IB Mathematics: Permutations and Combinations Calculation Methods | IB数学：排列与组合计算方法

📚 IB Mathematics: Permutations and Combinations Calculation Methods | IB数学：排列与组合计算方法

Counting the number of ways to arrange or choose objects is a central skill in IB Mathematics, in both Analysis and Approaches and Applications and Interpretation. Permutations and combinations provide structured methods for counting possibilities without listing them all, and they form the foundation for probability questions in the written papers.

计数排列或选样的方式是 IB 数学中非常核心的技能，无论 AA 还是 AI 课程都要求掌握。排列与组合提供了结构化的计数方法，无需逐一列举就能确定可能情况的总数，同时也是笔试中概率题的基础。

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

If one choice can be made in m ways and a second independent choice can be made in n ways, then the two choices together can be made in m × n ways. This rule extends naturally to any number of independent choices.

如果第一个选择有 m 种方式，第二个独立的选择有 n 种方式，那么两个选择共同发生共有 m × n 种方式。这个原理可以自然推广到任意多个相互独立的步骤。

total ways = m₁ × m₂ × … × mₖ

For example, choosing one pen from 3 and one notebook from 5 gives 3 × 5 = 15 possible pairs.

例如，从 3 支笔中选择 1 支，从 5 个笔记本中选择 1 个，则共有 3 × 5 = 15 种搭配。

• Example: a restaurant menu offers 4 starters and 6 main courses, so the number of possible starter–main menus is 4 × 6 = 24.
• 例：一家餐厅提供 4 种前菜和 6 种主菜，那么一餐共有 4 × 6 = 24 种前菜与主菜搭配。

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2. Factorial Notation | 阶乘符号

The factorial of a positive integer n is the product of every positive integer from n down to 1. By definition, 0! = 1.

正整数 n 的阶乘是从 n 一直乘到 1 的所有正整数的乘积。按定义，0! = 1。

n! = n × (n − 1) × (n − 2) × … × 2 × 1

Factorials grow extremely quickly. For instance, 5! = 5 × 4 × 3 × 2 × 1 = 120, and 10! = 3 628 800.

阶乘的增长速度非常快。例如，5! = 5 × 4 × 3 × 2 × 1 = 120，而 10! = 3 628 800。

This notation is the building block for every permutation and combination formula in this topic.

阶乘记号是排列与组合所有公式的基础构件。

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3. Permutations of Distinct Objects | 不同对象的排列

A permutation is an arrangement in which order matters. The number of ways to arrange r distinct objects chosen from n distinct objects is written as ⁿPᵣ.

排列是指“顺序有关”的安排方式。从 n 个不同对象中取出 r 个不同对象进行排列，其方法数记作 ⁿPᵣ。

ⁿPᵣ = n! / (n − r)!

When all n objects are arranged, the number of permutations is simply n! because ⁿPₙ = n! / 0! = n!.

当 n 个对象全部参与排列时，排列数就是 n!，因为 ⁿPₙ = n! / 0! = n!。

Example: from 5 different books, the number of ways to choose 3 and place them in a row on a shelf is ⁵P₃ = 5 × 4 × 3 = 60.

例：从 5 本不同的书中选出 3 本并在书架上排成一排，共有 ⁵P₃ = 5 × 4 × 3 = 60 种方式。

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4. Permutations with Repetition Allowed | 允许重复的排列

When each position may be filled from the same set of n options and repetition is allowed, the total number of arrangements of r positions is nʳ.

如果每个位置都可以从相同的 n 个选项中选择，并且允许重复，那么填满 r 个位置的总数为 nʳ。

number of arrangements with repetition = nʳ

A common IB exam example is a 4-digit PIN code using the digits 0 to 9. Since digits may repeat, the total number of PIN codes is 10⁴ = 10 000.

IB 考试中常见的例子是使用 0 到 9 的 4 位数字 PIN 码。由于数字可以重复，所以 PIN 码总数为 10⁴ = 10 000。

This is different from a permutation without repetition, where an object cannot be used more than once.

这与不允许重复的排列不同。在不允许重复的排列中，同一个对象不能被使用两次。

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5. Circular Permutations | 圆桌排列

When n distinct objects are arranged around a circle, rotations are regarded as the same arrangement. To remove this rotational symmetry, fix one object in place and arrange the remaining n − 1 objects.

当 n 个不同对象围成一圈时，旋转后的结果被视为同一种排列。为了消除旋转对称性，先固定一个对象，再排列其余 n − 1 个对象。

number of circular permutations = (n − 1)!

For example, 6 people can be seated around a circular table in 5! = 120 ways.

例如，6 个人围圆桌就座，共有 5! = 120 种方式。

If clockwise and counter-clockwise arrangements are considered identical, such as with some necklaces or bracelets, further divide by 2.

如果顺时针和逆时针被视为同一种情况，例如某些项链或手环，则需要再除以 2。

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6. Combinations | 组合

A combination is a selection in which order does not matter. The number of ways to choose r objects from n distinct objects is written as ⁿCᵣ or C(n, r).

组合是指“顺序无关”的选取方式。从 n 个不同对象中选取 r 个对象的方法数记作 ⁿCᵣ 或 C(n, r)。

ⁿCᵣ = n! / (r! (n − r)!)

The relationship between permutations and combinations is useful: each unordered selection can be arranged in r! ways, so ⁿCᵣ = ⁿPᵣ / r!.

排列与组合之间有重要关系：每一种无序选取都可以重排成 r! 种不同排列，因此 ⁿCᵣ = ⁿPᵣ / r!。

Example: choosing 3 class representatives from 5 students gives ⁵C₃ = 5! / (3! 2!) = 10.

例：从 5 名学生中选出 3 名班委，共有 ⁵C₃ = 5! / (3! 2!) = 10 种选法。

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7. Permutations with Identical Objects | 含相同对象的排列

If a set of n objects contains p identical objects of one kind, q identical objects of another kind, and so on, then the number of distinct arrangements is n! divided by the factorials of the repeated counts.

如果 n 个对象中有 p 个同类相同对象、q 个另一类相同对象……，那么不同的排列数为 n! 除以各重复数量的阶乘。

n! / (p! q! r! …)

For the word AAABBC, there are 6 letters with three A’s, two B’s and one C. The number of distinct arrangements is 6! / (3! 2!) = 720 / 12 = 60.

例如单词 AAABBC 中共有 6 个字母，其中 3 个 A、2 个 B、1 个 C。其不同排列数为 6! / (3! 2!) = 720 / 12 = 60。

A common mistake is treating identical letters as distinguishable; always divide by the repeated factorials.

常见错误是把相同字母当作不同对象来排列；一定要除以重复部分的阶乘。

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8. Restriction Techniques | 限制条件下的排列技巧

When certain objects must stay together, treat them as one block. For example, arranging 5 books in a row with two specified books always together gives 4! × 2! = 48 arrangements.

当某些对象必须相邻时，把它们看作一个整体块。例如，将 5 本书排成一排，且其中两本指定的书必须相邻，排列数为 4! × 2! = 48。

When certain objects must not be together, count the total unrestricted arrangements and subtract the arrangements where they are together. For 5 books with two specified books separated, the result is 5! − 4! × 2! = 120 − 48 = 72.

当某些对象不能相邻时，先计算无限制的排列数，再减去它们相邻的排列数。对于 5 本书且两本指定书不相邻，结果为 5! − 4! × 2! = 120 − 48 = 72。

Alternatively, use a gap method: arrange the other objects first, then place the restricted objects in the gaps between them.

另一种方法是“隔空法”：先排列其余对象，再把受限制的对象放入它们之间的空隙中。

• For r objects not together among n total objects, identify the number of gaps created by the remaining n − r objects.
• 若要 n 个对象中有 r 个互不相邻，先看其余 n − r 个对象形成了多少个空隙。

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9. The ‘At Least’ Trap and the Complement Method | “至少”问题与补集法

When a question asks for selections containing at least one item from a group, it is often easier to count the total number of selections and subtract the selections that do not satisfy the condition.

当题目要求选取结果中“至少”包含某个群体的一个对象时，通常更容易先计算无限制的总数，再减去不满足条件的选取数。

Example: choose a committee of 5 people from 6 boys and 4 girls. The number of committees with at least one girl is C(10, 5) − C(6, 5) = 252 − 6 = 246.

例：从 6 名男生和 4 名女生中选出 5 人组成委员会。至少有 1 名女生的选法数为 C(10, 5) − C(6, 5) = 252 − 6 = 246。

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