📚 Permutations | 排列
In IB Mathematics, permutations form a core part of combinatorics, enabling us to count the number of ways to arrange objects in a specific order. Understanding permutations is essential for solving problems in probability, algebra, and real-world decision-making. This article provides a comprehensive review of permutations, from basic definitions to advanced techniques, helping you master the topic for your IB exams.
在IB数学中，排列是组合数学的核心部分，用来计算将对象按特定顺序排列的方法数。理解排列对于解决概率、代数以及实际决策中的问题至关重要。本文将全面梳理排列的相关知识，从基本定义到高级技巧，帮助你为IB考试做好充分准备。
1. What is a Permutation? | 什么是排列？
A permutation is an ordered arrangement of a set of objects. For instance, if we have three distinct letters A, B, and C, the possible permutations are ABC, ACB, BAC, BCA, CAB, and CBA. Notice that the order is crucial – ABC is different from ACB.
排列是对一组对象进行的有序排列。例如，有三个不同的字母A、B和C，可能的排列有ABC、ACB、BAC、BCA、CAB和CBA。注意，顺序至关重要——ABC与ACB是不同的排列。
ects, the total number of permutations of all n objects is given by n factorial, denoted n!. In the previous example, 3! = 3 × 2 × 1 = 6.
一般地，如果有 n 个不同的对象，所有对象的全排列总数由 n 的阶乘给出，记作 n!。在上例中，3! = 3 × 2 × 1 = 6。
2. Factorial Notation | 阶乘符号
The factorial of a positive integer n, written as n!, is the product of all positive integers from n down to 1.
正整数n的阶乘，写作n!，是从n乘到1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! is defined as 1, which simplifies many combinatorial formulas.
例如，5! = 5 × 4 × 3 × 2 × 1 = 120。约定 0! = 1，这简化了许多组合公式。
3. Permutations of n Distinct Objects | n个不同对象的排列
When arranging all n distinct objects in a line, the number of permutations is simply n!. This is because there are n choices for the first position, (n − 1) for the second, and so on, until 1 choice remains for the last position.
当将 n 个不同的对象排成一列时，排列的总数就是 n!。这是因为第一个位置有 n 种选择，第二个位置有 (n − 1) 种选择，依此类推，直到最后一个位置只剩下 1 种选择。
For instance, the number of ways to arrange 4 different books on a shelf is 4! = 24.
例如，将 4 本不同的书排列在书架上的方法数为 4! = 24。
4. Permutations of r Objects from n (Partial Permutations) | 从n个对象中选取r个的排列（部分排列）
Often we need to arrange only r objects chosen from a set of n distinct objects. The number of such partial permutations is given by the notation nPr or P(n, r), and the formula is:
我们经常只需要从 n 个不同对象中选取 r 个进行排列。这种部分排列的数目记为 nPr 或 P(n, r)，其公式为：
nPr = n! / (n − r)!
This formula follows from the multiplication principle: there are n choices for the first object, (n − 1) for the second, …, and (n − r + 1) for the rth. This product equals n!/(n − r)!.
该公式来源于乘法原理：第一个对象有 n 种选择，第二个有 (n − 1) 种，……，第 r 个有 (n − r + 1) 种。这个乘积等于 n!/(n − r)!。
y ways can a president, vice-president, and secretary be elected from a club of 10 members? Here n = 10, r = 3, so 10P3 = 10! / 7! = 10 × 9 × 8 = 720.
示例：从 10 名俱乐部成员中选出一名主席、一名副主席和一名秘书，有多少种方法？这里 n = 10, r = 3，因此 10P3 = 10! / 7! = 10 × 9 × 8 = 720。
5. Permutations with Repeated Elements | 含重复元素的排列
When some objects are identical, the number of distinct permutations is reduced. If we have n items where there are p identical items of one type, q identical of another, and so on, the number of unique arrangements is:
当某些对象相同时，不同排列的数量就会减少。如果共有 n 个物品，其中一种有 p 个相同项，另一种有 q 个相同项，依此类推，那么唯一排列的数量为：
n! / (p! × q! × …)
For example, the number of distinct arrangements of the letters in the word 'MISSISSIPPI' is: total letters 11, with 4 S's, 4 I's, 2 P's. So the count = 11! / (4! × 4! × 2!) = 34650.
例如，单词 'MISSISSIPPI' 中字母的不同排列数为：总字母数 11，其中 S 有 4 个，I 有 4 个，P 有 2 个。因此排列数 = 11! / (4! × 4! × 2!) = 34650。
6. Circular Permutations | 环形排列
When objects are arranged in a circle (e.g., people around a round table), we must account for the fact that rotations produce the same arrangement. The number of circular permutations of n distinct objects is (n − 1)!. This is because we fix one object's position to break rotational symmetry, then arrange the remaining (n − 1) objects linearly.
当对象排成一个圆圈（例如，人们围坐在圆桌旁）时，我们必须考虑到旋转会产生相同的排列。 n 个不同对象的环形排列数为 (n − 1)!。这是因为我们通过固定一个对象的位置来打破旋转对称性，然后将剩下的 (n − 1) 个对象进行线性排列。
If the circle can be flipped (considered identical under reflection, as with a necklace), the number reduces further to (n − 1)!/2 for n > 2. This is called permutation under dihedral symmetry, though typically IB focuses on simpler circular permutations.
如果圆圈可以翻转（在反射下视为相同，例如项链），对于 n > 2，排列数会进一步减少到 (n − 1)!/2。这称为二面体对称下的排列，但 IB 通常侧重简单的环形排列。
7. Permutations under Restrictions | 带限制条件的排列
Many IB problems involve additional constraints, such as certain items must be placed together, or some items cannot be adjacent. For items that must be together, treat them as a single 'block'. Arrange the block and other items first, then arrange the items within the block.
许多 IB 题目包含额外限制条件，例如某些元素必须放在一起，或某些元素不能相邻。对于必须在一起的元素，可以将它们视为一个“块”。先排列这个块与其他元素，再排列块内部的元素。
n how many ways can 5 boys and 3 girls be seated in a row if all 3 girls must sit together? Treat the girls as one block; we then have 5 boys + 1 block = 6 units to arrange: 6! ways. Within the block, the girls can be arranged in 3! ways. Total = 6! × 3! = 720 × 6 = 4320.
示例：5 名男孩和 3 名女孩坐成一排，如果所有女孩必须坐在一起，有多少种安排方法？将 3 名女孩视为一个块；此时共有 5 名男孩 + 1 个块 = 6 个单元可排列：6! 种方式。在块内，女孩可按 3! 种方式排列。总数为 6! × 3! = 720 × 6 = 4320。
For restrictions such as 'two people cannot sit together', use the complement method: total arrangements minus arrangements where they are together.
对于“两人不能相邻”这样的限制条件，可使用补集法：总排列数减去他们在一起的排列数。
8. Permutations vs. Combinations | 排列与组合的区别
ons where order does not matter. The number of combinations of r objects from n is nCr = n! / [r! (n − r)!]. It is
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