Permutations, Combinations and Further Probability
AdvancedDistinguish order-sensitive permutations from order-free combinations and count with their formulas; in probability, move from the addition and multiplication laws to conditional probability, with Venn diagrams and tree diagrams unlocking the further probability problems.
- Categories add, stages multiply — the foundation of every counting problem
- Permutations nPr = n!/(n − r)! respect order; combinations nCr = n!/[r!(n − r)!] do not; nPr = nCr × r!
- Bundle for 'must be adjacent'; subtract from the total or insert gaps for 'must not be adjacent'
文體小知識：數學概念說明文——由計數的加法與乘法法則出發，建立排列與組合的概念及記法，再以集合記法、溫氏圖、樹形圖與條件概率處理進階概率問題。
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1Counting rests on two basic rules: if a task can be done in several non-overlapping categories, add the counts; if it consists of successive stages, multiply them. For example, travelling from Hong Kong to Guangzhou by through train or direct bus, with 3 and 5 departures respectively, gives 8 choices; but a journey via Shenzhen first, with 4 departures to Shenzhen and 5 onward to Guangzhou, gives 20 routes.
2Choosing r from n distinct objects and arranging them in a row — order matters — gives the permutations; choosing without regard to order gives the combinations. Both are computed with factorials, and the number of permutations equals the number of combinations times r factorial, because each combination can be internally ordered in r factorial ways. The formulas come in the steps; here the priority is the habit of asking first: does order matter?
3In probability, the addition law says the probability of A or B equals P(A) plus P(B) minus the probability of A and B together; if the two events are mutually exclusive — impossible together — that last term is zero and the probabilities simply add. The multiplication law says independent events multiply. For example, rolling a 2 or a 3 on a fair die has probability 16 + 16 = 13; rolling a six twice in two throws has probability 16 × 16 = 136.
4Conditional probability handles the probability of one event given that another has occurred. Tree diagrams are the perfect tool: each level of branches is one stage, each branch labelled with the probability of that step; multiplying along a path gives the probability of that outcome. For example, drawing two balls without replacement from 4 red and 3 white: P(first red) = 47; given the first is red, 3 red and 3 white remain, so P(second red) = 12.
5Counting and probability often appear together: count favourable outcomes and total outcomes with combination numbers, then divide. First judge whether order matters — ranked prizes need permutations, committee members need combinations; then judge whether sampling is with or without replacement — without replacement the probabilities change at each step, so use conditional probability or a tree diagram. Get these two judgements right and the question is mostly solved; what remains is careful arithmetic.
1Summary 重點摘要
Distinguish order-sensitive permutations from order-free combinations and count with their formulas; in probability, move from the addition and multiplication laws to conditional probability, with Venn diagrams and tree diagrams unlocking the further probability problems.
2Vocabulary cards Vocabulary
The number of ways to choose and order r objects from n distinct ones, written nPr.
The number of ways to choose r objects from n distinct ones without regard to order, written nCr.
The notation n! = n × (n − 1) × … × 1 with 0! = 1; factorials underpin the permutation and combination formulas.
If a task can be done in several non-overlapping categories, the total number of ways is the sum of the category counts.
If a task consists of successive stages, the total number of ways is the product of the stage counts.
Events that cannot occur together; the probability of their union is the sum of their probabilities.
Events whose occurrence does not affect each other's probability; their joint probability is the product.
The probability of A given that B has occurred, written P(A|B) = P(A∩B)/P(B).
A branching diagram of a multi-stage experiment with probabilities on the branches; multiply along paths and add matching outcomes.
A diagram of overlapping circles showing how events intersect and union, used with set notation A∩B, A∪B, A′ in probability.
3Step by step Steps
- Step 1: The addition and multiplication rulesFirst decide whether the task is 'by categories' or 'by stages'. Categories use addition: the categories are non-overlapping and any one completes the task, so totals add. Stages use multiplication: every stage requires a choice and none can be skipped, so totals multiply. For example, forming two-digit numbers without repetition from the digits 1, 2, 3: the tens place has 3 choices and the units place 2, giving 3 × 2 = 6 — stages; but asking 'how many of them contain the digit 1' requires discussing cases by position and adding — categories. Sketching the stages or categories before writing anything is the first move in every counting problem.
- Step 2: The concept and notation of permutationsChoosing r from n distinct objects and arranging them in order gives nPr = n!/(n − r)!. Intuition: the first position has n choices, the second n − 1, and so on through r positions. For example, 8 runners competing for gold, silver and bronze — order matters — give 8P3 = 8 × 7 × 6 = 336. The factorial n! = n × (n − 1) × … × 1, with 0! = 1 by convention. Watch for 'adjacent' questions: bundle the objects that must be adjacent into one unit, arrange it with the rest, then multiply by the internal arrangements — the standard permutation trick.
- Step 3: The concept and notation of combinationsChoosing without ordering uses combinations: nCr = n!/[r!(n − r)!]. Rule of thumb: 'select, form a team, committee, lottery without ranking' suggest combinations; 'queue, ranking, password, route order' suggest permutations. For example, choosing 3 officials from 10 students with no titles assigned: 10C3 = 120; but if the 3 take the posts of chair, secretary and treasurer, order matters and we multiply by 3! = 6, giving 720. The relation nPr = nCr × r! shows a permutation count is never smaller than its combination count, and the two formulas cross-check each other — when an answer looks wrong, first suspect a mix-up of order.
- Step 4: Classic techniques for counting problemsThree techniques recur: bundling for 'must be adjacent'; inserting gaps for 'must not be adjacent'; and grouped allocation for quota questions. For example, 6 people in a row with A and B adjacent: bundle A and B as one unit, arrange it with the other 4 — 5! = 120 — then multiply by the 2! = 2 internal orders, giving 240. If instead A and B must not be adjacent, subtract from the total: 6! − 240 = 720 − 240 = 480. Verification: recount by another method — check the 480 with the gap-insertion method — and only commit once the answers agree.
- Step 5: Set notation, Venn diagrams and the addition lawSet notation: A and B together is A∩B, at least one of them is A∪B, not-A is A′ with P(A′) = 1 − P(A). Addition law: P(A∪B) = P(A) + P(B) − P(A∩B); the subtracted term is exactly the overlap counted twice. For mutually exclusive events A∩B is empty and probabilities add directly. For example, in a class of 40, if 25 join the maths contest, 15 the physics contest and 8 join both, then 25 + 15 − 8 = 32 join at least one; a Venn diagram lays out the three regions at a glance, and such word problems appear regularly in extended papers.
- Step 6: Independence and conditional probabilityIf P(A∩B) = P(A) × P(B), the events A and B are independent: whether one occurs does not change the probability of the other. Conditional probability P(A|B) = P(A∩B)/P(B) recomputes the chance of A given that B has occurred. For example, throwing a fair die twice: the probability of a 6 on the second throw does not depend on the first, so the throws are independent; but drawing two cards from one deck without replacement, the chance that the second is an ace changes according to the first — not independent, and conditional probability is required. Independence is never judged by feel; verify with the equation. With replacement tends to mean independent; without replacement generally not.
- Step 7: Tree diagrams and further probability problemsA tree diagram draws a multi-stage experiment as branches: the first level carries the probabilities of the first stage, the second level the conditional probabilities given the first result; multiply along a path for that outcome, then add the probabilities of the paths that match the question. For example, 4 red and 3 white balls, two drawn without replacement — the probability both have the same colour: path red-red is 47 × 36 = 27, path white-white is 37 × 26 = 17, total 37. Workflow: decide with or without replacement, draw and label the tree, then 'multiply along paths, add matching outcomes'; for a reverse condition such as 'given the second ball is red, find the probability the first was red', list the path probabilities and take the appropriate ratio.
5Worked example Example
Example: (a) six students line up for a photo with A and B adjacent — find the number of arrangements; (b) from 4 boys and 3 girls, choose a committee of 3 — find the number of all-girl committees; (c) a bag holds 4 red and 3 white balls, two drawn one after another without replacement — given the first ball is red, find the probability the second is also red.
- (a) Bundle A and B as one unit, arranged with the other 4 people: 5! = 120 ways
- A and B can swap internally: 2! = 2 ways, so the total is 120 × 2 = 240
- (b) Choosing people ignores order, so use combinations: an all-girl committee means choosing all 3 from the 3 girls, 3C3 = 1 way
- (c) 'Given the first is red' is conditional probability: 3 red and 3 white balls remain, 6 in all
- So P(second red | first red) = 36 = 12; and the probability both are red is 47 × 12 = 27
- Answer: (a) 240 ways. (b) 1 way (choose all three girls). (c) P = 12; the probability both balls are red is 27. The keys: 'must be adjacent' calls for bundling; people chosen without titles call for combinations; 'given that …' is conditional probability — shrink both numerator and denominator to the given condition, and verify by multiplying along the tree diagram.
6Common mistakes Mistakes
7Quick check Quick check
Q1What distinguishes a permutation from a combination, and how are they related?
Q2In how many ways can 4 of 7 students be chosen for a debate competition?
Q3In how many ways can 5 people line up for a photo?
Q4Find the probability of rolling an even number or a 5 on a fair die.
Q5A bag holds 5 red and 2 white balls; two are drawn without replacement. Find the probability both are white.
8Recap Recap
- Categories add, stages multiply — the foundation of every counting problem
- Permutations nPr = n!/(n − r)! respect order; combinations nCr = n!/[r!(n − r)!] do not; nPr = nCr × r!
- Bundle for 'must be adjacent'; subtract from the total or insert gaps for 'must not be adjacent'
- Addition law P(A∪B) = P(A) + P(B) − P(A∩B); for mutually exclusive events the overlap is zero
- Conditional probability P(A|B) = P(A∩B)/P(B); on tree diagrams multiply along paths and add the paths that match