当前位置:网站首页>[combinatorics] polynomial theorem (polynomial theorem | polynomial theorem proof | polynomial theorem inference 1 item number is the number of non negative integer solutions | polynomial theorem infe
[combinatorics] polynomial theorem (polynomial theorem | polynomial theorem proof | polynomial theorem inference 1 item number is the number of non negative integer solutions | polynomial theorem infe
2022-07-03 16:35:00 【Programmer community】
List of articles
- One 、 The polynomial theorem
- Two 、 The polynomial theorem prove
- 3、 ... and 、 The polynomial theorem inference 1
- Four 、 The polynomial theorem inference 2
One 、 The polynomial theorem
The polynomial theorem :
set up
n
n
n As a positive integer ,
x
i
x_i
xi It's a real number ,
i
=
1
,
2
,
⋯
,
t
i=1,2,\cdots,t
i=1,2,⋯,t
(
x
1
+
x
2
+
⋯
+
x
t
)
n
\ \ \ \ (x_1 + x_2 + \cdots + x_t)^n
(x1+x2+⋯+xt)n
=
∑
full
foot
n
1
+
n
2
+
⋯
+
n
t
=
n
Not
negative
whole
Count
Explain
individual
Count
(
n
n
1
n
2
⋯
n
t
)
x
1
n
1
x
2
n
2
⋯
x
t
n
t
= \sum\limits_{ Satisfy n_1 + n_2 + \cdots + n_t = n Number of nonnegative integer solutions }\dbinom{n}{n_1 n_2 \cdots n_t}x_1^{n_1}x_2^{n_2}\cdots x_t^{n_t}
= full foot n1+n2+⋯+nt=n Not negative whole Count Explain individual Count ∑(n1n2⋯ntn)x1n1x2n2⋯xtnt
The above polynomials have
t
t
t The item , this
t
t
t Items added
n
n
n Power ;
Two 、 The polynomial theorem prove
In polynomials
(
x
1
+
x
2
+
⋯
+
x
t
)
n
(x_1 + x_2 + \cdots + x_t)^n
(x1+x2+⋯+xt)n :
Carry out the following processing step by step :
The first
1
1
1 Step : from
n
n
n In a factor , choose
n
1
n_1
n1 Personal factor , Each factor contributes
1
1
1 individual
x
1
x_1
x1 , Total contribution
n
1
n_1
n1 individual
x
1
x_1
x1 , The selection methods are
(
n
n
1
)
\dbinom{n}{n_1}
(n1n) Kind of ;
The first
2
2
2 Step : from
n
−
n
1
n-n_1
n−n1 In a factor , choose
n
2
n_2
n2 Personal factor , Each factor contributes
1
1
1 individual
x
2
x_2
x2 , Total contribution
n
2
n_2
n2 individual
x
2
x_2
x2 , The selection methods are
(
n
−
n
1
n
2
)
\dbinom{n-n_1}{n_2}
(n2n−n1) Kind of ;
⋮
\vdots
⋮
- The first
t
t
n
−
n
1
−
n
2
−
⋯
−
n
t
−
1
n-n_1-n_2 - \cdots -n_{t-1}
n−n1−n2−⋯−nt−1 In a factor , choose
n
t
n_t
nt Personal factor , Each factor contributes
1
1
1 individual
x
t
x_t
xt , Total contribution
n
t
n_t
nt individual
x
t
x_t
xt , The selection methods are
(
n
−
n
1
−
n
2
−
⋯
−
n
t
−
1
n
t
)
\dbinom{n-n_1-n_2 - \cdots -n_{t-1}}{n_t}
(ntn−n1−n2−⋯−nt−1) Kind of ;
t Step : from
According to the principle of step-by-step counting , product rule , Multiply the number of categories in each step above , Is the number of all kinds :
(
n
n
1
)
(
n
−
n
1
n
2
)
(
n
−
n
1
−
n
2
−
⋯
−
n
t
−
1
n
t
)
\ \ \ \ \dbinom{n}{n_1} \dbinom{n-n_1}{n_2} \dbinom{n-n_1-n_2 - \cdots -n_{t-1}}{n_t}
(n1n)(n2n−n1)(ntn−n1−n2−⋯−nt−1)
After deployment , Many can be asked out , The resulting :
=
n
!
n
1
!
n
2
!
⋯
n
t
!
=\cfrac{n!}{n_1! n_2! \cdots n_t!}
=n1!n2!⋯nt!n!
Note that the above formula is the total permutation number of multiple sets
=
(
n
n
1
n
2
⋯
n
t
)
=\dbinom{n}{n_1 n_2 \cdots n_t}
=(n1n2⋯ntn)
3、 ... and 、 The polynomial theorem inference 1
The polynomial theorem inference 1 :
In the above polynomial theorem , Different number of items It's the equation
n
1
+
n
2
+
⋯
+
n
t
=
n
n_1 + n_2 + \cdots + n_t = n
n1+n2+⋯+nt=n
Number of nonnegative integer solutions
C
(
n
+
t
−
1
,
n
)
C(n + t -1 , n)
C(n+t−1,n)
Proof process :
1 . The coefficient before each term
(
n
n
1
n
2
⋯
n
t
)
\dbinom{n}{n_1 n_2 \cdots n_t}
(n1n2⋯ntn) meaning :
n
1
n_1
n1 representative
x
1
x_1
x1 The index of ,
n
1
n_1
n1 Equivalent to how many formulas , When multiplying , Take it
x
1
x_1
x1
n
2
n_2
n2 representative
x
2
x_2
x2 The index of ,
n
2
n_2
n2 Equivalent to how many formulas , When multiplying , Take it
x
2
x_2
x2
⋮
\vdots
⋮
n
t
n_t
nt representative
x
t
x_t
xt The index of ,
n
t
n_t
nt Equivalent to how many formulas , When multiplying , Take it
x
t
x_t
xt
2 . One-to-one correspondence :
n
1
,
n
2
,
⋯
,
n
t
n_1, n_2, \cdots , n_t
n1,n2,⋯,nt A different set of choices , amount to
n
1
+
n
2
+
⋯
+
n
t
=
n
n_1 + n_2 + \cdots + n_t = n
n1+n2+⋯+nt=n
A solution to , Corresponding to different
x
1
,
x
2
,
⋯
,
x
n
x_1 , x_2, \cdots, x_n
x1,x2,⋯,xn The previous item ;
Three corresponding relationships :
Different solutions ,
n
1
,
n
2
,
⋯
,
n
t
n_1, n_2, \cdots , n_t
n1,n2,⋯,nt Different configuration , These items Number of different configurations ,
amount to
n
1
+
n
2
+
⋯
+
n
t
=
n
n_1 + n_2 + \cdots + n_t = n
n1+n2+⋯+nt=n Number of nonnegative integer solutions of ,
It is also equivalent to polynomial Expanded Number of items ;
So find
n
1
+
n
2
+
⋯
+
n
t
=
n
n_1 + n_2 + \cdots + n_t = n
n1+n2+⋯+nt=n Number of nonnegative integer solutions of ,
It corresponds to
n
1
,
n
2
,
⋯
,
n
t
n_1, n_2, \cdots , n_t
n1,n2,⋯,nt Number of different configurations ,
Corresponding Number of terms after polynomial expansion ,
The result is
C
(
n
+
t
−
1
,
n
)
C(n + t -1 , n)
C(n+t−1,n)
This number is also the combinatorial number of multiple sets
Derivation process Refer to the multiple set combination problem :
Multiple sets :S
=
{
n
1
⋅
a
1
,
n
2
⋅
a
2
,
⋯
,
n
k
⋅
a
k
}
,
0
≤
n
i
≤
+
∞
S = \{ n_1 \cdot a_1 , n_2 \cdot a_2 , \cdots , n_k \cdot a_k \} , \ \ \ 0 \leq n_i \leq +\infty
S={ n1⋅a1,n2⋅a2,⋯,nk⋅ak}, 0≤ni≤+∞
taker
r
r A combination of elements ,
r
≤
n
i
r \leq n_i
r≤ni , The derivation process is as follows :
stayk
k
k Of the elements , take
r
r
r Elements , Each element takes
0
∼
r
0 \sim r
0∼r Different elements ,
Usek
−
1
k-1
k−1 Split by lines
k
k
k The position of the elements ,
k
−
1
k - 1
k−1 A dividing line is equivalent to
k
k
k Boxes , Put... In each box
0
∼
r
0 \sim r
0∼r Different elements ,
The total number of elements placed isr
r
r individual , The number of dividing lines is
k
−
1
k-1
k−1 individual , Here comes a combinatorial problem , stay
k
−
1
k-1
k−1 A split line and
r
r
r Between elements , selection
r
r
r Elements , Namely Of multiple sets
r
≤
n
i
r \leq n_i
r≤ni Cases of Number of combinations ;
The result is :N
=
C
(
k
+
r
−
1
,
r
)
N= C(k + r - 1, r)
N=C(k+r−1,r)
Reference resources : 【 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 )
Four 、 The polynomial theorem inference 2
The polynomial theorem inference 3 :
∑
(
n
n
1
n
2
⋯
n
t
)
=
t
n
\sum\dbinom{n}{n_1 n_2 \cdots n_t} = t^n
∑(n1n2⋯ntn)=tn
Proof process :
Polynomial theorem
(
x
1
+
x
2
+
⋯
+
x
t
)
n
\ \ \ \ (x_1 + x_2 + \cdots + x_t)^n
(x1+x2+⋯+xt)n
=
∑
full
foot
n
1
+
n
2
+
⋯
+
n
t
=
n
Not
negative
whole
Count
Explain
individual
Count
(
n
n
1
n
2
⋯
n
t
)
x
1
n
1
x
2
n
2
⋯
x
t
n
t
= \sum\limits_{ Satisfy n_1 + n_2 + \cdots + n_t = n Number of nonnegative integer solutions }\dbinom{n}{n_1 n_2 \cdots n_t}x_1^{n_1}x_2^{n_2}\cdots x_t^{n_t}
= full foot n1+n2+⋯+nt=n Not negative whole Count Explain individual Count ∑(n1n2⋯ntn)x1n1x2n2⋯xtnt
If you will
x
1
=
x
2
=
⋯
=
x
t
=
1
x_1 = x_2 = \cdots = x_t = 1
x1=x2=⋯=xt=1 ,
You can get
(
1
+
1
+
⋯
+
1
)
n
\ \ \ \ (1 + 1 + \cdots + 1)^n
(1+1+⋯+1)n
=
∑
full
foot
n
1
+
n
2
+
⋯
+
n
t
=
n
Not
negative
whole
Count
Explain
individual
Count
(
n
n
1
n
2
⋯
n
t
)
1
n
1
1
n
2
⋯
1
n
t
= \sum\limits_{ Satisfy n_1 + n_2 + \cdots + n_t = n Number of nonnegative integer solutions }\dbinom{n}{n_1 n_2 \cdots n_t}1^{n_1}1^{n_2}\cdots 1^{n_t}
= full foot n1+n2+⋯+nt=n Not negative whole Count Explain individual Count ∑(n1n2⋯ntn)1n11n2⋯1nt
=
∑
full
foot
n
1
+
n
2
+
⋯
+
n
t
=
n
Not
negative
whole
Count
Explain
individual
Count
(
n
n
1
n
2
⋯
n
t
)
= \sum\limits_{ Satisfy n_1 + n_2 + \cdots + n_t = n Number of nonnegative integer solutions }\dbinom{n}{n_1 n_2 \cdots n_t}
= full foot n1+n2+⋯+nt=n Not negative whole Count Explain individual Count ∑(n1n2⋯ntn)
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