[combinatorics] permutation and combination (set permutation, step-by-step processing example)

Arrange and combine reference blogs :

• 【 Combinatorial mathematics 】 Basic counting principle ( The principle of addition | Multiplication principle )
• 【 Combinatorial mathematics 】 Examples of permutation and combination of sets ( array | Combine | Circular arrangement | binomial theorem )
• 【 Combinatorial mathematics 】 Permutation and combination ( Arrange and combine content summary | Select the question | Set arrangement | Set combination )
• 【 Combinatorial mathematics 】 Permutation and combination ( Examples of permutations )
• 【 Combinatorial mathematics 】 Permutation and combination ( Multiset arrangement | Full Permutation of multiple sets | Multiset incomplete permutation The repetition of all elements is greater than the number of permutations | Multiset incomplete permutation The repetition of some elements is less than the number of permutations )
• 【 Combinatorial mathematics 】 Permutation and combination ( The combinatorial number of multiple sets | The repetition of all elements is greater than the number of combinations | The combinatorial number of multiple sets deduction 1 Division line derivation | The combinatorial number of multiple sets deduction 2 Derivation of the number of nonnegative integer solutions of indefinite equations )
• 【 Combinatorial mathematics 】 Permutation and combination ( Example of the number of combinations of multiple sets | Three counting models | Select the question | Multiple set combinatorial problem | Nonnegative integer solutions of indefinite equations )
• 【 Combinatorial mathematics 】 Permutation and combination ( Two counting principles 、 Set arrangement example | Set arrangement 、 Example of circle arrangement )
• 【 Combinatorial mathematics 】 Permutation and combination ( Set combination 、 One to one correspondence model analysis example )

One 、 Set arrangement 、 Step by step example

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Yes

9

9

9 A different book ,

4

4

4 Ben hongpi ,

5

5

5 Ben Baipi ;

1.

9

9

9 The arrangement of this book :

9

9

9 This book , Every book is different , Elements do not repeat , Permutation refers to orderly selection ,

So here Elements do not repeat , Orderly selection , The corresponding is Arrangement of sets , Use the set arrangement formula ;

N

=

P

(

n

,

r

)

=

P

(

9

,

9

)

=

9

!

(

9

−

9

)

!

=

9

!

N = P(n,r) = P(9, 9) = \cfrac{9!}{(9-9)!} = 9!

N=P(n,r)=P(9,9)=(9−9)!9!​=9!

* Review of permutations and combinations :

• Number of permutations :$n$

  S , from

• Combinatorial number :$n$

  S , from

2. The arrangement of white papers together :

Step by step processing : Step by step processing is required , Arrange the white papers first , And then All white papers As an element , Sort with the red book ;

( 1 ) The first

1

1

1 Step :

5

5

5 This white paper is put together , The arrangement is Elements do not repeat Orderly selection , Is the arrangement of sets ;

N

=

P

(

n

,

r

)

=

P

(

5

,

5

)

=

5

!

(

5

−

5

)

!

=

5

!

N = P(n,r) = P(5, 5) = \cfrac{5!}{(5-5)!} = 5!

N=P(n,r)=P(5,5)=(5−5)!5!​=5!

( 2 ) The first

2

2

2 Step :

4

4

4 This red book , With a set of white papers Sort , Yes

5

5

5 Elements , Arrange them all ;

N

=

P

(

n

,

r

)

=

P

(

5

,

5

)

=

5

!

(

5

−

5

)

!

=

5

!

N = P(n,r) = P(5, 5) = \cfrac{5!}{(5-5)!} = 5!

N=P(n,r)=P(5,5)=(5−5)!5!​=5!

( 3 ) Step by step summary ( Multiplication principle ) : Multiply the number of arrangement schemes of the above two steps , It's the end result ;

N

=

5

!

 

5

!

N = 5! \ 5!

N=5! 5!

3. Put the white papers together , Put the red books together The arrangement of :

Step by step processing : Step by step processing is required ,

• Arrange the white papers first ;
• Then arrange the red books ;
• The final will be All white papers As an element , All red books are treated as an element , Arrange the above two elements ;

( 1 ) The first

1

1

1 Step :

5

5

5 This white paper is put together , The arrangement is Elements do not repeat Orderly selection , Is the arrangement of sets ;

N

=

P

(

n

,

r

)

=

P

(

5

,

5

)

=

5

!

(

5

−

5

)

!

=

5

!

N = P(n,r) = P(5, 5) = \cfrac{5!}{(5-5)!} = 5!

N=P(n,r)=P(5,5)=(5−5)!5!​=5!

( 2 ) The first

2

2

2 Step :

4

4

4 Put the red books together , The arrangement is Elements do not repeat Orderly selection , Is the arrangement of sets ;

N

=

P

(

n

,

r

)

=

P

(

4

,

4

)

=

4

!

(

4

−

4

)

!

=

4

!

N = P(n,r) = P(4, 4) = \cfrac{4!}{(4-4)!} = 4!

N=P(n,r)=P(4,4)=(4−4)!4!​=4!

( 3 ) The first

3

3

3 Step : The final will be All white papers As an element , All red books are treated as an element , Arrange the above two elements ;

N

=

P

(

n

,

r

)

=

P

(

2

,

2

)

=

2

!

(

2

−

2

)

!

=

2

!

N = P(n,r) = P(2, 2) = \cfrac{2!}{(2-2)!} = 2!

N=P(n,r)=P(2,2)=(2−2)!2!​=2!

( 4 ) Step by step summary ( Multiplication principle ) : Put the above

3

3

3 Multiply the number of arrangement schemes of steps , It's the end result ;

N

=

5

!

 

4

!

 

2

!

N = 5! \ 4! \ 2!

N=5! 4! 2!

4. White papers and red books are arranged alternately The arrangement of :

Step by step processing : Step by step processing is required ,

• Arrange the white papers first ;
• Then insert the red book into ;

( 1 ) The first

1

1

1 Step :

5

5

5 This white paper is put together , The arrangement is Elements do not repeat Orderly selection , Is the arrangement of sets ;

N

=

P

(

n

,

r

)

=

P

(

5

,

5

)

=

5

!

(

5

−

5

)

!

=

5

!

N = P(n,r) = P(5, 5) = \cfrac{5!}{(5-5)!} = 5!

N=P(n,r)=P(5,5)=(5−5)!5!​=5!

( 2 ) The first

2

2

2 Step :

5

5

5 This white paper is arranged to form

4

4

4 Empty space , Insert the red book into

4

4

4 A place , That is, the set is arranged completely ;

N

=

P

(

n

,

r

)

=

P

(

4

,

4

)

=

4

!

(

4

−

4

)

!

=

4

!

N = P(n,r) = P(4, 4) = \cfrac{4!}{(4-4)!} = 4!

N=P(n,r)=P(4,4)=(4−4)!4!​=4!

( 3 ) Step by step summary ( Multiplication principle ) : Put the above

2

2

2 Multiply the number of arrangement schemes of steps , It's the end result ;

N

=

5

!

 

4

!

N = 5! \ 4!

N=5! 4!