📚 The Selection of Members | 成员的选择：组合基础
In both pure and applied mathematics, the idea of selecting a group of members from a larger set lies at the heart of counting, probability, and algebraic expansions. Whether you are forming a committee, picking lottery numbers, or expanding (a + b)ⁿ, you are essentially making selections. This article revisits the key principles of combinations, permutations, and the binomial theorem, tailored for the Edexcel A‑Level Mathematics specification.
在纯数学和应用数学中，从较大的集合中选择成员的概念是计数、概率和代数展开的核心。无论是组成委员会、选择乐透号码，还是展开 (a + b)ⁿ，本质上都是在进行选择。本文回顾组合、排列和二项式定理的关键原理，内容与 Edexcel A‑Level 数学考试大纲紧密对接。
1. The Concept of Selection | 选择的概念
Many real‑world scenarios ask: “In how many ways can we choose some objects from a given set?” A selection may be ordered (the sequence matters) or unordered (only the chosen members matter). Recognising which type applies is the first step in any counting problem.
许多现实场景会问：“从一个给定的集合中选择一些对象，有多少种不同的选法？”这种选择可以是有序的（顺序重要），也可以是无序的（只关心哪些成员被选中）。识别问题属于哪一种，是解决任何计数问题的第一步。
For example, choosing a captain and vice‑captain from a team of 11 is an ordered selection — the two roles are different. But selecting 5 players to form a basketball squad is unordered, because the squad is just a set of members.
例如，从一支 11 人的队伍中选出队长和副队长是一个有序的选择——两个角色不同。而选择 5 名球员组成篮球队则是无序的，因为球队只是一个成员的集合。
2. The Fundamental Counting Principle | 基本计数原理
her can be performed in m × n ways. This simple idea underpins all counting of selections.
如果一个操作可以用 m 种方式完成，而另一个独立操作可以用 n 种方式完成，那么这两个操作合在一起可以用 m × n 种方式完成。这个简单的想法支撑了所有选择问题的计数。
m 4 types and one drink from 3 types. The total number of meal combinations is 4 × 3 = 12.
考虑一个套餐，你从 4 种三明治中选一种，从 3 种饮料中选一种。套餐的总组合数为 4 × 3 = 12。
When we select several members one after another, the principle often leads to factorials. For instance, arranging n distinct members in a line gives n! ways.
当我们需要逐个选择多个成员时，该原理通常会引出阶乘。例如，将 n 个不同的成员排成一列，有 n! 种排列方式。
3. Permutations: Ordered Selections | 排列：有序的选择
A permutation is an arrangement of members where the order is important. The number of ways to arrange r members chosen from n distinct members is denoted by nPr and is given by:
排列是指顺序重要的成员排列。从 n 个不同成员中选择 r 个进行排列的方法数用 nPr 表示，计算公式为：
nPr = n! / (n − r)!
This formula arises because there are n choices for the first member, (n − 1) for the second, and so on, down to (n − r + 1). Multiplying these together yields the numerator n! divided by (n − r)!.
这个公式的产生是因为第一个成员有 n 种选择，第二个有 (n − 1) 种，依此类推，直到第 r 个有 (n − r + 1) 种。将这些相乘得到分子 n! 除以 (n − r)!。
A typical exam question: “How many different 4‑digit PINs can be formed from the digits 1–9 if no digit is repeated?” Here, n = 9, r = 4, so 9P4 = 9! / 5! = 9 × 8 × 7 × 6 = 3024.
典型的考题：“如果不允许数字重复，从数字 1–9 中可以形成多少个不同的四位 PIN 码？”这里 n = 9, r = 4，所以 9P4 = 9!/5! = 9 × 8 × 7 × 6 = 3024。
4. Combinations: Unordered Selections | 组合：无序的选择
When the order does not matter, a selection is called a combination. The number of ways to choose r members from n distinct members is written nCr (or n r as a binomial coefficient) and is given by:
当顺序不重要时，这种选择称为组合。从 n 个不同成员中选出 r 个的方法数记作 nCr（或写为二项式系数 n r），计算公式为：
nCr = n! / [r! (n − r)!]
Dividing the permutation count by r! removes the order, because each set of r members can be arranged in r! different orders. Thus nCr = nPr / r!.
将排列数除以 r! 就去掉了顺序，因为每 r 个成员可以排列成 r! 种不同的顺序。因此 nCr = nPr / r!。
In a lottery, 6 numbers are drawn from 49. The number of possible tickets is 49C6 = 49!/(6!·43!) = 13,983,816. That is why the odds of winning the jackpot are so low.
在乐透中，从 49 个号码中抽出 6 个。可能彩票的组合数是 49C6 = 49!/(6!·43!) = 13,983,816。这就是为什么中头奖的概率极低。
5. Calculating Combinations – The Formula | 计算组合 – 公式
When using the formula, cancellation often simplifies the work. For 10C3, write 10!/(3!·7! 8)/(3 × 2 × 1) = 120. Always check whether it is quicker to use the complementary property nCr = nCn−r.
使用公式时，约分常常能简化计算。对于 10C3，可以写成 10!/(3!·7!) = (10 × 9 × 8)/(3 × 2 × 1) = 120。时常检查使用互补性质 nCr = nCn−r 是否更快。
This symmetry reflects the fact that choosing r members to include is the same as choosing n − r to exclude. For instance, 10C7 = 10C3 = 120.
这种对称性反映了这样一个事实：选择 r 个成员包含在内，等同于选择 n − r 个排除在外。例如，10C7 = 10C3 = 120。
Be careful with special cases: nC0 = 1 (there is one way to choose nothing), nC1 = n, and nCn = 1. These often appear in binomial expansions.
注意特殊情况：nC0 = 1（选择 0 个成员只有一种方式），nC1 = n，nCn = 1。这些在二项式展开中经常出现。
6. Notation and Properties of nCr | 组合数的记法与性质
The binomial coefficient can also be written as ( n r ) or C(n, r). Key properties include the additive identity: nCr−1 + nCr = n+1Cr, which is the basis of Pascal’s triangle.
二项式系数也可以写成 ( n r ) 或 C(n, r)。关键性质包括加法恒等式：nCr−1 + nCr = n+1Cr，这是帕斯卡三角形的基础。
Pascal’s triangle provides a quick reference for small n. Row n gives the coefficients nC0, nC1, ..., nCn. For example, row 4 is 1, 4, 6, 4, 1, matching 4Cr for r = 0 to 4.
帕斯卡三角形为较小的 n 提供了快速参考。第 n 行给出系数 nC0, nC1, …, nCn。例如，第 4 行是 1, 4, 6, 4, 1，对应于 r = 0 到 4 时的 4Cr。
In algebraic problems, you may be asked to find the coefficient of a term in a binomial expansion without fully expanding. This relies entirely on nCr.
在代数题中，你可能需要在不完全展开的情况下求出二项展开中某项的系数。这完全依赖于 nCr。
7. Selecting Members with Restrictions | 受限情况下的成员选择
Exam questions frequently add restrictions such as “the committee must include at least one woman” or “the chosen digits must contain exactly two even numbers”. These are tackled by splitting the problem into cases or by subtracting the forbidden arrangements from the total.
考题常常会附加限制条件，例如“委员会中必须包含至少一名女性”或“选出的数字中恰好包含两个偶数”。解决这类问题的方法是把问题拆分成多个情况，或者从总数中减去不允许的排列方式。
two cases: exactly 4 women (and 2 men) or exactly 5 women (and 1 man). Number of ways = (5C4 × 8C2) + (5C5 × 8C1) = (5 × 28) + (1 × 8) = 148.
比如，从 8 名男性和 5 名女性中选出一个 6 人小组，要求至少包含 4 名女性。我们考虑两种情况：恰好 4 名女性（同时 2 名男性），或恰好 5 名女性（同时 1 名男性）。选法数量 = (5C4 × 8C2) + (5C5 × 8C1) = (5 × 28) + (1 × 8) = 148。
Always check whether the complement method is quicker: total ways without restriction minus the number of ways that violate the condition. This avoids multi‑case working.
务必检查补集法是否更快：无限制下的总方法数减去违反条件的方法数。这样可以避免处理多种情况。
8. Selections and the Binomial Theorem | 组合与二项式定理
The binomial theorem states that for positive integer n:
(a + b)ⁿ = Σr=0ⁿ nCr an−r br
Here nCr is exactly the number of ways to choose r members from n. In the expansion, it counts how many ways we can pick r factors that contribute a b when multiplying out the n brackets (a + b)(a + b)…(a + b).
对于正整数 n，二项式定理为：(a + b)ⁿ = Σr=0ⁿ nCr an−r br。这里的 nCr 正是从 n 个成员中选择 r 个的方法数。在展开式中，它计算了在将 n 个括号 (a + b) 相乘时，有多少种方式选取 r 个提供 b 的因子。
e term in x² corresponds to r = 3 (since a = 2x, b = 3, we need a² b³). The coefficient is 5C3 80. Understanding the connection with selections makes the binomial theorem less abstract.
例如，在 (2x + 3)⁵ 中，含 x² 的项对应于 r = 3（因为 a = 2x, b = 3，我们需要 a² b³）。其系数为 5C3 × (2)² × 3³ = 10 × 4 × 27 = 1080。理解它与选择之间的联系，使二项式定理不再那么抽象。
9. Probability of Selecting Members | 选择成员的概率
When every member has an equal chance of being chosen, the probability of a particular unordered selection is 1 divided by the total number of equally likely combinations. Often questions ask for the probability that a randomly formed group satisfies a condition.
当每个成员被选中的几率相等时，某个特定无序选择的概率等于 1 除以所有等可能组合的总数。题目经常会问随机组成的群体满足某个条件的概率。
mber of ways to choose a group satisfying the condition) / (total number of ways to choose the group). This is simply favourable combinations over total combinations.
基本公式是 P(成功) = (满足条件的组的选择方式数) / (选择组的总方式数)。这不过是有利组合数除以总组合数。
For example, a class has 12 students: 7 boys and 5 girls. A group of 4 is selected at random. The probability that it contains exactly 2 girls is (5C2 × 7C2) / 12C4 = (10 × 21) / 495 = 210/495 = 14/33. Always simplify fractions unless told otherwise.
例如，一个班级有 12 名学生：7 名男生、5 名女生。随机选出 4 人小组。恰好包含 2 名女生的概率为 (5C2 × 7C2) / 12C4 = (10 × 21) / 495 = 210/495 = 14/33。除非题目另有要求，否则必须化简分数。
10. Tips for the Exam | 考试技巧
Always decide if order matters: if “captain and vice‑captain” is different to “two leaders”, only the former is a permutation. Look for keywords like “arrange”, “list”, “order”, or “sequence” for permutations; “choose”, “select”, “group”, or “committee” for combinations.
始终判断顺序是否重要：如果“队长与副队长”和“两位负责人”不同，只有前者是排列问题。寻找排列问题中的关键词，如“arrange”、“list”、“order”、“sequence”；组合问题中的关键词，如“choose”、“select”、“group”、“committee”。
When using a calculator, the nCr and nPr functions are reliable. After obtaining an answer, check with a smaller case or mental verification. For instance, 6C2 = 15; if your result is far from this pattern, you may have pressed the wrong key.
使用计算器时，nCr 和 nPr 函数很可靠。得出答案后，用一个较小的情形或心算进行检验。例如，6C2 = 15；如果你的结果与此模式相差甚远，可能按错了键。
Finally, be comfortable relating combinations back to binomial expansions: nCr is the coefficient of br in (1 + b)ⁿ, and more generally in (a + b)ⁿ. This connection often saves time in pure mathematics questions.
最后，要能熟练地将组合联系回二项展开：nCr 是 (1 + b)ⁿ 中 br 的系数，更一般地也在 (a + b)ⁿ 中。这种联系往往能在纯数学题中节省时间。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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