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[combinatorics] exponential generating function (example 2 of solving multiple set permutation with exponential generating function)
2022-07-03 18:20:00 【Programmer community】
List of articles
- One 、 Example of solving multiple set permutation with exponential generating function 2
Reference blog : Look in order
- 【 Combinatorial mathematics 】 Generating function Brief introduction ( Generating function definition | Newton's binomial coefficient | Common generating functions | Related to constants | Related to binomial coefficient | Related to polynomial coefficients )
- 【 Combinatorial mathematics 】 Generating function ( Linear properties | Product properties )
- 【 Combinatorial mathematics 】 Generating function ( Shift property )
- 【 Combinatorial mathematics 】 Generating function ( The nature of summation )
- 【 Combinatorial mathematics 】 Generating function ( Commutative properties | Derivative property | Integral properties )
- 【 Combinatorial mathematics 】 Generating function ( Summary of nature | Important generating functions ) *
- 【 Combinatorial mathematics 】 Generating function ( Generate function examples | Given the general term formula, find the generating function | Given the generating function, find the general term formula )
- 【 Combinatorial mathematics 】 Generating function ( Generate function application scenarios | Solving recursive equations using generating functions )
- 【 Combinatorial mathematics 】 Generating function ( Use the generating function to solve multiple sets r Combinatorial number )
- 【 Combinatorial mathematics 】 Generating function ( Use generating function to solve the number of solutions of indefinite equation )
- 【 Combinatorial mathematics 】 Generating function ( Examples of using generating functions to solve the number of solutions of indefinite equations )
- 【 Combinatorial mathematics 】 Generating function ( Examples of using generating functions to solve the number of solutions of indefinite equations 2 | Extended to integer solutions )
- 【 Combinatorial mathematics 】 Generating function ( Positive integer split | disorder | Orderly | Allow repetition | No repetition | Unordered and unrepeated splitting | Unordered repeated split )
- 【 Combinatorial mathematics 】 Generating function ( Positive integer split | Unordered non repeated split example )
- 【 Combinatorial mathematics 】 Generating function ( Positive integer split | Basic model of positive integer splitting | Disorderly splitting with restrictions )
- 【 Combinatorial mathematics 】 Generating function ( Positive integer split | Repeated ordered splitting | Do not repeat orderly splitting | Proof of the number of repeated ordered splitting schemes )
- 【 Combinatorial mathematics 】 Exponential generating function ( The concept of exponential generating function | Permutation number exponential generating function = General generating function of combinatorial number | Example of exponential generating function )
- 【 Combinatorial mathematics 】 Exponential generating function ( Properties of exponential generating function | The exponential generating function solves the arrangement of multiple sets )
- 【 Combinatorial mathematics 】 Exponential generating function ( Example of solving multiple set permutation with exponential generating function )
One 、 Example of solving multiple set permutation with exponential generating function 2
Use white Red Blue Coloring
n
n
n Lattice , The number of white colors is even , Find the number of coloring schemes
This is a The problem of permutation , When different squares are colored and exchanged , It becomes a different solution ,
Red , Blue Coloring , There is no limit to , The number of colors can be
0
,
1
,
2
,
3
,
4
,
⋯
0, 1,2,3,4,\cdots
0,1,2,3,4,⋯
white Coloring , The number of colors is even , The number of colors is
0
,
2
,
4
,
6
,
8
,
⋯
0, 2, 4, 6, 8 , \cdots
0,2,4,6,8,⋯
Red , Blue Number of colors
0
,
1
,
2
,
3
,
4
,
⋯
0, 1,2,3,4,\cdots
0,1,2,3,4,⋯ Sequence , The corresponding generating function item is :
x
0
0
!
+
x
1
1
!
+
x
2
2
!
⋯
=
1
+
x
+
x
2
2
!
+
⋯
\cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} \cdots = 1 + x + \cfrac{x^2}{2!} + \cdots
0!x0+1!x1+2!x2⋯=1+x+2!x2+⋯
white Number of colors
0
,
2
,
4
,
6
,
8
,
⋯
0, 2, 4, 6, 8 , \cdots
0,2,4,6,8,⋯ Sequence , The corresponding generating function item is :
x
0
0
!
+
x
2
2
!
+
x
4
4
!
⋯
=
1
+
x
2
2
!
+
x
4
4
!
+
⋯
\cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} \cdots = 1+ \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots
0!x0+2!x2+4!x4⋯=1+2!x2+4!x4+⋯
The exponential generating function of the number of the above coloring schemes is :
G
e
(
x
)
=
(
1
+
x
+
x
2
2
!
+
⋯
)
(
1
+
x
+
x
2
2
!
+
⋯
)
(
1
+
x
2
2
!
+
x
4
4
!
+
⋯
)
G_e(x) = (1 + x + \cfrac{x^2}{2!} + \cdots) (1 + x + \cfrac{x^2}{2!} + \cdots) (1+ \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots)
Ge(x)=(1+x+2!x2+⋯)(1+x+2!x2+⋯)(1+2!x2+4!x4+⋯)
among
1
+
x
+
x
2
2
!
+
⋯
1 + x + \cfrac{x^2}{2!} + \cdots
1+x+2!x2+⋯ Sure It's written in
e
x
e^x
ex form ;
among
1
+
x
2
2
!
+
x
4
4
!
+
⋯
1+ \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots
1+2!x2+4!x4+⋯ We can write it in the following form :
1
+
x
2
2
!
+
x
4
4
!
+
⋯
=
1
2
(
e
x
+
e
−
x
)
1+ \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots = \cfrac{1}{2}(e^x + e^{-x})
1+2!x2+4!x4+⋯=21(ex+e−x)
e
x
+
e
−
x
e^x + e^{-x}
ex+e−x Add up , The sign of odd power is opposite , Direct appointment , Even power Double the original , So multiply it outside
1
2
\cfrac{1}{2}
21 ;
Put the above
e
x
e^x
ex and
1
2
(
e
x
+
e
−
x
)
\cfrac{1}{2}(e^x + e^{-x})
21(ex+e−x) Replace the Exponential generating function ;
G
e
(
x
)
=
(
1
+
x
+
x
2
2
!
+
⋯
)
(
1
+
x
+
x
2
2
!
+
⋯
)
(
1
+
x
2
2
!
+
x
4
4
!
+
⋯
)
G_e(x) = (1 + x + \cfrac{x^2}{2!} + \cdots) (1 + x + \cfrac{x^2}{2!} + \cdots) (1+ \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots)
Ge(x)=(1+x+2!x2+⋯)(1+x+2!x2+⋯)(1+2!x2+4!x4+⋯)
=
1
2
(
e
x
+
e
−
x
)
(
e
x
)
(
e
x
)
\ \ \ \ \ \ \ \ \ \ \ \, =\cfrac{1}{2}(e^x + e^{-x})(e^x )(e^x)
=21(ex+e−x)(ex)(ex)
=
1
2
e
3
x
+
1
2
e
x
\ \ \ \ \ \ \ \ \ \ \ \, =\cfrac{1}{2}e^{3x} + \cfrac{1}{2}e^{x}
=21e3x+21ex
take
1
2
e
x
\cfrac{1}{2}e^{x}
21ex Expand to
1
2
(
1
+
x
+
x
2
2
!
+
⋯
)
=
1
2
∑
n
=
0
∞
x
n
n
!
\cfrac{1}{2}(1 + x + \cfrac{x^2}{2!} + \cdots)=\cfrac{1}{2}\sum\limits_{n=0}^\infty \cfrac{x^n}{n!}
21(1+x+2!x2+⋯)=21n=0∑∞n!xn
take
1
2
e
3
x
\cfrac{1}{2}e^{3x}
21e3x Expand to
1
2
(
1
+
3
x
+
(
3
x
)
2
2
!
+
⋯
)
=
1
2
∑
n
=
0
∞
3
n
x
n
n
!
\cfrac{1}{2}(1 + 3x + \cfrac{(3x)^2}{2!} + \cdots)=\cfrac{1}{2}\sum\limits_{n=0}^\infty \cfrac{3^nx^n}{n!}
21(1+3x+2!(3x)2+⋯)=21n=0∑∞n!3nxn
=
1
2
∑
n
=
0
∞
3
n
x
n
n
!
+
1
2
∑
n
=
0
∞
x
n
n
!
\ \ \ \ \ \ \ \ \ \ \ \, =\cfrac{1}{2}\sum\limits_{n=0}^\infty \cfrac{3^nx^n}{n!} + \cfrac{1}{2}\sum\limits_{n=0}^\infty \cfrac{x^n}{n!}
=21n=0∑∞n!3nxn+21n=0∑∞n!xn
=
∑
n
=
0
∞
3
n
+
1
2
⋅
x
n
n
!
\ \ \ \ \ \ \ \ \ \ \ \, =\sum\limits_{n=0}^\infty \cfrac{3^n + 1}{2} \cdot \cfrac{x^n}{n!}
=n=0∑∞23n+1⋅n!xn
x
n
n
!
\cfrac{x^n}{n!}
n!xn The coefficient before is
3
n
+
1
2
\cfrac{3^n + 1}{2}
23n+1
therefore white Red Blue Coloring
n
n
n Lattice , White is even , The coloring scheme has
3
n
+
1
2
\cfrac{3^n + 1}{2}
23n+1 Kind of ;
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