8C Answers: Combinations, Pascal’s Triangle and Binomial Expansion | 8C 答案:组合、帕斯卡三角与二项式展开

📚 8C Answers: Combinations, Pascal’s Triangle and Binomial Expansion | 8C 答案:组合、帕斯卡三角与二项式展开

In mathematics, the notation 8C_r (read as ‘8 choose r’) gives the number of ways to select r items from a set of 8 distinct items, where the order of selection does not matter. These values appear everywhere in combinatorics, probability and the binomial expansion of (a + b)⁸. This article provides a complete set of 8C answers, explains how they are calculated, and shows how to use them confidently in exam questions.

在数学中,记号 8C_r(读作“8 选 r”)表示从 8 个不同对象中选出 r 个对象的方法数,并且选择顺序不重要。这些数值在组合数学、概率以及 (a + b)⁸ 的二项式展开中随处可见。本文给出完整的 8C 答案,解释它们的计算方法,并展示如何在考试题目中自信地使用这些答案。


1. What Does 8C Mean? | 8C 表示什么?

The symbol 8C_r is a binomial coefficient. It counts the number of unordered subsets of size r that can be formed from a set of 8 objects. For example, 8C₃ tells you how many different groups of 3 can be chosen from 8 people.

符号 8C_r 是一个二项式系数。它计算的是从 8 个对象中能组成多少个大小为 r 的无序子集。例如,8C₃ 告诉你从 8 个人中可以选出多少个不同的 3 人小组。

Because the order does not matter, choosing {A, B, C} is the same as choosing {C, A, B}. This is different from permutations, where order is counted separately.

因为顺序不重要,选择 {A, B, C} 与选择 {C, A, B} 是相同的。这与排列不同,排列会单独计算每一种顺序。


2. The Combination Formula for 8C_r | 8C_r 的组合公式

For any integer r from 0 to 8, the value of 8C_r is given by the factorial formula:

对于 0 到 8 之间的任意整数 r,8C_r 的值由阶乘公式给出:

8C_r = 8! / [r! × (8 – r)!]

Here, n! means n factorial, the product of all positive integers from 1 to n. For example, 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320.

这里 n! 表示 n 的阶乘,即从 1 到 n 的所有正整数的乘积。例如,8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320。

Using this formula, you can calculate any 8C_r value exactly. It also shows why 8C₀ = 1 and 8C₈ = 1, because 0! is defined as 1.

使用这个公式,你可以精确计算任意 8C_r 的值。它也能解释为什么 8C₀ = 1 且 8C₈ = 1,因为 0! 被定义为 1。


3. Full List of 8C_r Values | 8C_r 完整数值表

The complete table of 8C_r answers is shown below. These values are symmetric and appear as a row in Pascal’s triangle.

下面是 8C_r 的完整答案表。这些数值具有对称性,并出现在帕斯卡三角的一行中。

  • 8C₀ = 1
  • 8C₁ = 8
  • 8C₂ = 28
  • 8C₃ = 56
  • 8C₄ = 70
  • 8C₅ = 56
  • 8C₆ = 28
  • 8C₇ = 8
  • 8C₈ = 1

Notice that the largest value is 8C₄ = 70. This makes sense: there are more ways to choose exactly half of the objects than to choose very few or very many.

请注意,最大值是 8C₄ = 70。这是合理的:恰好选择一半对象的方法数,比选择很少或很多对象的方法数要多。


4. Symmetry Property of 8C_r | 8C_r 的对称性

One important property is symmetry: 8C_r = 8C_{8-r}. This means 8C₁ = 8C₇, 8C₂ = 8C₆, and 8C₃ = 8C₅.

一个重要的性质是对称性:8C_r = 8C_{8-r}。这意味着 8C₁ = 8C₇,8C₂ = 8C₆,8C₃ = 8C₅。

This occurs because choosing r objects to include is equivalent to choosing 8 – r objects to exclude. You can use this property to save time in calculations.

这是因为选出 r 个对象包含在内,等价于选出 8 – r 个对象排除在外。你可以利用这个性质节省计算时间。


5. Pascal’s Triangle Row 8 | 帕斯卡三角第 8 行

The values 1, 8, 28, 56, 70, 56, 28, 8, 1 form the 8th row of Pascal’s triangle. Each number is the sum of the two numbers directly above it.

数值 1、8、28、56、70、56、28、8、1 构成帕斯卡三角的第 8 行。每个数字都是其正上方两个数字之和。

Pascal’s triangle provides a quick way to read off 8C answers without using the formula repeatedly. It also links directly to binomial coefficients in expansion.

帕斯卡三角提供了一种快速读取 8C 答案的方法,而不需要反复使用公式。它还与二项式展开中的系数直接相关。


6. Using 8C_r in Probability | 在概率中使用 8C_r

In probability, 8C_r often appears as the number of favourable outcomes or total outcomes. For example, if you flip a fair coin 8 times, the number of sequences with exactly 3 heads is 8C₃ = 56.

在概率中,8C_r 经常作为有利结果数或总结果数出现。例如,如果你抛一枚公平硬币 8 次,恰好出现 3 次正面的序列数是 8C₃ = 56。

The total number of possible outcomes is 2⁸ = 256, so the probability of exactly 3 heads is 56/256 = 7/32.

总可能结果数为 2⁸ = 256,因此恰好出现 3 次正面的概率为 56/256 = 7/32。


7. Binomial Expansion of (a + b)⁸ | (a + b)⁸ 的二项式展开

The binomial theorem states that (a + b)⁸ expands using 8C_r as the coefficients. The full expansion is:

二项式定理表明 (a + b)⁸ 的展开使用 8C_r 作为系数。完整展开式为:

(a + b)⁸ = a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸

Each term has the form 8C_r × a^(8-r) × b^r. The powers of a decrease from 8 to 0 while the powers of b increase from 0 to 8.

每一项的形式为 8C_r × a^(8-r) × b^r。a 的幂从 8 递减到 0,而 b 的幂从 0 递增到 8。


8. Finding a Specific Coefficient | 求特定项的系数

To find a specific term without expanding the whole bracket, use the general term formula:

要在不展开整个括号的情况下找到特定项,可使用通项公式:

T_{r+1} = 8C_r × a^(8-r) × b^r

For example, the coefficient of a⁵b³ in (a + b)⁸ is 8C₃ = 56. The coefficient of a²b⁶ is 8C₆ = 28.

例如,(a + b)⁸ 中 a⁵b³ 的系数是 8C₃ = 56。a²b⁶ 的系数是 8C₆ = 28。

This method is much faster than expanding the whole expression, especially in exam settings.

这种方法比展开整个表达式快得多,尤其是在考试情境中。


9. Common Mistakes with 8C Answers | 8C 答案常见错误

One common mistake is confusing combinations with permutations. Remember that 8C_r does not count different orders, while 8P_r does.

一个常见错误是混淆组合与排列。请记住,8C_r 不计算不同顺序,而 8P_r 会计算不同顺序。

Another mistake is forgetting that 0! = 1, which leads to incorrect values for 8C₀ and 8C₈. Also, check that your r value is never negative or greater than 8.

另一个错误是忘记 0! = 1,这会导致 8C₀ 和 8C₈ 的值计算错误。此外,还要检查 r 值不为负数且不超过 8。


10. Worked Example: Choosing a Committee | 例题:选择委员会

A committee of 4 people is to be chosen from 8 candidates. How many different committees are possible?

一个由 4 人组成的委员会需要从 8 名候选人中选出。共有多少种不同的委员会?

Number of committees = 8C₄ = 8! / (4! × 4!) = 40320 / (24 × 24) = 70

So there are 70 possible committees. If the committee had a president, secretary, treasurer and member, order matters, so you would use 8P₄ = 1680 instead.

因此有 70 种可能的委员会。如果委员会有主席、秘书、财务和成员等职位,顺序就有影响,此时应改用 8P₄ = 1680。


11. Exam-Style Practice Questions | 考试型练习题

Try these quick questions to test your understanding of 8C answers:

尝试以下快速练习题,检验你对 8C 答案的理解:

  • Evaluate 8C₂ + 8C₃. | 计算 8C₂ + 8C₃。
  • Find the coefficient of x³y⁵ in (x + y)⁸. | 求 (x + y)⁸ 中 x³y⁵ 的系数。
  • A bag has 8 marbles. How many ways can you choose 5 marbles? | 一个袋子里有 8 个弹珠。选出 5 个弹珠有多少种方法?
  • What is 8C₅ – 8C₃? | 8C₅ – 8C₃ 等于多少?

Check your answers: 8C₂ + 8C₃ = 28 + 56 = 84; the coefficient is 8C₅ = 56; choosing 5 marbles gives 8C₅ = 56; and 8C₅ – 8C₃ = 56 – 56 = 0.

核对答案:8C₂ + 8C₃ = 28 + 56 = 84;系数为 8C₅ = 56;选出 5 个弹珠有 8C₅ = 56 种方法;8C₅ – 8C₃ = 56 – 56 = 0。


12. Quick Reference Summary | 快速参考总结

The nine 8C answers are 1, 8, 28, 56, 70, 56, 28, 8, 1. They form Pascal’s triangle row 8 and serve as coefficients in (a + b)⁸.

九个 8C 答案分别是 1、8、28、56、70、56、28、8、1。它们构成帕斯卡三角第 8 行,并作为 (a + b)⁸ 中的系数。

Use the formula 8C_r = 8! / [r!(8-r)!] when you need an exact value, and remember the symmetry property 8C_r = 8C_{8-r} to speed up calculations.

当你需要精确值时,使用公式 8C_r = 8! / [r!(8-r)!],并记住对称性 8C_r = 8C_{8-r} 以加快计算速度。

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