排列

排列是一组中对象的集合或组合，其中所选对象的顺序或排列确实很重要。换句话说，排列是对象按确定顺序的排列。因此，在深入研究置换之前，让我们先对阶乘进行简要讨论。

阶乘

• 自然数n的阶乘由符号n！表示。
• n！是从1到n(包括1和n)的所有自然数的乘积。
  即n×(n-1)×(n-2)×(n-3)。 。 。 ×1

例子：

If n = 3
3! = 3 × 2 × 1 = 6

If n = 5
5! = 5 × 4 × 3 × 2 × 1 = 120

If n = 1
1! = 1

Note: The factorial of 0 is defined as 1 by convention i.e. 0! = 1

为什么要使用阶乘?

阶乘的主要用途是计算排列的数量(排列某些对象的方式的数量)。让我们通过示例来理解这一点。

例子

问题1.一个班级只有3个职位空缺。三个人P，A和R同时到达。在这3个空位上，P，A和R可以通过几种方式排列?

解决方案：

For the very first seat, we have 3 choices i.e. P, A and R.
Let us randomly select A for the first seat.

For the second seat, we have 2 choices i.e. P and R
Let us randomly select R for the second seat.

For the third seat, we have 1 choice i.e. P

To summarize, we did the following:
Placed a person on seat 1 and placed a person on seat 2 and placed a person on seat 3.
 

Usage of and comes from the fact that occupation of all 3 seats was mandatory.

In mathematics, and is related with multiplication, hence we can say that total choices = 3 × 2 × 1 = 3!

If we change the seating order to P on the first seat, A on the second seat, and R on the third, does that change the total number of choices?

No, it does not. This is because equal importance is given to all three P, A, and R.

问题2：如果其中2个人始终坐在一起，可以找到安排5个人的方式数量?

解决方案：

Let us consider the 2 people as a unit and the remaining 3
person as 3 separate units, So we have total 4 units.

The number of ways of arranging these 4 units is 4!
(just the way we proved in previous problem)
The number of ways of arranging the 2 person amongst themselves is 2!

In conclusion, the number of ways of arranging the 4 units and 2 person amongst themselves is 4! × 2!

问题3.找到以元音开头和结尾的所有三个字母的单词。鉴于不允许重复字母。

解决方案：

Total vowels in english = 7 ( a, e, i, o, u, y, w)
Total consonants in english = 26 – 7 = 19

1.The choices for the first letter are 7
2.The choices for the third letter are 6
(since 1 vowel was placed as first letter)
3.The choices for the middle letter are 19 + (7 – 2) = 24
(19 consonants + the vowels which were not placed)

Hence total permutations are 7 × 6 × 24 = 1008

Do observe that here we first satisfied the vowel condition
for first and third letter and then placed the middle letter.

排列

• 如果有n个对象可用并且我们安排了所有对象，则每种可能的安排都称为置换。
• 如果在n个可用对象中，我们选择r并对其进行排列。然后，每种可能的排列都称为r –排列。
• 在排列中，对象的顺序很重要。

在置换中，我们主要处理四种问题

1. 重复排列
2. 排列无重复
3. r –无需重复的排列
4. 重复排列

重复排列

这是最简单的。在此类问题中，对象可能会重复。让我们通过一些例子来理解这些问题。

例子

问题1.使用3、4、5和7可以形成多少个大于500的3位数字?

解决方案：

Since a three-digit number, greater than 500 will have either 5 or 7 at its hundredth place, we have 2 choices for this place.

There is no restriction on repetition of the digits, hence for the remaining 2 digits we have 4 choices each

So the total permutations are: 2 × 4 × 4 = 32

问题2.使用数字1、2、4、5和9可以形成1000到2000之间的偶数个数?

解决方案：

Since the number is supposed to be even, the digits at units place must either be 2 or 4 leaving us with 2 choices for the digit at units place.

The number is supposed to lie between 1000 and 2000, So the digits at thousand’s place must be 1, we thus have 
1 choice for the digit at thousands place.

Rest of the 2 digits can be any one from 1, 2, 4, 5 and 9 i.e. 5 choices each

So the total permutations are: 2 × 5 × 5 × 1 = 50

排列无重复

在此类问题中，不允许重复对象。让我们通过一些例子来理解这些问题。

例子

问题1.用数字2、4、6和8可以无重复地形成3个可被3整除的3位数?

解决方案：

For a number to be divisible by 3, the sum of it digits must be divisible by 3

From the given set, various arrangements like 444 can be formed but since repetition isn’t allowed we won’t be considering them.

We are left with just 2 cases i.e. 2, 4, 6 and 4, 6, 8

The number of arrangements are 3! in each case

Hence the total number of permutations are: 3! + 3! = 12

问题2.如果不允许重复数字，可以使用0、3、5、7和9形成多少个可被5整除的4位数字?

解决方案：

For the number to be divisible by 5, the digit at units place must either be 0 or 5, hence we have 2 possibilities.

Case 1. The digit at units place is 0
There are 4 choices for 10³ place (all numbers except 0)
There are 3 choices for the 10² place (1 got used up at 10³ place)
There are 2 choices for the 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 0 at units place are
4 × 3 × 2 = 24

Case 2. The digit at units place is 5
There are 3 choices for 10³ place (all except 0 and 5)
There are 2 choices for 10² place (1 got used up at 10³ place)
There is 1 choice for 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 5 at units place are 3 × 2 × 1 = 6

Total arrangements = Number of arrangements in case 1 + Number of arrangements in case 2
                                = 24 + 6 = 30

r –无重复排列

这是当我们仅在n个对象中不重复地排列r个对象时。让我们通过一个例子来理解这一点。

例子。一家冰淇淋店有10种口味的冰淇淋。找到制备具有3种不同口味的冰淇淋蛋筒的方法有多少种?

解决方案：

Let us consider n = 10 (total number of flavors) and r = 3 (number of different flavors needed)

For first flavor we have 10 choices
For second flavor we have 10 – 1 choices
For third flavor we have 10 – 2 choices and this is same as (n – r + 1)

The numbers of arrangement would be: 10 × (10 – 1) × (10 – 3 + 1) = 720

From this we can generalize that, the number of ways of arranging r objects out of n different objects is:
n × (n – 1)  . . . (n – r + 1) = ⁿP_r

置换公式( ⁿ P _r )

From the previous example, we understood that_r
ⁿP_r = n × (n – 1)  . . .  (n – r + 1) 

Multiplying and Dividing by (n – r)!
ⁿP_r = n × (n – 1) × (n – 2) × . . . × (n – r + 1) × (n – r)! / (n – r)!

ⁿP_r = n × (n – 1) × (n – 2) × . . . × (n – r + 1) × (n – r) × (n – r – 1) × . . . × 1 / (n – r)!

ⁿP_r = n! / (n – r)! where  0 ≤ r ≤ n 

问题1.找到⁶ P ₃ ?

解决方案：

As per the above formula:
⁶P₃ = 6! / (3!) = 6 × 5 × 4 = 120

问题2。10名奥运会选手参加比赛。查找1^日，2^日和3^第三名可能的不同安排?

解决方案：

We have to find different arrangements of 10 taken 3 at time.
Here n = 10
         r = 3

¹⁰P₃ = 10! / (7!) = 10 × 9 × 8 = 720

问题3.^如果n P ₂ = 12，求n?

解决方案：

ⁿP_r = n! / (n – r)!
ⁿP₂ = n! / (n – 2)!
      = n × (n – 1) × (n – 2)! / (n – 2)!
      = n × (n – 1)
      = n² – n

∴ n² – n = 12

Solving the equation,
n² – n – 12 = 0
n (n – 4) + 3 (n – 4) = 0
(n + 3) (n – 4) = 0
∴ n = -3 or n = 4

∵ n ≥ 0, n = 4

r –重复排列

可以认为这是将n个对象分布到r个盒子中，其中允许重复对象，并且任何盒子都可以容纳任意数量的对象。

The 1^st  box can hold n objects
The 2^nd box can hold n objects
The 3^rd box can hold n objects
.                                          . 
.                                          . 
.                                          . 
The r^th box can hold n objects

Hence total number of arrangements are:
n × n × n . . . (r times)
= n^r

问题1.一名警官每周3次访问犯罪现场进行调查。如果每天的访问次数没有限制，请问如何安排他的访问时间?

解决方案：

The number of ways to schedule first visit is 7 (any of the 7 days)
The number of ways to schedule second visit is 7 (any of the 7 days)
The number of ways to schedule third visit is 7 (any of the 7 days)

Hence, the number of ways to schedule first and second and third visit is 7 × 7 × 7 = 7³ = 343

问题2.如果一个囚室中可以容纳任意数量的囚犯，可以将多少名囚犯放在4个牢房中?

解决方案：

The 1^st prisoner can be sent to any of the 4 cells
The 2^nd prisoner can be sent to any of the 4 cells
.                                                                   .
.                                                                   .
.                                                                   .
The 6^th prisoner can be sent to any of the 4 cells

The total arrangements are:
4 × 4 × 4 . . . (6 times) = 4⁶

排列组合关系

Permutation	Combination
A permutation is a way of arranging some objects.	A Combination is a way of selecting objects.
In permutation the order matters i.e. ABC and BCA are counted separately	In Combination, the order does not matter i.e. ABC and BCA are counted as the same.
The permutation of n objects taken r at a time is denoted by nPr	The combination of n objects taken r at a time is denoted by nCr

ⁿ P _r和ⁿ C _r之间的关系

We can understand ⁿC_r through the following analogy

Consider that we have n distinct boxes and r identical
balls. (n > r)

The task is to place all the r balls into boxes such that no
box contains more than 1 ball.

Had the balls been distinct, this was a problem of 
r – permutation without repetition and then the 
answer was ⁿP_r as discussed earlier.

But since all r objects are same, the r! ways of arranging them
can be considered as a single way.

To group all r! ways of arranging, we divide ⁿP_r by r!

ⁿC_r = $_{}^{n}\textrm{P}_{r} \over r!$ $=\frac{n!}{(n - r)! \times r!}$    

^因此，n P _r和ⁿ C _r之间的关系为：

ⁿ C _r = $_{}^{n}\textrm{P}_{r} \over r!$

例子

问题1.如果不允许重复字母，则可以在单词“ SATURDAY”中形成多少个4个字母的含意或无意的单词?

解决方案：

The word SATURDAY has 8 letters i.e. S, A, T, U, R, D, A, and Y

To form 4-letter words, we first have to select 4 letters from these 8 letters

The ways of selecting 4 letters from 8 letters disregarding the order is ⁸C₄ .

After selection, there are 4! arrangements.

Hence, total number of words formed are: ⁸C₄ × 4!

Note: 
Selecting r objects out of n objects and then arranging them is same as r-permutation of n objects.
So the above result can be directly obtained using ⁿP_r formula where n = 8 and r = 4

问题2.考虑到选择每种颜色必须有2个球，找到从4个红色，6个蓝色和5个白色中选择6个球的方法的数目吗?

解决方案：

We need to select 2 balls each of color red, blue and white as per the given condition.

The number of ways of selecting 2 red balls is ⁴C₂
The number of ways of selecting 2 blue balls is ⁶C₂
The number of ways of selecting 2 white balls is ⁵C₂

Hence, the total ways of selection are ⁴C₂ × ⁶C₂ × ⁵C₂  = 900