📚 Probabilities using Permutations and Combinations | 利用排列组合计算概率
Probability questions often ask us to count the number of ways certain events can occur, then divide by the total number of possible outcomes. When the outcomes are equally likely, permutations and combinations give us a precise language for these counts. Mastering this topic opens up a wide range of IB‑style problems, from card hands to committee selections and arrangement puzzles.
概率问题常常要求我们计算某一事件发生的不同方式数，再除以可能结果的总数。当所有结果等可能时，排列与组合为我们提供了精确计数的工具。掌握这一主题能够处理从扑克牌到委员会选举和排列谜题等各类IB风格的题目。
1. Fundamental Counting Principle | 基本计数原理
If a first operation can be performed in m independent ways and a second operation in n independent ways, then the total number of ways to perform both operations in sequence is m × n. This rule extends to any finite number of operations and is the backbone of all counting arguments.
如果第一个操作有m种独立方式，第二个操作有n种独立方式，那么依次完成两个操作的总方式数为 m × n。这一法则可以推广到有限个操作，是所有计数推理的基础。
2. Permutations: Definition and Formula | 排列：定义与公式
A permutation is an ordered arrangement of objects. The number of permutations of n distinct items taken r at a time, denoted P(n, r) or ⁿPᵣ, is given by n! / (n – r)!. When r = n, this becomes n!; when r = 0, P(n, 0) = 1 by convention.
排列是指对象的有序安排。从n个不同物品中取出r个的排列数记作 P(n, r) 或 ⁿPᵣ，其计算公式为 n! / (n – r)!。当 r = n 时即为全排列 n!；规定 P(n, 0) = 1。
3. Factorial Notation | 阶乘记号
The factorial n! is defined as n × (n – 1) × (n – 2) × … × 3 × 2 × 1, with 0! = 1. Factorials grow very fast; already 10! we often keep factorials in fraction form until the last moment to simplify cancellation.
阶乘 n! 定义为 n × (n – 1) × (n – 2) × … × 3 × 2 × 1，且规定 0! = 1。阶乘增长极快，如 10! = 3 628 800。在概率计算中，我们常将阶乘保留为分数形式，直到最后一步再进行化简约分。
4. Combinations: Definition and Formula | 组合：定义与公式
A combination is a selection of objects where the order does not matter. The number of combinations of n distinct items taken r at a time, written C(n, r), ⁿCᵣ or (ⁿᵣ), is n! / [r!(n – r)!]. This is exactly the binomial coefficient and is symmetric: C(n, r) = C(n, n – r).
组合是指不考虑顺序的对象选择。从n个不同物品中取出r个的组合数记作 C(n, r)、ⁿCᵣ 或 (ⁿᵣ)，计算公式为 n! / [r!(n – r)!]。它正是二项式系数，并具有对称性：C(n, r) = C(n, n – r)。
5. Probability as a Ratio of Counts | 概率作为计数之比
When a sample space consists of equally likely outcomes, the probability of an event A is P(A) = n(A) / n(S), where n(A) is the number of outcomes favourable to A and n(S) is the total number of outcomes in the sample space. Both numbers are often found using permutations or combinations.
当样本空间由等可能结果构成时，事件A的概率为 P(A) = n(A) / n(S)，其中 n(A) 是A包含的有利结果数，n(S) 是样本空间的结果总数。这两个数目通常通过排列或组合求得。
6. Probability Problems Using Permutations | 使用排列计算概率问题
Consider a race with 8 runners. If all finishing orders are equally likely, what is the probability that a specific runner finishes in the top 3? The total number of ordered top‑3 finishes is P (3 choices), then choose and order the remaining 2 finishers from the other 7 runners: 3 × P(7, 2) = 3 × 7 × 6 = 126. Hence P(top 3) = 126 / 336 = 3/8.
考虑一场有 8 名选手的比赛。假设所有完赛顺序等可能，求某指定选手进入前三名的概率。前三名有序排列的总数为 P(8, 3) = 8 × 7 × 6 = 336。有利结果：将该选手置于 3 个奖牌位置之一（3 种选择），然后从剩余 7 人中选出并排列其余两名选手：3 × P(7, 2) = 3 × 7 × 6 = 126。因此进入前三的概率为 126 / 336 = 3/8。
7. Probability Problems Using Combinations | 使用组合计算概率问题
A committee of 4 is to be chosen from 10 men and 8 women. Find the probability that exactly 2 women are selected. The sample space consists of all unordered groups of 4 from 18 people: n(S) = C(18, 4) = 18! / (4! × 14!) = 3060. Favourable selections: choose 2 women from 8, C(8, 2) = 28, and 2 men from 10, C(10, 2) = 45. By the fundamental counting principle, n(A) = 28 × 45 = 1260. The required probability is 1260 / 3060 = 21/51 ≈ 0.412.
从 10 名男士和 8 名女士中随机选择 4 人组成委员会。求恰好选出 2 名女士的概率。样本空间为从 18 人中无序选取 4 人的所有组：n(S) = C(18, 4) = 18! / (4! × 14!) = 3060。有利选择：从 8 名女士中选 2 人，C(8, 2) = 28，从 10 名男士中选 2 人，C(10, 2) = 45。由基本计数原理，n(A) = 28 × 45 = 1260。所求概率为 1260 / 3060 = 21/51 ≈ 0.412。
8. Distinguishing Permutations and Combinations | 区分排列与组合
The key test is: “If I swap two selected items, do I get a different outcome?” If yes, the order matters and we use permutations. If swapping gives the same group, order does not matter and we use combinations. A password is a permutation; a lottery ticket is a combination. Always identify whether the arrangement is ordered before applying a formula.
关键的判断是：“如果我交换两个选中的元素，是否会得到不同的结果？”如果是，则顺序重要，应使用排列；如果交换后仍是同一组，则顺序不重要，应使用组合。密码属于排列问题；彩票号码属于组合问题。在使用公式前务必先判断排列是否有序。
9. Permutations with Repeated Items | 含重复项的排列
When a set contains identical items, the number of distinct permutations of n objects, where there are n₁ identical items of type 1, n₂ of type 2, …, is n! / (n₁! × n₂! × …). For instance, the 11 letters in "MISSISSIPPI" contain 1 M, 4 I’s, 4 S’s and 2 P’s. The number of distinct arrangements is 11! / (4! × 4! × 2!) = 34650.
当集合中包含相同物品时，n个物体的不同排列数为 n! / (n₁! × n₂! × …)，其中 n₁ 为第一类相同物品的个数，n₂ 为第二类相同物品的个数，依此类推。例如，“MISSISSIPPI”的 11 个字母包含 1 个 M、4 个 I、4 个 S 和 2 个 P。其互异排列数为 11! / (4! × 4! × 2!) = 34650。
10. Probability Involving Repeated Items | 涉及重复物品的概率
utation of the letters A, A, B, C has the two A’s together. Total distinct arrangements: 4! / 2! = 12. Treat the two A’s as a single block ‘AA’. Then we arrange the block AA, B, C: 3! dentical items are present, always use the formula for repeated permutations to avoid over‑counting.
求字母 A、A、B、C 的随机排列中两个 A 相邻的概率。不同排列总数为 4! / 2! = 12。将两个 A 视为一个整体块“AA”，则需排列 AA、B、C 这三样：共 3! = 6 种方式。因此所求概率为 6 / 12 = 1/2。当存在相同物品时，务必使用重复排列公式以避免重复计数。
11. Conditional Probability with Counting | 利用计数的条件概率
Often a condition reduces the sample space. Suppose a hand of 2 cards is dealt from a standard deck. What is the probability that both are aces, given that at least one is an ace? Using combinations: S = all 2‑card hands, size C(52, 2) = 1326. Condition B = “at least one ace”: n(B) = total hands – hands with no ace = 1326 – C(48, 2) = 1326 – 1128 = 198. Event A ∩ B = “both aces”: n(A∩B) = C(4, 2) = 6. Hence P(A|B) = 6 / 198 = 1/33. Counting helps us handle the reduced space precisely.
条件往往会缩小样本空间。设从一副标准扑克牌中抽 2 张。已知至少有一张 A，求两张都是 A 的概率。使用组合：样本空间 S 为所有 2 张牌组，大小为 C(52, 2) = 1326。条件 B 为“至少一张 A”：n(B) = 总手牌数 – 无 A 手牌数 = 1326 – C(48, 2) = 1326 – 1128 = 198。事件 A ∩ B 为“两张都是 A”：n(A∩B) = C(4, 2) = 6。因此 P(A|B) = 6 / 198 = 1/33。计数帮助我们精确处理缩小后的样本空间。
12. IB Exam Tips and Common Pitfalls | IB 考试技巧与常见误区
Watch for “at least one” scenarios – using the complement 1 – P(none) often makes the counting much easier. Also, never mix up permutations and combinations; explicitly decide whether order matters before calculating. When items are identical, remember to divide by the factorials of identical counts. Finally, always check that your probability lies between 0 and 1, and simplify fractions fully. Past papers regularly test these ideas in contexts like letters in a word, team selections, or keys arranged on a ring.
警惕“至少一个”的情境——利用补集 1 – P(无一) 通常能极大简化计数。此外，切勿混淆排列与组合；计算前务必明确顺序是否重要。当物品相同时，记得除以相同个数的阶乘。最后，始终检查概率值是否在 0 到 1 之间，并将分数彻底化简。历年真题常常在字母排列、队伍选拔或钥匙环排列等场景中考查这些知识点。
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