[combinatorics] combinatorial identities (recursive combinatorial identities | sum of variable terms | simple combinatorial identities and | sum of variable terms | staggered sums of combinatorial ide

One 、 Combinatorial identity ( recursion )

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Combinatorial identity ( recursion ) :

1 .

(

n

k

)

=

(

n

n

−

k

)

\dbinom{n}{k} = \dbinom{n}{n-k}

(kn​)=(n−kn​) , effect : Simplification

2 .

(

n

k

)

=

n

k

(

n

−

1

k

−

1

)

\dbinom{n}{k} = \dfrac{n}{k} \dbinom{n - 1}{k - 1}

(kn​)=kn​(k−1n−1​) , effect : The coefficient of variation is eliminated when summing ;

3 .

(

n

k

)

=

(

n

−

1

k

)

+

(

n

−

1

k

−

1

)

\dbinom{n}{k} = \dbinom{n - 1}{k} + \dbinom{n - 1}{k - 1}

(kn​)=(kn−1​)+(k−1n−1​) , effect : Disassemble items when summing , Split a combined number into the sum of two terms , Or the difference between the two , Then merge ;

Two 、 Combinatorial identity ( Change the next term to sum ) Simple and

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Simple and :

∑

k

=

0

n

(

n

k

)

=

2

n

\sum_{k=0}^{n}\dbinom{n}{k} = 2^n

k=0∑n​(kn​)=2n

1. prove ( binomial theorem ) : It can be proved by binomial theorem ,

(

x

+

y

)

n

=

∑

k

=

0

n

(

n

k

)

x

k

y

n

−

k

(x + y)^n = \sum\limits_{k=0}^n \dbinom{n}{k}x^k y^{n-k}

(x+y)n=k=0∑n​(kn​)xkyn−k in , send

x

=

y

=

1

x=y=1

x=y=1 , You can get the above Simple and Combinatorial identity ;

2. prove ( Combinatorial analysis ) : Put the equal sign On the left and On the right See each as someone Solution of combinatorial counting problem ,

( 1 ) left Combination counting problem :

∑

k

=

0

n

(

n

k

)

\sum\limits_{k=0}^{n}\dbinom{n}{k}

k=0∑n​(kn​) Can be seen as

n

n

n Number of all subsets of elements ; ( This is also the number of power sets in the set ) ;

This is the classification count , Finally, add the number of all classes , Which contains

0

0

0 The number of elements , contain

1

1

1 Number of element subsets ,

⋯

\cdots

⋯ , contain

n

n

n Number of element subsets ;

( 2 ) On the right side Combination counting problem :

n

n

n Of the elements , Each of the elements Put into subset , Don't put it in the subset , Two options , Then the choices of all elements are ,

2

×

2

×

⋯

×

2

⏟

n

individual

=

2

n

\begin{matrix} \underbrace{ 2 \times 2 \times \cdots \times 2 } \\ n individual \end{matrix} = 2^n

2×2×⋯×2​n individual ​=2n A choice , This is a Multiplication rule of step counting ,

This is a step count , Finally, multiply all the step-by-step results , That is to say

1

1

1 Select the number of elements , The first

2

2

2 Select the number of elements ,

⋯

\cdots

⋯ , The first

n

n

n Select the number of elements ;

3. Application scenarios : Use in sequence summation scenarios ;

Two 、 Combinatorial identity ( Change the next term to sum ) Staggered and

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Staggered and :

∑

k

=

0

n

(

−

1

)

k

(

n

k

)

=

0

\sum_{k=0}^{n} (-1)^k \dbinom{n}{k} = 0

k=0∑n​(−1)k(kn​)=0

1. prove ( binomial theorem ) : It can be proved by binomial theorem ,

(

x

+

y

)

n

=

∑

k

=

0

n

(

n

k

)

x

k

y

n

−

k

(x + y)^n = \sum_{k=0}^n \dbinom{n}{k}x^k y^{n-k}

(x+y)n=∑k=0n​(kn​)xkyn−k in , send

x

=

−

1

,

y

=

1

x= -1 , y=1

x=−1,y=1 , You can get the above Staggered and Combinatorial identity ;

2. prove ( Combinatorial analysis ) : Put the equal sign On the left and On the right See each as someone Solution of combinatorial counting problem , Expand the above combination number completely , Here we need to move items first , take

k

k

k In the case of odd numbers ,

(

−

1

)

k

(-1)^k

(−1)k by

−

1

-1

−1 , Move the sub item of this situation to the right , We have the following formula :

∑

k

=

0

accidentally

Count

(

n

k

)

=

∑

k

=

1

p.

Count

(

n

k

)

\sum_{k=0}^{ even numbers } \dbinom{n}{k} = \sum_{k=1}^{ Odd number } \dbinom{n}{k}

k=0∑ accidentally Count ​(kn​)=k=1∑ p. Count ​(kn​)

( 1 ) left Combination counting problem :

∑

k

=

0

accidentally

Count

(

n

k

)

\sum_{k=0}^{ even numbers } \dbinom{n}{k}

∑k=0 accidentally Count ​(kn​) Can be seen as

n

n

n All of the elements Even number The number of subsets ;

( 2 ) On the right side Combination counting problem :

∑

k

=

1

p.

Count

(

n

k

)

\sum_{k=1}^{ Odd number } \dbinom{n}{k}

∑k=1 p. Count ​(kn​) Can be seen as

n

n

n All of the elements An odd number The number of subsets ;

Above Number of odd subsets And Number of even subsets They are equal. ;

3. Application scenarios : Use in sequence summation scenarios ;