📚 IB Mathematics: Concepts and Calculations of Permutations | IB数学：排列的概念与计算
Permutations are a fundamental topic in IB Mathematics, forming the basis of counting techniques and probability. This article explains the concept of permutations, introduces key formulas, and walks through worked examples aligned with IB-level questions.
排列是IB数学中的基础内容，也是计数原理与概率计算的基石。本文将解释排列的概念、介绍关键公式，并通过与IB考试风格一致的例题详细讲解。
1. What Is a Permutation? | 什么是排列？
A permutation is an arrangement of objects in a specific order. When the order changes, the permutation changes. For example, the arrangements ABC and ACB are different permutations of the same three letters.
排列是指将若干对象按照特定顺序进行安排。顺序改变，排列也随之改变。例如，ABC 和 ACB 是三个相同字母的不同排列。
The key idea is that permutations care about order. If you are selecting a president, vice-president and secretary from a group, the roles matter, so it is a permutation problem.
关键在于排列注重顺序。如果从一组人中选出会长、副会长和秘书，不同的职位对应不同的人，这就是排列问题。
2. Factorial Notation | 阶乘符号
Before calculating permutations, you need to understand factorial notation. The factorial of a positive integer n, written as n!, is the product of all positive integers from 1 to n.
在计算排列之前，需要理解阶乘符号。正整数 n 的阶乘写作 n!，表示从 1 到 n 的所有正整数的乘积。
n! = n × (n − 1) × (n − 2) × … × 2 × 1
For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. By convention, 0! = 1 and 1! = 1. These special cases appear frequently in permutation formulas.
例如，5! = 5 × 4 × 3 × 2 × 1 = 120。按照约定，0! = 1，1! = 1。这些特殊值在排列公式中经常出现。
3. The Permutation Formula ⁿPᵣ | 排列公式 ⁿPᵣ
The number of ways to arrange r objects selected from n distinct objects is denoted by ⁿPᵣ or P(n, r). The formula is:
从 n 个不同对象中选取 r 个对象进行排列的方法数记为 ⁿPᵣ 或 P(n, r)。其公式为：
ⁿPᵣ = n! / (n − r)!
This formula counts ordered arrangements. For example, selecting and arranging 3 letters from 5 distinct letters gives ⁵P₃ = 5! / 2! = 60.
该公式计算有序排列的数量。例如，从5个不同字母中选取并排列3个字母，得到 ⁵P₃ = 5! / 2! = 60。
Notice that when r = n, the formula becomes ⁿPₙ = n! / 0! = n!. This is the number of ways to arrange all n objects in a line.
注意当 r = n 时，公式变为 ⁿPₙ = n! / 0! = n!。这就是将 n 个对象全部排成一行的方法数。
4. Permutations of All Objects | 全排列
The simplest type of permutation is arranging all n distinct objects in a row. The number of arrangements is n!.
最简单的排列是将 n 个不同对象排成一行。排列总数为 n!。
For example, the letters A, B, C, D can be arranged in 4! = 24 different orders. Each order is a distinct permutation.
例如，字母 A、B、C、D 可以有 4! = 24 种不同顺序。每种顺序都是一个不同的排列。
In IB questions, you may be asked to find the number of ways to arrange books on a shelf, students in a line, or digits in a code. In each case, if all objects are distinct and all are used, the answer is simply n!.
在IB题目中，可能会要求计算书架上摆书、学生排队或数字组成密码的方法数。如果所有对象互不相同且全部使用，答案就是 n!。
5. Permutations with Repeated Objects | 含重复对象的排列
When some objects are identical, the number of distinct arrangements is reduced. If n objects contain groups of identical objects with sizes n₁, n₂, …, nₖ, then the number of distinct permutations is:
当某些对象相同时，不同排列的数量会减少。如果 n 个对象中包含若干组相同对象，组大小分别为 n₁、n₂、…、nₖ，则不同排列的数量为：
n! / (n₁! × n₂! × … × nₖ!)
For example, the word “IBMATHS” has all distinct letters, so 7! arrangements. But the word “TUTOR” has two T’s, so the number of distinct arrangements is 5! / 2! = 60.
例如，单词”IBMATHS”中所有字母都不相同，因此有7!种排列。但单词”TUTOR”中有两个T，因此不同排列数为 5! / 2! = 60。
This formula is often tested in IB Paper 1. Always divide by the factorial of each repeated group’s size.
这个公式在IB Paper 1中经常考到。一定要除以每个重复组大小的阶乘。
6. Circular Permutations | 圆排列
In a circular arrangement, rotations are considered the same. For n distinct objects arranged around a circle, the number of distinct arrangements is (n − 1)!.
在圆形排列中，旋转后被视作相同。n 个不同对象围成一圈，不同排列数为 (n − 1)!。
Why? Fix one object to break the symmetry, then arrange the remaining n − 1 objects in a line relative to the fixed object. This gives (n − 1)! arrangements.
为什么？固定一个对象以打破对称性，然后将其余 n − 1 个对象相对于固定对象排成一行，得到 (n − 1)! 种排列。
If clockwise and anticlockwise orders are considered the same, such as for necklaces or bracelets, the number becomes (n − 1)! / 2.
如果顺时针与逆时针顺序视为相同，例如项链或手链，则排列数变为 (n − 1)! / 2。
7. Permutations with Restrictions | 有限制条件的排列
Many IB problems impose restrictions, such as certain objects must be together, or certain objects must not be together. These require careful counting strategies.
许多IB问题会设置限制条件，例如某些对象必须相邻，或某些对象不能相邻。这需要仔细的计数策略。
If certain objects must be together, treat them as a single block. Then arrange the block with the other objects, and multiply by the internal arrangements of the block.
如果某些对象必须相邻，可将它们视为一个整体。然后将这个整体与其他对象一起排列，并乘以整体内部的排列数。
For example, arrange 5 people A, B, C, D, E in a row such that A and B are together. Treat AB as one block: 4! ways to arrange the block and C, D, E, and 2! ways to arrange A and B inside. Total = 4! × 2! = 48.
例如，将5个人A、B、C、D、E排成一排，要求A和B相邻。将AB视为一个整体：排列这个整体与C、D、E共有4!种方式，A和B内部有2!种排列。总数 = 4! × 2! = 48。
8. Permutations with “Not Together” Restriction | “不相邻”限制
When certain objects must not be together, subtract the “together” arrangements from the total arrangements. This is a common complementary counting method.
当某些对象不能相邻时，从总排列数中减去“相邻”的排列数。这是常见的补集计数方法。
Example: Arrange 4 people A, B, C, D in a row so that A and B are not next to each other. Total arrangements = 4! = 24. Arrangements with A and B together = 3! × 2! = 12. Therefore, not together = 24 − 12 = 12.
例如：将4个人A、B、C、D排成一排，要求A和B不相邻。总排列数 = 4! = 24。A和B相邻的排列数 = 3! × 2! = 12。因此，不相邻排列数 = 24 − 12 = 12。
Alternatively, you can arrange the other objects first, then place the restricted objects in the gaps. This gap method is especially useful for larger sets.
另一种方法是先排列其他对象，再把有限制的对象放入空隙中。这种间隔法在处理较大集合时尤为有效。
9. Permutations vs Combinations | 排列与组合的区别
The main difference is whether order matters. Permutations count ordered arrangements; combinations count unordered selections.
主要区别在于顺序是否重要。排列计数的有序排列，组合计数无序选择。
For example, choosing 3 representatives from 10 students is a combination: ¹⁰C₃. But assigning 3 different positions to 3 selected students is a permutation: ¹⁰P₃.
例如，从10名学生中选出3名代表是组合问题：¹⁰C₃。但将3个不同职位分配给选出的3名学生则是排列问题：¹⁰P₃。
| Permutation | Combination |
| Order matters | Order does not matter |
| Formula: ⁿPᵣ = n! / (n − r)! | Formula: ⁿCᵣ = n! / [r! (n − r)!] |
| Example: arranging books | Example: choosing books |
Notice that ⁿPᵣ = ⁿCᵣ × r!. This relationship links the two concepts.
注意 ⁿPᵣ = ⁿCᵣ × r!。这个关系将两个概念联系起来。
10. Worked Example: Arranging Letters | 例题：字母排列
How many distinct arrangements can be made from the letters of the word “MATHEMATICS”?
单词”MATHEMATICS”的字母可以组成多少种不同排列？
The word has 11 letters: M appears 2 times, A appears 2 times, T appears 2 times, and H, E, I, C, S each appear once. Using the repeated-object formula:
该单词共有11个字母：M出现2次，A出现2次，T出现2次，H、E、I、C、S各出现1次。使用重复对象公式：
11! / (2! × 2! × 2!) = 4,989,600
This is the number of distinct letter arrangements, assuming only the letters themselves are distinguished.
这就是不同字母排列的数量，前提是仅字母本身被区分。
11. Worked Example: Seating Arrangement | 例题：座位安排
Eight students are to be seated in a row. How many ways can they be seated if two particular students, Alice and Bob, must sit next to each other?
8名学生排成一排就座。如果其中两位特定学生Alice和Bob必须相邻，共有多少种就座方式？
Treat Alice and Bob as one block. Then we have 7 items to arrange: the block plus the other 6 students. This gives 7! arrangements. Inside the block, Alice and Bob can switch places, giving 2! arrangements. Total:
将Alice和Bob视为一个整体。这样共有7个对象需要排列：这个整体加上其他6名学生，得到7!种排列。整体内部Alice和Bob可以互换位置，有2!种排列。总数：
7! × 2! = 5040 × 2 = 10,080
If instead they must not sit together, we would subtract 10,080 from the total 8! = 40,320, giving 30,240.
如果他们不能相邻，则从总数8! = 40,320中减去10,080，得到30,240。
12. Common IB Exam Tips | IB考试常见提示
Always read the question carefully to decide if order matters. Look for keywords like “arrange”, “order”, “line” for permutations, and “choose”, “select”, “committee” for combinations.
务必仔细审题，判断顺序是否重要。注意关键词：出现”arrange”、”order”、”line”时使用排列；出现”choose”、”select”、”committee”时使用组合。
When using the permutation formula, check whether objects are distinct. If repetition is allowed in arrangements, use the multiplication principle nʳ instead of ⁿPᵣ.
使用排列公式时，检查对象是否互不相同。如果允许重复选择，则应使用乘法原理 nʳ，而不是 ⁿPᵣ。
For restricted arrangements, decide whether to use the block method, the gap method, or complementary counting. Practice each method with past paper questions to become confident.
对于有限制条件的排列，判断应使用整体法、间隔法还是补集计数法。用历年真题练习每种方法，增强信心。
Finally, remember that 0! = 1. This seemingly small fact is essential when r = n in the formula ⁿPᵣ.
最后，记住0! = 1。这个看似微小的知识点在 r = n 时对公式 ⁿPᵣ 至关重要。
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