Most counting problems, in exams and outside them, are solved by a small number of moves used again and again. The formulas of the previous chapters are only half the work; the other half is seeing which move a problem is asking for. This chapter collects the nine moves that crack most problems. Each one is described in the same five parts: what it is, why it works, the sign in the problem that calls for it, an example, and the trap that usually catches people. At the end there is a routine for attacking any problem, a list of classic traps, and two practice rounds.
By the end you can
- Recognise from the wording which counting move a problem needs.
- Use the complement, the block, the gaps and "choose the places first" with confidence.
- Spot an overcount and divide it out, or split into cases when a restriction affects another.
- Follow a fixed routine to set up any counting problem.
- Avoid the classic traps: double counting, forgotten inner orders, identical against different items.
What it is. Before any formula, swap two of the chosen items in one result and ask whether the result changed.
Why it works. Arrangements count ordered results and combinations count groups; the only difference between them is whether a swap makes a new result. The test asks exactly that question.
When to use it. Always, as the first step. Words like president, first, position, code, podium point to order; words like committee, group, team, set, ticket, hand point to no order.
Example. Choosing 2 class representatives with equal roles from 30 students: swapping the two gives the same pair, so C(30, 2) = 435. Choosing a representative and a deputy: the swap gives a new result, so A(30, 2) = 870.
The trap. The word "choose" does not mean combination. "Choose a president and a treasurer" is an arrangement.
What it is. In the slot method, fill the positions with conditions before the free ones.
Why it works. The multiplication principle needs each stage to have a fixed number of options. A free position has the same number of options whatever happens; a restricted one may not, if the free ones have already used up what it needs.
When to use it. "Cannot start with 0", "must end in a vowel", "must be odd", "the first place goes to…".
Example. Anagrams of LIVRO that start and end with a vowel: the ends first, I and O in 2 orders; then the 3 consonants in the middle, 3! = 6. Total 12.
The trap. Two restrictions that compete for the same items (the last digit even and the first digit not 0) cannot both go first. Then use Trick 3.
What it is. Divide the results into separate cases, count each one, and add.
Why it works. The addition principle: if every result falls in exactly one case, the total is the sum of the cases.
When to use it. When one choice changes the number of options of another (is the last digit 0 or not?), or when the question itself lists cases ("4 or 5 digits", "exactly 2 or exactly 3").
Example. Even 3-digit numbers with different digits: ending in 0, 9 · 8 = 72; ending in 2, 4, 6 or 8, 8 · 8 · 4 = 256. Total 328.
The trap. The cases must not overlap and must cover everything. Check with a small example that no result belongs to two cases.
What it is. Count everything, count what you do not want, and subtract.
Why it works. Every result either has the property or does not, so wanted = total − unwanted. It pays off when the unwanted side is one simple case.
When to use it. "At least one", "not all", "not side by side", "some repeated digit".
Example. 4-digit PINs with at least one repeated digit: all PINs, 10⁴ = 10 000, minus those with all digits different, 10 · 9 · 8 · 7 = 5040. Total 4960.
The trap. The opposite of "at least one" is "none", not "exactly one".
What it is. Items that must stay together are glued into one block, counted as a single item; then the block's inner orders are multiplied in.
Why it works. Every order with the items together is exactly one order of the block among the others, times one order inside the block.
When to use it. "Together", "side by side", "next to each other", "in a row without separating".
Example. Anagrams of PERNAMBUCO with the vowels E, A, U, O together, in any order: the block plus P, R, N, M, B, C is 7 items, 7! = 5040, times 4! = 24 inside the block: 120 960.
The trap. Forgetting the inner orders, or multiplying them when the order inside is fixed ("together and in this order").
What it is. To keep some items apart, first place the others; then put the separated items into the gaps between them, at most one per gap.
Why it works. m items in a row leave m + 1 gaps (the two ends included). Items put into different gaps can never touch.
When to use it. "No two side by side", "never together", for more than two items, where the complement would need inclusion–exclusion.
The trap. Forgetting the two ends as gaps, or using the gaps for items that only need to be apart from one particular item.
What it is. Count as if everything were different and ordered, then divide by how many times each real result was counted.
Why it works. If every real result was counted the same number of times m, the true count is the rough count divided by m. That is exactly where k! in combinations, a!b! in repeated letters and n in circular tables come from.
When to use it. Identical items, rotations of a circle, groups without names, or a symmetric condition such as "A before B" (exactly half of all orders).
Example. In how many orders of 6 runners does Ana finish ahead of Bia? Swapping the two pairs every order with one where Bia is ahead: 6!/2 = 360.
The trap. Dividing only works when every result was counted the same number of times. In "at least one man" by choosing "the" man first, a committee with one man is counted once and one with two men twice, so no single division fixes it.
What it is. Decide which positions a kind of item will occupy (a combination of positions), then fill each kind of position.
Why it works. A result is fixed by where each kind goes and by the order within each kind; the two choices are stages.
When to use it. Strings with a fixed number of some symbol, items that keep a fixed order among themselves, shortest paths on a grid.
Example. Anagrams of CAPITULO in which the vowels A, I, U, O appear in this order (not necessarily together): choose the 4 places of the vowels, C(8, 4) = 70; they go in the fixed order, 1 way; the 4 consonants fill the rest, 4! = 24. Total 1680, which is 8!/4!, as Trick 7 also says.
The trap. Choosing the places is a combination (the places are a set); filling them is a permutation of the items, unless the items are identical or in a fixed order.
What it is. Any problem where identical items are shared among different boxes is k stars and n − 1 bars: C(n + k − 1, k).
Why it works. Identical items have no identity, so a sharing is fixed by how many go into each box, which is a row of stars and bars.
When to use it. Identical sweets, coins or votes; repeats allowed and order ignored; whole-number solutions of x + y + z = s.
Example. 10 identical votes among 3 candidates, each candidate may get none: C(12, 2) = 66 possible tallies.
The trap. If the items are different (10 different people voting), each item chooses its box: 3¹⁰, not stars and bars.
| Step | Ask |
|---|---|
| 1 | What exactly is one result? Write one down, as a concrete example. |
| 2 | Does order matter (swap test)? May items repeat? Are the items different or identical? |
| 3 | Are there restrictions? Fill them first, or split into cases. |
| 4 | Is the opposite simpler? Then count the complement. |
| 5 | Did I count some result more than once? Divide, or rethink the method. |
| 6 | Check with a smaller version you can list by hand. |
| Wrong | Right | Why |
|---|---|---|
| "at least one man": choose the man, then the rest | all minus no man | committees with several men are counted several times |
| two together: 5! for 6 people | 5! · 2! | the inner order of the block was forgotten |
| 10 different people vote for 3 candidates: C(12, 2) | 3¹⁰ | different voters are not identical stars |
| 3 equal prizes to 10 people: A(10, 3) | C(10, 3) | equal prizes have no order |
| 6 people into 3 unnamed pairs: 90 | 90/3! = 15 | groups without names must not be ordered |
| "and then" counted with + | · | stages multiply, separate cases add |
A round of 10 questions
Answer as many as you can. Mistakes come back at the end for a review round.
Knowing a trick is half of it; the other half is seeing when it applies. Each question is a problem and what it asks for. Pick the trick that opens it. A wrong answer shows the right trick and one sentence on why it works there. The questions come in a new order every round.
A round of 10 questions
Answer as many as you can. Mistakes come back at the end for a review round.
Every exercise is made on the spot, so a new round is never the same. The computer picks the answer first and builds the question around it, working with exact fractions, so the answer key is always right. Answers are read exactly: 0.75, 3/4 and 6/8 count as the same number (where lowest terms are asked for, 6/8 just gets a reminder). A wrong answer shows the worked solution line by line, in the order the lesson does it; the new number in each line is coloured.
Key ideas
- Swap test first: a new result after swapping means order matters.
- Restricted positions first; when restrictions compete, split into cases and add.
- "At least one" and "not together" are usually total minus the opposite.
- Together: a block, times its inner orders. Apart: place the others, then use the m + 1 gaps.
- Count as if all were different and ordered, then divide by how often each result was counted.
- Choose the places of a kind of item first; identical items into boxes are stars and bars.
- Write one concrete result before counting, and check the method on a small case.