box. That visual difference is the fastest way to keep the two apart.

Restriction technique 1: deal with the difficulty first

Real questions add conditions: a digit cannot be zero, a particular person must sit at one end, two people must be together. The master rule is to deal with the most restricted box first, while it still has its restriction, and fill the free boxes afterwards. If you fill the easy boxes first you can use up an object that the restricted box needed, and miscount.

The classic case is the leading digit. How many three-digit numbers (no repeated digits) can be made from $0, 1, 2, \dots, 9$? The hundreds digit cannot be $0$, or the "number" is really two digits, so deal with it first: $9$ choices ($1$ to $9$). The tens digit may now be $0$ but not the hundreds digit already used, so $9$ choices, and the units digit has $8$ left. The count is $9 \times 9 \times 8$, with the restricted box handled before the others.

Restriction technique 2: fix a forced position

When an object is forced into a specific position, place it first in the $1$ way it is allowed, then count the remaining free positions normally. Fixing a position simply removes one box (it contributes a factor of $1$) and shrinks the pool for the rest.

For example, $6$ people queue and Maya must stand at the front. Put Maya in front ($1$ way), then arrange the other $5$ people in the remaining $5$ places: $1 \times 5! = 120$. As a sanity check, Maya is equally likely to be in any of the $6$ places, so exactly $\tfrac{1}{6}$ of the $720$ unrestricted queues have her at the front, and $\tfrac{1}{6} \times 720 = 120$.

Restriction technique 3: keep items together as a block

When certain objects must be together, glue them into a single block and arrange the block among the other items. Then, separately, arrange the objects inside the block, because they can still be ordered among themselves. The two counts multiply.

Take $6$ people in a row where two friends, $A$ and $B$, insist on standing next to each other. Treat $A$ and $B$ as one block, leaving $5$ items (the block plus the other $4$ people) to arrange in $5!$ ways. Inside the block, $A$ and $B$ can be in $2!$ orders ($AB$ or $BA$). By the multiplication principle the answer is $5! \times 2! = 120 \times 2 = 240$.

If a block of $k$ items must stay together, the internal factor is $k!$. The same idea, run the other way, also handles "must be separated": that is usually fastest by the complement, next.

Restriction technique 4: "at least one" via the complement

"At least one" almost never wants a direct count, because it splits into awkward cases (exactly one, exactly two, ...). Instead count the complement, the arrangements that have none of the wanted feature, and subtract from the total:

How many four-digit numbers from the digits $1$ to $9$ (repetition allowed) contain at least one $7$? The total is $9^4 = 6561$. The numbers with no $7$ use only the other $8$ digits, so there are $8^4 = 4096$ of them. Therefore the numbers with at least one $7$ number

The complement also cracks "must be separated": count all arrangements, subtract the ones where the forbidden items are together (found by the block method). The trigger words are at least one, at most, not all, and must be apart.

How exam questions ask about this dot point

The phrasing tells you which tool to reach for. Match the wording to the move:

• "How many outfits / meals / outcomes...", several independent choices. Multiplication principle: one box per stage, multiply across.
• "...digits/letters may be repeated", "with replacement", "a PIN", "a binary string". Ordered with repetition: each box keeps all $n$ choices, count $n^r$. Read off $n$ = choices per box, $r$ = number of boxes.
• "...no digit repeated", "arrange $r$ of the $n$ in order", "first, second, third", "without replacement". Permutation $^{n}P_{r} = \dfrac{n!}{(n-r)!}$; each box drops by one.
• "In how many ways can $n$ people/books be arranged in a row?" All-objects permutation $^{n}P_{n} = n!$.
• "...the first digit cannot be $0$", "must not start with...". Deal with the difficulty first: fill the restricted box before the free ones.
• "...$X$ must sit at the end / be first". Fix that position ($1$ way), then arrange the rest.
• "...must sit together / be next to each other / as a group". Block method: arrange the block among the others ($\times$), then arrange inside the block ($k!$).
• "...at least one...", "at most...", "not all...", "must be separated". Count the complement and subtract from the total.