置换公式

在数学中，排列涉及将一个组的所有成员组织成某个系列或设计的方法。更进一步来说，如果组已经完成，则其组件的重定向称为置换方法。几乎每个数学领域都以更好或稍微有效的方法进行排列。它们通常发生在监控详细受限站点上的不同方向时。

排列公式

它是一次一个、一个或几个或所有所提供的关联数字的单独排列。例如，如果我们有两个元素 A 和 B，那么就有两种可能的解释，AB 和 BA。

当“r”个元素排列在总共“n”个元素中时的整数排列是，

ⁿ P _r = n！ /（n - r）！。

例如，

Let n = 2 (A and B) and r = 1 (All permutations of size 1). The answer is 2!/(2 – 1)! = 2. The two permutations are AB, and BA.

置换公式的解释

A permutation is a kind of arrangement that shows how to permute. If there are three separate integers 1, 2, and 3, and if somebody is interested to permute the integers taking 2 at a point, it offers (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be performed in 6 ways.

Here, (1, 2) and (2, 1) are separate. Again, if these 3 integers shall be set enduring all at a time, then the arrangements 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 known, n separate items can be selected accepting r (r < n) at a time in n(n – 1)(n – 2) … (n – r + 1) ways. In particular, the first item can be any of the n items. Now, after selecting the first item, the second item will be any of the remaining n – 1 thing. Similarly, the third item can be any of the remaining n – 2 things. Alike, the r^th item can be any of the remaining n – (r – 1) things.

Therefore, the total numeral of permutations of n separate items taking r at a time is n(n – 1)(n – 2) … [n – (r – 1)] which is noted as ⁿP_r. Or, in other words,

ⁿP_r = n!/(n – r)!

示例问题

问题1：排列的类型有哪些？

解决方案：

The permutation of a collection of things or components in order relies on three conditions:

1. When recurrence of essences is not allowed
2. When recurrence of essences is allowed
3. When the components of a group are not different

问题 2：计算 n = 5 和 r = 2 的排列数。

解决方案：

Given,

n = 5

r = 2

Using the formula given above:

Permutation: ⁿP_r = (n!) / (n – r)!

= (5!) / (5 – 2)!

= 5! / 3! = (5 × 4 × 3! )/ 3!

= 20

问题 3：在不允许重复字母的情况下，可以从 POEM 单词的字母中创建多少个有目的或无目的的 3 字母短语？

解决方案：

Here n = 4, as the word POEM has 4 letters. Since we have to create 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

⇒ P(n, r) = 4!/(4 − 3)!

= 4 × 3 × 2 × 1/1

= 24

问题 4：当允许单词重复时，可以从单词 KANHA 的字母中创建多少个有目的或无目的的 4 字母短语？

解决方案：

The number of letters, in this case, is 5, as the word KANHA has 5 alphabets.

And r = 4, as a 4-letter term has to be selected.

Thus, the permutation will be:

Permutation (when repetition is permitted) = 5⁴= 625

问题5：要求4男3女排成一排，让女方排位。有多少这样的配置是可行的？

解决方案：

We are given that there are 4 men and 3 women.

i.e. there are 7 positions.

The even positions are: 2nd, 4th, and the 6th places

These three places can be occupied by 3 women in P(3, 3) ways = 3!  

= 3 × 2 × 1  

= 6 ways

The remaining 4 positions can be occupied by 4 men in P(4, 4) = 4!  

= 4 × 3 × 2 × 1  

= 24 ways

Therefore, by the Fundamental Counting Principle,  

Total number of ways of seating arrangements = 24 ×  6 

= 144

问题 6：找出可以由单词“TABLE”的字母组成的短语的数量，无论是否有意义。

解决方案：

‘TABLE’ contains 5 letters.

Thus, the numeral of phrases that can be formed with these 5 letters = 5! = 5 × 4 × 3 × 2 × 1 = 120.

问题 7：找出短语主语字母的排列数，使得元音始终出现在奇数位置。

解决方案：

The word ‘SUBJECT’ has 7 letters.

There are 6 consonants and 1 vowels in it.

No. of ways 1 vowels can occur in 7 different places = ⁷P₁ = 7 ways.

After 1 vowels take 1 place, no. of ways 6 consonants can take 6 places = ⁶P₆ = 6! = 720 ways.

Therefore, total number of permutations possible = 720 × 720 = 518,400 ways.