从 6 个项目中抽取 5 个项目而不放回，可以形成多少个排列？

排列被称为按顺序组织组、主体或数字的过程，从集合中选择主体或数字，被称为组合，其中数字的顺序无关紧要。

在数学中，排列也被称为组织一个群的过程，其中一个群的所有成员都被排列成某种顺序或顺序。如果组已经排列，则置换过程称为对其组件的重新定位。排列发生在几乎所有数学领域。它们大多出现在考虑某些有限集合上的不同命令时。

置换公式

在排列中，从一组 n 个事物中挑选出 r 个事物，没有任何替换。在这个挑选的顺序。

ⁿP_r = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group

r = subset size, the number of things to be selected from the group

组合

组合是从集合中选择数字的函数，这样（不像排列）选择的顺序无关紧要。在较小的情况下，可以计算组合的数量。这种组合被称为一次合并n个事物而不重复。组合起来，顺序无关紧要，您可以按任何顺序选择项目。对于那些允许重复出现的组合，经常使用术语 k-selection 或 k-combination with replication。

组合配方

组合 r 个东西是从一组 n 个东西中挑选出来的，挑选的顺序无关紧要。

ⁿC_r =n! ⁄ ((n-r)! r!)

Here, 

n = Number of items in set

r = Number of things picked from the group

从 6 个项目中抽取 5 个项目而不放回，可以形成多少个排列？

解决方案：

Consider the question above, but repetition is not allowed. if A={1,2,3,4,5,6} and k=5, there are 720  different possibilities:

In common, we can say that there are k place in the selected list: 

(place 1, place 2, …, place k). There are n options for the first place , (n−1) 

options for the second place (since one component has already been allocated to the 

first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1) 

options for the kth place. Thus, when ordering matters and repetition is not allowed, 

the total number of ways to choose k objects from a set with n component is n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

ⁿP_k = n×(n−1)×…×(n−k+1).

Now, we have 720 possibility, the permutation can be formed by sampling 5 items from 6.

like A = {1,2,3,4,5,6}

1.(1,2,3,4,5)  11.(1,2,4,5,6) 21.(1,2,6,3,5)  31.(1,3,4,2,5) 41.(1,3,5,4,6)                

2.(1,2,3,5,4)  12.(1,2,3,6,5) 22.(1,2,6,5,3)  32.(1,3,4,5,2) 42.(1,3,5,6,4)

3.(1,2,3,4,6)  13.(1,2,5,3,6) 23.(1,2,6,4,5)  33.(1,3,4,2,6) 43.(1,3,6,2,4)

4.(1,2,3,6,4)  14.(1,2,5,6,3) 24.(1,2,6,5,4)  34.(1,3,4,6,2) 44.(1,3,6,4,2)

5.(1,2,3,5,6)  15.(1,2,5,3,4) 25.(1,3,2,4,5)  35.(1,3,4,5,6) 45.(1,3,6,2,5)

6.(1,2,3,6,5)  16.(1,2,5,4,3) 26.(1,3,2,5,4)  36.(1,3,4,6,5) 46.(1,3,6,5,2)

7.(1,2,4,3,5)  17.(1,2,5,4,6) 27.(1,3,2,4,6)  37.(1,3,5,2,4) 47.(1,3,6,4,5)

8.(1,2,4,5,3)  18.(1,2,5,6,4) 28.(1,3,2,6,4)  38.(1,3,5,4,2) 48.(1,3,6,5,4)and so on….

9.(1,2,4,3,6)  19.(1,2,6,3,4) 29.(1,3,2,5,6)  39.(1,3,5,2,6)

10.(1,2,4,6,3) 20.(1,2,6,4,3) 30.(1,3,2,6,5)  40.(1,3,5,6,2)

Likewise when we put 1 on the first position we have 120 possibilities on the same way when we put 2 on the first position we have 120 possibilities same for 3,4,5 and 6.

So, when we do the total of all the possibilities we get = 120+120+120+120+120+120

                                                                                    = 720

So, 720 permutations can be formed by sampling 5 items from 6 without replacement

类似问题

问题1：生日问题是什么？

Let us consider this example to understand birthday problem

There are total 30 people in the room

What is the probability that at least two people have the same birthday or what is the probability that someone in the room share His/Her birthday with at least someone else

White color = p(someone shares with at least                                someone else) or p(s)                         Green color = p(all 30 people have different            birthday) or p(d)

p(s) + p(d) = 1 or 100%

p(s) = 100% – p(d)

If we have two person:

Lets see, person one, their birthday could be 365 days out of 365 days

Now person two could be born on any day that person was not born on

So, 365⁄365 (1st person birthday) 364⁄365 (2nd person birthday)

= 365 × 364 ⁄ 365² = (365! ⁄ (365-2)!) ⁄ 365² = (365! ⁄ 363!) ⁄ 365²

If we have three person:

So, 365⁄365 (1st person birthday) 364⁄365 (2nd person birthday) 363⁄365 (3rd person)

= 365 × 364 × 363 ⁄ 365³ = (365! ⁄ (365-3)!) ⁄ 365³ = (365! ⁄ 362!) ⁄ 365³

Now, if we have 30 people, the probability that no one shares the same birthday

= 365× 364×363×……×336 ⁄ 365³⁰ = (365! ⁄ (365-30)!) ⁄ 365³⁰ = (365! ⁄ 335!) ⁄ 365³⁰ = .2936 or 29.36%

p(d) = .2936 or 29.36%

p(s) = 100% – p(d)

      = 100% – 29.36% or 1 – .2936

      = .7063 ≈ 70.6%

问题2：从4个项目中抽取3个项目而不放回，可以形成多少个排列？

解决方案：

Consider the question above, but repetition is not allowed. if A={1,2,3,4} and k=3, there are 24 different possibilities:

In common, we can say that there are k place in the selected list:

(place 1, place 2, …, place k). There are n options for the first place, (n−1) options for the second place (since one component has already been allocated to the first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1) options for the kth place. Thus, when ordering matters and repetition is not allowed,

The total number of ways to choose k objects from a set with n component is

n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

ⁿP_k = n×(n−1)×…×(n−k+1)

Now, we have 24 possibility, the permutation can be formed by sampling 3 items from 4.

like A = {1,2,3,4}

1.(1,2,3)              11.(2,4,3)               21.(4,1,3)                  

2.(1,3,2)              12.(2,3,4)               22.(4,3,1)  

3.(1,2,4)              13.(3,1,2)               23.(4,2,3)  

4.(1,4,2)              14.(3,2,1)               24.(4,3,2)  

5.(1,3,4)              15.(3,1,4) 

6.(1,4,3)              16.(3,4,1) 

7.(2,1,3)              17.(3,2,4) 

8.(2,3,1)              18.(3,4,2) 

9.(2,1,4)              19.(4,1,2) 

10.(2,4,1)             20.(4,2,1) 

Likewise when we put 1 on the first position we have 6 possibilities, in the same way when we put 2 on the first position we have 6 possibilities same for 3 and 4.

So, when we do the total of all the possibilities we get = 6+6+6+6

                                                                                    = 24

SO, 24 permutations can be formed by sampling 3 items from 4 without replacement

问题 3：从 5 个项目中抽取 4 个项目而不放回，可以形成多少个排列？

解决方案：

Consider the question above, but repetition is not allowed. if A={1,2,3,4,5} and k=4, there are 120  different possibilities:

In common, we can say that there are k place in the selected list:

(place 1, place 2, …, place k). There are n options for the first place , (n−1) options for the second place (since one component has already been allocated to the first place and cannot be chosen here), (n−2) options for the third place, … (n−k+1) options for the kth place. 

Thus, when ordering matters and repetition is not allowed, the total number of ways to choose k objects from a set with n component is

n×(n−1)×…×(n−k+1).

Any of the selected lists in the above setting (select k component, ordered and no repetition) is called a k-permutation of the component in set A. We use the following notation to show the number of k-permutations of an n-component set:

ⁿP_k = n×(n−1)×…×(n−k+1).

Now, we have 120 possibilities, the permutation can be formed by sampling 4 items from 5.

like A = {1,2,3,4,5}

1.(1,2,3,4)              11.(1,3,4,5)               21.(1,5,2,4)                  

2.(1,2,4,3)              12.(1,3,5,4)               22.(1,5,4,2)  

3.(1,2,3,5)              13.(1,4,2,3)               23.(1,5,3,4)  

4.(1,2,5,3)              14.(1,4,3,2)               24.(1,5,4,3) and so on….

5.(1,2,4,5)              15.(1,4,2,5)

6.(1,2,5,4)              16.(1,4,5,2)

7.(1,3,2,4)              17.(1,4,3,5)

8.(1,3,4,2)              18.(1,4,5,3)

9.(1,3,2,5)              19.(1,5,2,3)

10.(1,3,5,2)             20.(1,5,3,2)

Likewise when we put 1 on the first position we have 24 possibilities on the same way when we put 2 on the first position we have 24 possibilities same for 3, 4 and 5.

so, when we do the total of all the possibilities we get = 24+24+24+24+24

                                                                                    = 120

So, 120 permutations can be formed by sampling 4 items from 5 without replacement.