Using Permutations and Combinations to Find Probability in IB Math | IB数学：利用排列组合求概率的方法

📚 Using Permutations and Combinations to Find Probability in IB Math | IB数学：利用排列组合求概率的方法

Probability is a measure of how likely an event is to occur. In many IB Mathematics problems, the sample space and the favorable outcomes are not listed one by one, but are counted using permutations and combinations. This article explains these counting techniques and how to apply them directly to probability calculations.

概率是衡量事件发生可能性大小的量。在IB数学许多题目中，样本空间与有利结果并非逐一列出，而是通过排列与组合来计数。本文将解释这些计数方法，并说明如何将它们直接应用于概率计算。

1. The Fundamental Counting Principle | 基本计数原理

If one task can be done in m ways and a second independent task can be done in n ways, then the two tasks together can be done in m × n ways. This is the multiplication rule.

若完成第一件事有 m 种方法，完成第二件独立的事有 n 种方法，则依次完成这两件事共有 m × n 种方法。这就是乘法原理。

For example, if a restaurant offers 3 starters and 4 main courses, there are 3 × 4 = 12 possible meals. This principle underlies all permutation and combination formulas.

例如，某餐厅有3种前菜和4种主菜，则共有 3 × 4 = 12 种套餐选择。该原理是一切排列组合公式的基础。

2. Factorial Notation | 阶乘记号

For a positive integer n, the factorial n! is defined as n × (n − 1) × … × 2 × 1. By convention, 0! = 1.

对正整数 n，阶乘 n! 定义为 n × (n − 1) × … × 2 × 1。按规定 0! = 1。

n! = n × (n − 1) × … × 2 × 1

Factorials count the number of ways to arrange a set of distinct objects in a line. For instance, 3! = 6 is the number of arrangements of the letters A, B, C.

阶乘用于计算一组互异物体排成一行的方法数。例如，3! = 6 即字母 A、B、C 的排列总数。

3. Permutations: Arrangements of Objects | 排列：物体的有序安排

A permutation is an ordered arrangement of objects. The number of ways to choose and arrange r objects from a set of n distinct objects is denoted by nPr and is given by:

排列是有序的安排。从 n 个互异物体中选出 r 个并排成一列的方法数记为 nPr，计算公式为：

nPr = n! / (n − r)!

The key idea is that order matters. For example, the number of ways to arrange 3 books chosen from 5 on a shelf is 5P3 = 5! / 2! = 5 × 4 × 3 = 60.

关键在“顺序有关”。例如，从5本不同的书中选3本放到书架上的方法数为 5P3 = 5! / 2! = 5 × 4 × 3 = 60。

4. Permutations with Identical Objects | 含相同物体的排列

If some objects are identical, the number of distinct arrangements is reduced. The number of distinct permutations of n objects where there are a identical objects of one type, b of another, and so on, is:

当物体中存在相同项时，互不相同的排列数会减少。n 个物体中，若有 a 个同类相同物、b 个另一类相同物……则互异排列数为：

n! / (a! × b! × …)

For the word “MISSISSIPPI”, there are 11 letters: M appears once, I appears 4 times, S appears 4 times, P appears 2 times. The number of distinct arrangements is 11! / (1! × 4! × 4! × 2!) = 34 650.

以单词“MISSISSIPPI”为例，共11个字母：M出现1次，I出现4次，S出现4次，P出现2次。互异排列数为 11! / (1! × 4! × 4! × 2!) = 34 650。

5. Combinations: Selections without Order | 组合：不计顺序的选择

A combination is a selection of objects in which order does not matter. The number of ways to choose r objects from n distinct objects is denoted by nCr and is given by:

组合是“不计顺序”的选择。从 n 个互异物体中选出 r 个的方法数记为 nCr，计算公式为：

nCr = n! / [r! (n − r)!]

For example, the number of ways to select 2 students from a group of 10 is 10C2 = 10! / (2! × 8!) = 45. Notice that choosing Alice then Bob is the same as choosing Bob then Alice.

例如，从10名学生中选2人的方法数为 10C2 = 10! / (2! × 8!) = 45。注意，先选 Alice 再选 Bob 与先选 Bob 再选 Alice 是同一选择。

6. Permutation or Combination? | 排列还是组合？

The most common mistake is confusing the two. The table below summarises the difference:

最常见的错误是混淆二者。下表总结了它们的区别：

Scenario / 情境	Order matters? / 顺序有关？	Use / 使用
Arranging books on a shelf / 书架上摆书	Yes / 是	Permutation / 排列
Selecting a committee / 选委员会	No / 否	Combination / 组合
Arranging letters of a word / 重排单词字母	Yes / 是	Permutation / 排列
Drawing 5 cards from a deck / 从牌堆抽5张	No / 否	Combination / 组合

When calculating probability, ask: “Does the order of the chosen items affect the outcome described?” If yes, use permutations; otherwise, use combinations.

计算概率时问自己：“所选项目的顺序是否影响所描述的结果？”若影响则用排列，否则用组合。

7. Probability as a Ratio of Counts | 概率为计数之比

For a finite sample space where every outcome is equally likely, the probability of event A is:

对有限样本空间且每个结果等可能的情形，事件A的概率为：

P(A) = Number of favorable outcomes / Total number of outcomes

Both the numerator and the denominator are counted using permutations or combinations, depending on what is being selected or arranged.

分子与分母都用排列或组合来计算，具体取决于所选取或排列的对象。

8. Example: Committee Selection | 例题：委员会选择

A class of 20 students contains 12 girls and 8 boys. If 4 students are selected at random to form a committee, find the probability that the committee contains exactly 2 girls.

某班有20名学生，其中12名女生、8名男生。若随机选出4人组成委员会，求委员会中恰好有2名女生的概率。

Total outcomes: choose any 4 from 20, which is 20C4 = 4845.

总结果数：从20人中任选4人，即 20C4 = 4845。

Favorable outcomes: choose 2 girls from 12 and 2 boys from 8. By the multiplication rule, the number is 12C2 × 8C2 = 66 × 28 = 1848.

有利结果数：从12名女生中选2人，从8名男生中选2人。由乘法原理，其数量为 12C2 × 8C2 = 66 × 28 = 1848。

P = 1848 / 4845 ≈ 0.381

Thus the probability is approximately 0.381, or 38.1%.

因此概率约为 0.381，即 38.1%。

9. Example: Arrangements in a Row | 例题：排成一行

Seven digits 0, 1, 2, 3, 4, 5, 6 are arranged randomly to form a 7-digit number. What is the probability that the number is even?

将数字 0, 1, 2, 3, 4, 5, 6 随机排列成一个七位数。求该数为偶数的概率。

The first digit cannot be 0, so total arrangements: choose first digit in 6 ways (1–6), then arrange the remaining 6 digits in 6! ways. Total = 6 × 6! = 4320.

首位不能为0，因此总排列数：首位有6种选择（1–6），其余6个数字任意排列为6!种。总数 = 6 × 6! = 4320。

For an even number, the last digit must be 0, 2, 4, or 6. We can count by cases:

偶数要求末位为 0、2、4 或 6。我们分情况计数：

Last digit 0: first digit can be any of 1–6 (6 ways), then arrange the middle 5 digits in 5! ways. Count = 6 × 5! = 720.
Last digit 2, 4 or 6: for each of these, the first digit cannot be 0 and cannot be the same as the last digit. There are 5 choices for the first digit (excluding 0 and the chosen last digit), then arrange the remaining 5 digits in 5! ways. Count per choice = 5 × 5! = 600. There are 3 such cases, giving 3 × 600 = 1800.

末位为0：首位可从1–6中选6种，中间5位排列5!种，共有 6 × 5! = 720 种。

末位为2、4或6：对每个末位，首位不能为0且不能等于末位，有5种选择，剩下5位排列5!种。每个末位有 5 × 5! = 600 种，共3个末位，即 3 × 600 = 1800 种。

Favorable total = 720 + 1800 = 2520.

有利总数 = 720 + 1800 = 2520。

P = 2520 / 4320 = 7 / 12 ≈ 0.583

The probability is 7/12.

该概率为 7/12。

10. Example: Drawing Cards | 例题：抽取卡片

From a standard deck of 52 cards, 5 cards are drawn without replacement. Find the probability of getting exactly 3 hearts.

从一副52张的标准扑克牌中无放回地抽取5张。求恰好抽到3张红心的概率。

The total number of 5-card hands is 52C5 = 2 598 960.

5张牌的组合总数为 52C5 = 2 598 960。

There are 13 hearts and 39 non-hearts. Favorable hands: choose 3 hearts from 13 and 2 non-hearts from 39.

一副牌有13张红心、39张非红心。有利情况：从13张红心选3张，从39张非红心选2张。

Favorable = 13C3 × 39C2 = 286 × 741 = 211 926

Therefore P = 211 926 / 2 598 960 ≈ 0.0815.

因此 P = 211 926 / 2 598 960 ≈ 0.0815。

11. Common Pitfalls | 常见易错点

Several mistakes appear frequently in IB exams:

IB考试中以下错误经常出现：

Using permutations when order is irrelevant, for example in committee problems.
Forgetting that 0! = 1 when simplifying factorials.
Assuming items are distinct when they are actually identical.
Not checking whether the event description uses “at least” or “exactly”; these require different counting strategies.

在不需要考虑顺序的问题中误用排列，比如委员会问题。

化简阶乘时忘记 0! = 1。

把本质相同的物体当作互异物体。

没有区分事件是“至少”还是“恰好”，二者需要不同的计数策略。

For “at least” problems, it is often easier to subtract the complementary probability from 1: P(at least one) = 1 − P(none).

对于“至少”型问题，通常用补集计算更方便：P(至少一个) = 1 − P(一个都没有)。

12. Summary and Final Tips | 总结与最后建议

To solve probability problems using permutations and combinations:

利用排列组合求解概率问题时应遵循：

Decide whether the situation involves ordered arrangements (permutation) or unordered selections (combination).
Count the total number of equally likely outcomes in the sample space.
Count the number of favorable outcomes, often using the multiplication rule to combine smaller selections or arrangements.
Write the probability as the ratio of the favorable count to the total count.
Simplify the fraction if possible, and express the answer as a fraction, decimal, or percentage as requested.

判断问题是有序排列（排列）还是无序选择（组合）。

计数样本空间中所有等可能结果的总数。

计数有利结果数，通常用乘法原理将若干较小选择或排列相乘。

将概率写为有利数与总数的比。

尽可能化简分数，并按题目要求写成分数、小数或百分比。

Practice with past-paper questions to build confidence. The more familiar you are with common wording, the faster you will identify whether to use nPr or nCr.

多做真题练习以增强信心。对常见表述越熟悉，你就能越快判断该用 nPr 还是 nCr。

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Using Permutations and Combinations to Find Probability in IB Math | IB数学：利用排列组合求概率的方法