排列组合公式

在数学中，排列与将一方的所有合作伙伴收集成某种序列或格式的过程有关。换句话说，如果党已经被执行，那么其成员的重定向称为置换过程。几乎每一个数学社区都以或多或少的重要方式发生排列。当观察到对特定有限区域的不同管理时，它们通常会发生。

排列

它是对所提供的多个组件的不同解释，一次一个，或一些，或所有。例如，如果我们有两个组件 A 和 B，那么就有两种可能的表现，AB 和 BA。

当“r”个组件位于总共“n”个组件中时，排列的数字是ⁿ P _r = n！ /（n - r）！ .例如，让 n = 3（A、B 和 C）和 r = 2（所有大小为 2 的排列）。答案是 3！/(3 – 2)！ = 6. 六个排列是 AB、AC、BA、BC、CA 和 CB。

置换公式的解释

A permutation is a type of performance that indicates how to permute. If there are three different numerals 1, 2 and 3 and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be accomplished in 6 methods. 

Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways. 

In general, n distinct things can be set taking r (r < n) at a time in n(n – 1)(n – 2)…(n – r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n – 1 things. Likewise, the third thing can be any of the remaining n – 2 things. Alike, the r^th thing can be any of the remaining n – (r – 1) things. 

Hence, the entire number of permutations of n distinct things carrying r at a time is n(n – 1)(n – 2)…[n – (r – 1)] which is written as ⁿ P_r. Or, in other words, 

ⁿ P_r = n!/(n – r)!

组合

它是共享数量的组件的不同部分，一次一个，或一些，或全部携带。例如，如果有两个组件 A 和 B，那么只有一种方法可以选择两个事物，同时选择它们。

从总共“n”个组件中选择“r”个组件时的组合数是， ⁿ C _r = n！ / [(r!) x (n – r)! ]。例如，让 n = 3（A、B 和 C）和 r = 2（大小为 2 的所有组合）。答案是 3!/((3 – 2)! × 2!) = 3。六种组合是 AB、AC 和 BC。

ⁿ C _r = ⁿ C _{(n – r)}

Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items but for selecting, AB and BA are the same.

组合公式说明

Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3). 

Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as ³C₂. In general, the number of combinations of n distinct things taken r at a time is, 

ⁿ C_r = n! /[r! × (n – r)!] = ⁿP_r/r!

示例问题

问题 1：找出 n = 9 和 r = 3 的排列和组合的数量。

解决方案：

Given,

n = 9

r = 3

Using the formula given above:

Permutation:

ⁿP_r = (n!) / (n – r)! 

= (9!) / (9 – 3)! 

= 9! / 6! = (9 × 8 × 7 × 6! )/ 6! 

= 504

Combination:

ⁿC_r = n!/r!(n − r)!

= 9!/3!(9 − 3)!

= 9!/3!(6)!

= 9 × 8 × 7 × 6!/3!(6)!

= 84

问题 2：一个由 4 男 2 女组成的委员会有多少种方式可以从 6 男 5 女中选出？

解决方案：

Choose 4 men out of 6 men = ⁶C₄ ways = 15 ways

Choose 2 women out of 5 women = ⁵C₂ ways = 10 ways

The committee can be chosen in ⁶C₄ × ⁵C₂  = 150 ways.

问题 3：使用“LOVE”一词中的 2 个字母可以创建多少相当大的单词？

解决方案：

The term “LOVE” has 4 distinct letters.

Therefore, required number of words = ⁴P₂ = 4! / (4 – 2)!

Required number of words = 4! / 2! = 24 / 2 = 12

问题4：5个辅音3个元音，3个辅音2个元音能组成多少个词？

解决方案：

Number of ways of choosing 3 consonants from 5.

= ⁵C₃

Number of ways of choosing 2 vowels from 3.

= ³C₂

Number of ways of choosing 3 consonants from 2 and 2 vowels from 3.

= ⁵C₃ × ³C₂

= 10 × 3

= 30

It means we can have 30 groups where each group contains a total of 5 letters (3 consonants and 2 vowels).

Number of ways of arranging 5 letters among themselves

= 5! = 5 × 4 × 3 × 2 × 1 = 120

Hence, the required number of ways

= 30 × 120

= 3600

问题 5：如果你有 5 个项目并选择 4 个，你会得到多少种不同的组合？

解决方案：

Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (5 in this example); “r” is the number of items you’re choosing (4 in this example):

C(n, r) = n! / r! (n – r)! 

= 5! / 4! (5 – 4)!

= (5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 1)

= 120/24 

= 5

The solution is 5.

问题 6：6 个辅音和 3 个元音， 2 个辅音和 1 个元音的表达式可以创建多少个？

解决方案：

Number of ways of selecting 2 consonants from 6.

= ⁶C₂

Number of ways of selecting 1 vowels from 3.

= ³C₁

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4.

= ⁶C₂ × ³C₁

= 15 × 3

= 45

It means we can have 45 groups where each group contains a total of 3 letters (2 consonants and 1 vowels).

Number of ways of arranging 3 letters among themselves.

= 3! = 3 × 2 × 1

= 6

Hence, the required number of ways.

= 45 × 6

= 270

问题 7： “PHONE”一词的字母可以有多少种不同的形式，以使元音一致地出现？

解决方案：

The word ‘PHONE’ has 5 letters. It has the vowels ‘O’,’ E’, in it and these 2 vowels should consistently come jointly. Thus these two vowels can be grouped and viewed as a single letter. That is, PHN(OE).

Therefore we can take total letters like 4 and all these letters are distinct.

Number of methods to organize these letters.

= 4! = 4 × 3 × 2 × 1

= 24

All the 2 vowels (OE) are distinct.

Number of ways to arrange these vowels among themselves.

= 2! = 2 × 1

= 2

Hence, the required number of ways.

= 24 × 2

= 48.