Counting problems reduce to two rules. The Product Rule: if a task has independent stages with $m$, $n$, ... options, the total is $m \times n \times \dots$ — use it whenever the problem says "AND" (one choice and another). The Sum Rule: if choices are mutually exclusive alternatives, add them — use it for "OR".

Example: A meal deal has 7 starters and 12 mains: a two-course meal is $7 \times 12 = 84$ (AND). Choosing a pen or a pencil from 4 pens and 5 pencils is $4+5=9$ (OR).

Ask: if I select the same items in a different order, is that a genuinely different outcome? If yes, order matters — it's a permutation (arranging, codes, PINs, "1st/2nd/3rd"). If no — the outcome is the same set either way — it's a combination (choosing a committee, a team, a handshake).

Without repetition (an item can't be reused), the number of choices shrinks by one each step: arranging 4 distinct letters is $4\times3\times2\times1 = 4! = 24$. With repetition allowed (e.g. a 4-digit PIN from 0–9), the choices stay constant: $10\times10\times10\times10=10^4$.

For identical items (not all objects distinct), find the total arrangements as if distinct, then divide by the factorial of each repeated group: arranging "STATISTICS" (10 letters: S×3, T×3, I×2) gives $\dfrac{10!}{3!\,3!\,2!} = 50\,400$.

Calculate the permutation first (as if order mattered), then divide by the number of ways the chosen group could be reordered, since those are all the same combination.

Choosing 2 doubles players from 5: ordered pairs $= 5\times4=20$; each pair is counted twice (A-then-B, B-then-A), so unique pairs $=\dfrac{20}{2}=10$. The same logic gives "$n$ choose 2" $=\dfrac{n(n-1)}{2}$ — handy for handshakes or round-robin matches.

"Must be together" — glue the required items into one block, arrange the block with everything else, then multiply by the arrangements inside the block. 3 history books kept together among 4 science books: arrange 5 "items" ($5!=120$), then the 3 books inside their block ($3!=6$): total $120\times6=720$.

"Must NOT be together" — find the complement: (total arrangements) $-$ (arrangements with them together).

"Must satisfy a restriction" (e.g. even, or $>5000$) — fill the most restricted slot first (often the last digit, for "even"), then fill the remaining slots left to right.

Even 6-digit number, no zeros, Ed is thinking of: last digit must be from $\{2,4,6,8\}$ (4 choices); the other 5 digits are unrestricted from $\{1,...,9\}$ (9 choices each): $9^5 \times 4 = 236\,196$.

$P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ — and counting principles are exactly how you find both the numerator and denominator for compound events, especially when listing every outcome by hand isn't practical.

Choosing 2 students at random from 14 boys and 16 girls — probability both are girls: total ways to choose 2 from 30 is $\frac{30\times29}{2}=435$; ways to choose 2 girls from 16 is $\frac{16\times15}{2}=120$; so $P = \frac{120}{435} = \frac{8}{29}$.

In how many ways can 4 boys and 4 girls be arranged in a line if no two boys stand next to each other?

1. Arrange the 4 girls first: $4! = 24$ ways — this creates 5 gaps (including the two ends) around them: GGGG_.
2. Choose 4 of those 5 gaps for the boys and arrange the boys within them (order matters, no repeats): $5\times4\times3\times2 = 120$.
3. Multiply: $24 \times 120 = 2880$ total arrangements.

This "arrange the unrestricted group first to create gaps" trick is the standard method whenever a question demands that certain items stay apart.