8个人组成4个人委员会有几种方式

“组合”一词被随意使用，而且通常以错误的方式出现。比如，“嘿，手提箱锁密码是什么？”有人说但真正应该说的是“嘿，手提箱锁排列是什么？”那么有什么区别呢？究竟什么是排列组合？它们是两个非常不同的术语。让我们详细了解它们，

排列

排列是根据给定条件以特定顺序排列对象或给定数量的行为，可能是一组对象或集合中的数字。示例 – 有多少 2 个字母的单词可以使用 LATE 一词中的字母组成？答案是⁴ P ₂ （发音为 4 p 2）= 4！/（4 - 2）！ = 4!/(2)! = (4 × 3 × 2 × 1)/(2 × 1) = 24/2 = 12。

组合

组合是从一组对象或集合中选择对象或给定数量的方式，或者从一组对象或集合中选择集合，这样对象的顺序无关紧要。例如——从4个人中可以选出多少个2人一组？答案 - ⁴ C ₂ （发音为 4 C 2）= 4！/（4 - 2）！（2）！ = 4!/(2)!(2)! = (4 × 3 × 2 × 1)/(2 × 1)(2 × 1) = 24/4 = 6。

排列组合公式

• The formula for permutations is: ⁿP_r = n!/(n – r)!
• The formula for combinations is: ⁿC_r = n!/[r! (n – r)!]

找出由 8 人组成的 4 人委员会的数量。

回答：

Here, analyzing the problem given, the hint is to select 4 people from 8 people. The biggest confusion is whether to apply permutation or combination? So start thinking way that whether the order of people will make a difference? if yes then go for permutation otherwise it’s a combination that will solve the question.

In this question let’s first select p₁, p₂, p₃, p₄ in the written order and then select p₂, p₁, p₄, p₃. Does that make a different group here? The answer is no. Because irrespective of order the first, second, third, fourth person have been selected. So here, go with the combination as the order doesn’t matter.

Applying combination using formula, ⁿC_r = n!/[r! (n – r)!]

Here, n = 8 and r = 4

So ⁸C₄ = 8!/(8 – 4)!(4)! = 8!/(4)!(4)! = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1)(4 × 3 × 2 × 1) = 70

类似问题

问题 1：找出由 5 人组成的 3 人委员会的数量。

回答：

⁵C₃ =  5!/(5 – 3)!(3)! = 5!/(2)!(3)! = (5 × 4 × 3 × 2 × 1) / (2 × 1)(3 × 2 × 1) = 10

问题 2：可以从一组 5 人中选出多少个由 3 名成员组成的不同委员会，以便始终选择一个特定的人？

回答：

Since one particular person is always to be taken from the available 5 people in the committee of the 3. So in fact, choose 2 persons from the remaining 4 and that can be done in C(4, 2) = ⁴C₂ = 4!/2! 2! = 6 number of ways.

问题 3：8 男 6 女可以组成多少个 5 人委员会，由 3 男 2 女组成？

回答：

Well, one can form 8 choose 3 groups of men, and for each of those, one can choose any of the 6 choose 2 groups of women.

ⁿC_r = n!/[r! (n – r)!] 

= ⁶C₂ = (6)!/((2!)(6 – 2)!) = 15 

= ⁸C₃ = (8)!/((3!)(8 – 3)!) = 56

So, 56 × 15 = 840 possible combinations, assuming one doesn’t care about anything other than the number of men, the number of women.

If one of those people is to be the chair, then there are for each possible group, 5 possible chairs, so multiply that by 5, 840 × 5 = 4200