Use the addition and multiplication principles, permutations and combinations to count arrangements and selections, including cases with restrictions

What this dot point is asking

SEAB wants you to count arrangements and selections using the addition and multiplication principles, distinguish permutations (order matters) from combinations (order does not), and handle restrictions, blocks, and repeated objects.

The answer

The two counting principles

• Multiplication principle: if a task has stages with $n_1, n_2, \ldots$ independent choices, the total number of ways is the product $n_1 \times n_2 \times \cdots$.
• Addition principle: if outcomes split into mutually exclusive cases, add the counts of each case.

Permutations: order matters

The number of ways to arrange $r$ objects chosen from $n$ distinct objects, where order matters, is

Arranging all $n$ distinct objects gives $n!$. Arrangements with identical objects divide by the factorials of the repeats.

Combinations: order does not matter

The number of ways to select $r$ objects from $n$, where order is irrelevant, is

A combination is a permutation with the $r!$ orderings collapsed.

Restrictions and techniques

• Block method: items that must stay together are glued into one block, arranged, then the block's internal order counted.
• Gap method: items that must be separated are placed in the gaps between others.
• Complement: count the total and subtract the unwanted cases when "at least" conditions appear.

Counting circular arrangements

Arranging objects in a circle differs from a row, because rotating the whole circle does not create a new arrangement. Fixing one object's position removes this rotational duplication, so $n$ distinct objects arranged in a circle give $(n - 1)!$ arrangements rather than $n!$. For example, $5$ people around a round table can be seated in $(5 - 1)! = 4! = 24$ ways. If reflections (clockwise versus anticlockwise being the same, as for a bracelet) are also considered identical, divide by a further $2$. Recognising when a problem is circular, and fixing a reference position to kill the rotational symmetry, is a standard H2 refinement of the basic permutation count.

Selecting then arranging in one problem

Many counting problems combine a combination and a permutation: first choose which objects, then arrange them. Because the two stages are independent, multiply the counts. To choose $3$ of $8$ books and then arrange them on a shelf, compute $\binom{8}{3} \times 3! = 56 \times 6 = 336$, which equals $^8P_3$ as a check. Spotting that "choose then order" is a combination multiplied by a permutation, and that it reproduces the direct permutation, both structures the working and provides a built-in verification of the answer.

Examples in context

Example 1. Lottery odds. Choosing $6$ numbers from $49$ where order does not matter is $\binom{49}{6} = 13{,}983{,}816$ combinations, which is why the chance of a single ticket winning is so small.

Example 2. Seating with a rule. Seating a family where two siblings must not sit together uses the complement: total arrangements minus the block arrangements where they are together, the standard "separated" counting trick.

Try this

Q1. How many ways can $6$ different people stand in a queue? [1 mark]

• Cue. $6! = 720$.

Q2. How many ways can a team of $3$ be chosen from $10$ players? [2 marks]

• Cue. $\binom{10}{3} = 120$.

Q3. State whether choosing a president and a secretary from a club is a permutation or combination. [1 mark]

• Cue. A permutation, because the two roles are distinct so order matters.