📚 Combinations | 组合
A combination is a selection of items from a larger set where the order of selection does not matter. In IB Mathematics, combinations form the backbone of counting principles used in probability, binomial expansions, and many real-world problems. Understanding when and how to apply combination formulas is essential for success in both Analysis & Approaches and Applications & Interpretation.
组合是指从一个更大的集合中选取若干项，而不考虑选取顺序。在IB数学中，组合是计数原理的核心基础，广泛应用于概率、二项式展开和许多实际问题。准确判断何时以及如何运用组合公式，对在数学分析与方法、数学应用与解释两门课程中取得好成绩至关重要。
1. What is a Combination? | 什么是组合？
A combination is a way of selecting r objects from a set of n distinct objects, where the arrangement or order of the chosen objects is irrelevant. If you are choosing a team of 3 players from a squad of 10, the selection {Alice, Bob, Carol} is exactly the same as {Carol, Alice, Bob}.
组合是从 n 个不同物体中选取 r 个物体的一种方式，所选物体的排列或顺序无关紧要。如果你从一支10人的队伍中选出3名球员，那么选择 {Alice, Bob, Carol} 和 {Carol, Alice, Bob} 是完全相同的。
We denote the number of combinations of n objects taken r at a time as C(n, r), nCr, or sometimes using binomial coefficient notation (nr). All these notations represent the same quantity.
我们将从 n 个不同物体中一次选取 r 个的组合数记作 C(n, r)、nCr，有时也使用二项式系数符号 (nr)。所有这些记号都代表同一个数值。
2. Distinction Between Permutations and Combinations | 排列与组合的区别
Permutations count the number of ways to arrange objects in a specific order, while combinations ignore order. In permutations, AB and BA are considered different arrangements, but in combinations they are the same selection. The key question to ask is: 'Does the order matter?' If yes, use permutations. If no, use combinations.
排列计数的是物体按特定顺序排列的方式数，而组合则不考虑顺序。在排列中，AB 和 BA 被视为不同的排列，但在组合中它们是同一个选择。关键是要问：“顺序是否重要？”如果重要，就用排列；如果不重要，就用组合。
For example, when forming a 3-digit code using digits 1 to 5, order matters (123 is different from 321), so we use permutations. When selecting 3 toppings for a pizza from 10 options, order does not matter, so we use combinations.
例如，用数字1到5组成一个三位数密码时，顺序是重要的（123 不同于 321），因此我们使用排列。而从10种可选配料中为比萨挑选3种时，顺序不重要，因此我们使用组合。
3. The Formula for Combinations | 组合数公式
The number of combinations is given by the formula:
组合数的计算公式如下：
C(n, r) = n! / [r! (n − r)!]
where n! (read as 'n factorial') equals n × (n−1) × ... × 2 × 1, and 0! = 1. This formula arises from dividing the number of permutations P(n, r) = n! / (n−r)! by the r! ways of arranging the r selected objects among themselves.
其中 n!（读作“n的阶乘”）等于 n × (n−1) × ... × 2 × 1，且 0! = 1。这个公式来源于：排列数 P(n, r) = n! / (n−r)! 除以所选 r 个物体在内部排列的 r! 种方式。
For instance, C(5, 2) = 5! / (2! × 3!) = (5 × 4 × 3!) / (2 × 1 × 3!) = 20/2 = 10. There are 10 ways to pick 2 items from 5 distinct items.
例如，C(5, 2) = 5! / (2! × 3!) = (5 × 4 × 3!) / (2 × 1 × 3!) = 20/2 = 10。从5个不同物品中选取2个有10种方式。
4. Fundamental Properties of Combinations | 组合的基本性质
Symmetry: C(n, r) = C(n, n − r). Selecting r objects is equivalent to leaving n − r objects behind. For example, C(10, 3) = C(10, 7).
对称性：C(n, r) = C(n, n − r)。选取 r 个物体等同于留下 n − r 个物体。例如，C(10, 3) = C(10, 7)。
Edge cases: n, 0) = 1 (there is exactly one way to choose nothing), and C(n, 1) = n. Also C(n, n) = 1.
边界情形：C(n, 0) = 1（一种都不选的方式恰好有一种），且 C(n, 1) = n。同样 C(n, n) = 1。
Bounds for r: r) = 0; you cannot select a negative number of objects or more objects than are available.
r 的取值范围：若 r < 0 或 r > n，则定义 C(n, r) = 0；你不可能选取负数个物体，也不能选取超过总数的物体。
5. Pascal's Triangle and the Recurrence Relation | 帕斯卡三角形与递推关系
Combination numbers satisfy the recurrence relation known as Pascal's rule:
组合数满足被称为帕斯卡法则的递推关系：
C(n, r) = C(n − 1, r − 1) + C(n − 1, r)
This relationship is visually represented by Pascal's triangle, where each entry is the sum of the two entries directly above it. The 0-th row is 1; the 1st row is 1 1; the 2nd row is 1 2 1; and the 3rd row is 1 3 3 1, corresponding to C(3,0), C(3,1), C(3,2), C(3,3).
这一关系可以通过帕斯卡三角形直观地表示，其中每一项是其正上方两项之和。第0行是1；第1行是1 1；第2行是1 2 1；第3行是1 3 3 1，分别对应 C(3,0), C(3,1), C(3,2), C(3,3)。
Pascal's triangle provides a quick way to compute small combination values without using the factorial formula. The recurrence is also the foundation for many combinatorial proofs.
帕斯卡三角形提供了快速计算较小组合数值的方法，无需使用阶乘公式。这一递推也是许多组合恒等式证明的基础。
6. Binomial Theorem | 二项式定理
The binomial theorem states that for any non-negative integer n:
二项式定理指出，对于任意非负整数 n：
(a + b)n = ∑r=0n C(n, r) an−r br
The coefficients C(n, r) are precisely the binomial coefficients. For example, (x + 2)4 = C(4,0)x4 + C(4,1)x3×2 + C(4,2)x2×22 + C(4,3)x×23 + C(4,4)×24 = x4 + 8x3 + 24x2 + 32x + 16.
其中的系数 C(n, r) 正好是二项式系数。例如，(x + 2)4 = C(4,0)x4 + C(4,1)x3×2 + C(4,2)x2×22 + C(4,3)x×23 + C(4,4)×24 = x4 + 8x3 + 24x2 + 32x + 16。
This connection shows why combinations are fundamental in algebra, probability, and calculus. The sum of all binomial coefficients for a given n is 2n, since (1+1)n = 2n.
这一联系表明了组合在代数、概率和微积分中的基础地位。对于一个给定的 n，所有二项式系数之和为 2n，因为 (1+1)n = 2n。
7. Combinatorial Identities | 组合恒等式
Several elegant identities arise from combinations:
组合能衍生出若干优美的恒等式：
- Sum of row in Pascal's triangle: ∑r=0n C(n, r) = 2n.
- Alternating sum: ∑r=0n (−1)r C(n, r) = 0 for n ≥ 1.
- Hockey-stick identity: ∑r=kn C(r, k) = C(n+1, k+1).
- Vandermonde's identity: ∑k=0r C(m, k) C(n, r−k) = C(m+n, r).
- 帕斯卡三角形的行和：∑r=0n C(n, r) = 2n。
- 交错和：对于 n ≥ 1，∑r=0n (−1)r C(n, r) = 0。
- 冰球棍恒等式：∑r=kn C(r, k) = C(n+1, k+1)。
- 范德蒙恒等式：∑k=0r C(m, k) C(n, r−k) = C(m+n, r)。
These identities are not only useful in simplifying algebraic expressions but also appear in probability theory and statistical mechanics. Knowing how to apply them saves time in examinations.
这些恒等式不仅在简化代数表达式时有用，而且出现在概率论和统计力学中。知道如何运用它们可以在考试中节省时间。
8. Combinations with Restrictions | 带限制条件的组合
IB questions often introduce constraints, such as 'at least one', 'at most', or 'specific items must be included or excluded'. A sound approach is to break the problem into cases or use complementary counting.
IB试题常常引入限制条件，比如“至少一个”、“至多”、“特定项必须包含或排除”。一种可靠的方法是分情况讨论或使用补集计数。
Example: From a group of 6 girls and 5 boys, how many ways can we form a committee of 4 that includes at least 2 girls? We can add combinations with exactly 2 girls, exactly 3 girls, exactly 4 girls: C(6,2)×C(5,2) + C(6,3)×C(5,1) + C(6,4)×C(5,0).
例题：从6名女生和5名男生中选取一个4人委员会，要求至少包含2名女生，共有多少种选法？我们可以将恰好2名女生、恰好3名女生、恰好4名女生的组合数相加：C(6,2)×C(5,2) + C(6,3)×C(5,1) + C(6,4)×C(5,0)。
Alternatively, total unrestricted committees minus those with 0 or 1 girl gives the same result. This complementary method often reduces calculation errors.
另一种方法是，无限制委员会总数减去只有0或1名女生的委员会数，也能得到相同结果。这种补集法常常能减少计算错误。
9. Applications in Probability | 概率中的应用
In probability, combinations are used to count equally likely outcomes. If an experiment involves selecting a sample of size r from a population of size n at random, the probability of a specific event is the ratio of favorable combinations to total combinations.
在概率中，组合被用来计数等可能的结果。如果一个试验是从大小为 n 的总体中随机抽取容量为 r 的样本，那么某一特定事件的概率就是有利组合数与总组合数之比。
) × C(6,1)] / C(10,3). Combinations ensure each subset of 3 balls is equally likely.
例如，一个袋子里有4个红球和6个蓝球。如果不放回地随机抽出3个球，那么恰好2个红球的概率 P = [C(4,2) × C(6,1)] / C(10,3)。组合保证了每3个球组成的子集等可能被抽到。
The hypergeometric distribution and the binomial distribution both rely heavily on combination formulas. Mastering combinations therefore directly supports understanding probability distributions in the IB curriculum.
超几何分布和二项分布都高度依赖组合公式。因此，掌握组合能直接帮助理解IB课程中的概率分布。
10. Common Mistakes and Tips | 常见错误与提示
Mistake 1: Using permutations instead of combinations when order does not matter. Always ask: if I swap two chosen items, does the outcome change? If not, use combinations.
错误1：在顺序不重要时错误地使用排列。始终要问：如果交换两个已选物品，结果会改变吗？如果不会，就使用组合。
Mistake 2: Forgetting that combinations are only valid for distinct objects. If items are identical, you need a different approach such as stars and bars.
错误2：忘记组合仅适用于不同物体。如果物品是相同的，就需要使用不同的方法，比如星棒模型。
Mistake 3: Misapplying restrictions. When counting with 'at least', check whether cases overlap and always consider using the complement: 'at least one' can be found via 1 − P(none).
错误3：错误应用限制条件。当计数“至少”时，要检查不同情况是否有重叠，并始终考虑使用补集：“至少一个”可以通过 1 − P(零) 来求得。
Practice writing out cases explicitly before using formulas, and visualise selections with tree diagrams or Venn diagrams when learning the concept.
在使用公式之前，先练习明确列出各种情况，并在学习概念时借助树状图或维恩图来可视化选择过程。
11. Worked Examples | 例题解析
Example 1: A student must answer 5 out of 8 questions on an exam, but must choose at least 3 from the first 4 questions. How many ways can the selection be made?
例1：一个学生必须在考试中从8道题里选5道回答，但至少要从前面4题中选3道。共有多少种选择方式？
. Total = C(4,3)×C(4,2) + C(4,4)×C(4,1) = 4×6 + 1×4 = 24 + 4 = 28.
我们分两种情况：(i) 从前面4题选3道，再从后面4题选2道；(ii) 前面4题全选，再从后面4题选1道。总数为 C(4,3)×C(4,2) + C(4,4)×C(4,1) = 4×6 + 1×4 = 24 + 4 = 28。
Example 2: How many different hands of 5 cards from a standard deck contain exactly 2 aces?
例2：从一副标准扑克牌中抽出5张牌，恰好含有2张A的牌型有多少种？
ng 3 cards from the 48 non-aces: C(48,3)=17296. Multiply: 6 × 17296 = 103,776 hands.
从4张A中选2张：C(4,2)=6。从剩下的48张非A牌中选3张：C(48,3)=17296。相乘：6 × 17296 = 103,776 种牌型。
12. Summary | 总结
ombinations are used when the order of selection is not important. The formula C(n, r) = n! / [r!(n−r)!] sits at the core of counting theory. Properties such as symmetry, Pascal' lem calls for combinations, carefully handling restrictions, and connecting combinations to probability and algebraic expansion.
当选取顺序不重要时，就使用组合。公式 C(n, r) = n! / [r!(n−r)!] 是计数理论的核心。对称性、帕斯卡递推以及与二项式定理的联系使组合成为一种多功能的工具。在IB考试中，成功的关键在于识别题目何时需要使用组合、仔细处理限制条件，并将组合与概率、代数展开紧密相连。
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