组合数学 (Spring 2013)/Sieve methods: Difference between revisions

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Created page with "== Principle of Inclusion-Exclusion == Let <math>A</math> and <math>B</math> be two finite sets. The cardinality of their union is :<math>|A\cup B|=|A|+|B|-{\color{Blue}|A\cap B|…"
 
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Let <math>S_k=\sum_{|I|=k}|A_I|\,</math>. Conventionally, <math>S_0=|A_\emptyset|=|U|</math>. The principle of inclusion-exclusion can be expressed as
Let <math>S_k=\sum_{|I|=k}|A_I|\,</math>. Conventionally, <math>S_0=|A_\emptyset|=|U|</math>. The principle of inclusion-exclusion can be expressed as
{{Equation|<math>
{{Equation|<math>
\left|\bar{A_1}\cap\bar{A_2}\cap\cdots\cap\bar{A_n}\right|=
S_0-S_1+S_2+\cdots+(-1)^nS_n.
S_0-S_1+S_2+\cdots+(-1)^nS_n.
</math>
</math>
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Therefore, we have a one-to-one correspondence between surjective mappings from an <math>n</math>-set to an <math>m</math>-set and the ordered <math>m</math>-partitions of an <math>n</math>-set.
Therefore, we have a one-to-one correspondence between surjective mappings from an <math>n</math>-set to an <math>m</math>-set and the ordered <math>m</math>-partitions of an <math>n</math>-set.


The Stirling number of the second kind <math>S(n,m)</math> is the number of <math>m</math>-partitions of an <math>n</math>-set. There are <math>m!</math> ways to order an <math>m</math>-partition, thus the number of surjective mappings <math>f:[n]\rightarrow[m]</math> is <math>m! S(n,m)</math>. Combining with what we have proved for surjections, we give the following result for the Stirling number of the second kind.
The Stirling number of the second kind <math>\left\{{n\atop m}\right\}</math> is the number of <math>m</math>-partitions of an <math>n</math>-set. There are <math>m!</math> ways to order an <math>m</math>-partition, thus the number of surjective mappings <math>f:[n]\rightarrow[m]</math> is <math>m! \left\{{n\atop m}\right\}</math>. Combining with what we have proved for surjections, we give the following result for the Stirling number of the second kind.


{{Theorem|Proposition|
{{Theorem|Proposition|
:<math>S(n,m)=\frac{1}{m!}\sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n</math>.
:<math>\left\{{n\atop m}\right\}=\frac{1}{m!}\sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n</math>.
}}
}}


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</math>
</math>
}}
}}
== Möbius inversion ==
=== Posets ===
A '''partially ordered set''' or '''poset''' for short is a set <math>P</math> together with a binary relation denoted <math>\le_P</math> (or just <math>\le</math> if no confusion is caused), satisfying
* (''reflexivity'') For all <math>x\in P, x\le x</math>.
* (''antisymmetry'') If <math>x\le y</math> and <math>y\le x</math>, then <math>x=y</math>.
* (''transitivity'') If <math>x\le y</math> and <math>y\le z</math>, then <math>x\le z</math>.
We say two elements <math>x</math> and <math>y</math> are '''comparable''' if <math>x\le y</math> or <math>y\le x</math>; otherwise <math>x</math> and <math>y</math> are '''incomparable'''.
;Notation
* <math>x\ge y</math> means <math>y\le x</math>.
* <math>x<y</math> means <math>x\le y</math> and <math>x\neq y</math>.
* <math>x>y</math> means <math>y<x</math>.
=== The Möbius function===
Let <math>P</math> be a finite poset. Consider functions in form of <math>\alpha:P\times P\rightarrow\mathbb{R}</math> defined over domain <math>P\times P</math>. It is convenient to treat such functions as matrices whose rows and columns are indexed by <math>P</math>.
;Incidence algebra of poset
:Let
::<math>I(P)=\{\alpha:P\times P\rightarrow\mathbb{R}\mid \alpha(x,y)=0\text{ for all }x\not\le_P y\}</math>
:be the class of <math>\alpha</math> such that <math>\alpha(x,y)</math> is non-zero only for <math>x\le_P y</math>.
:Treating <math>\alpha</math> as matrix, it is trivial to see that <math>I(P)</math> is closed under addition and scalar multiplication, that is,
:* if <math>\alpha,\beta\in I(P)</math> then <math>\alpha+\beta\in I(P)</math>;
:* if <math>\alpha\in I(P)</math> then <math>c\alpha\in I(P)</math> for any <math>c\in\mathbb{R}</math>;
:where <math>\alpha,\beta</math> are treated as matrices.
:With this spirit, it is natural to define the matrix multiplication in <math>I(P)</math>. For <math>\alpha,\beta\in I(P)</math>,
::<math>(\alpha\beta)(x,y)=\sum_{z\in P}\alpha(x,z)\beta(z,y)=\sum_{x\le z\le y}\alpha(x,z)\beta(z,y)</math>.
:The second equation is due to that for <math>\alpha,\beta\in I(P)</math>, for all <math>z</math> other than <math>x\le z\le y</math>, <math>\alpha(x,z)\beta(z,y)</math> is zero.
:By the transitivity of relation <math>\le_P</math>, it is also easy to prove that <math>I(P)</math> is closed under matrix multiplication (the detailed proof is left as an exercise). Therefore, <math>I(P)</math> is closed under addition, scalar multiplication and matrix multiplication, so we have an algebra <math>I(P)</math>, called '''incidence algebra''', over functions on <math>P\times P</math>.
;Zeta function and Möbius function
:A special function in <math>I(P)</math> is the so-called '''zeta function''' <math>\zeta</math>, defined as
::<math>\zeta(x,y)=\begin{cases}1&\text{if }x\le_P y,\\0 &\text{otherwise.}\end{cases}</math>
:As a matrix (or more accurately, as an element of the incidence algebra), <math>\zeta</math> is invertible and its inversion, denoted by <math>\mu</math>, is called the '''Möbius function'''. More precisely, <math>\mu</math> is also in the incidence algebra <math>I(P)</math>, and <math>\mu\zeta=I</math> where <math>I</math> is the identity matrix (the identity of the incidence algebra <math>I(P)</math>).
There is an equivalent explicit definition of Möbius function.
{{Theorem|Definition (Möbius function)|
:<math>\mu(x,y)=\begin{cases}
-\sum_{x\le z< y}\mu(x,z)&\text{if }x<y,\\
1&\text{if }x=y,\\
0&\text{if }x\not\le y.
\end{cases}
</math>
}}
To see the equivalence between this definition and the inversion of zeta function, we may have the following proposition, which is proved by directly evaluating <math>\mu\zeta</math>.
{{Theorem|Proposition|
:For any <math>x,y\in P</math>,
::<math>\sum_{x\le z\le y}\mu(x,z)=\begin{cases}1 &\text{if }x=y,\\
0 &\text{otherwise.}\end{cases}</math>
}}
{{Proof|
It holds that
:<math>(\mu\zeta)(x,y)=\sum_{x\le z\le y}\mu(x,z)\zeta(z,y)=\sum_{x\le z\le y}\mu(x,z)</math>.
On the other hand, <math>\mu\zeta=I</math>, i.e.
:<math>(\mu\zeta)(x,y)=\begin{cases}1 &\text{if }x=y,\\
0 &\text{otherwise.}\end{cases}</math>
The proposition follows.
}}
Note that <math>\mu(x,y)=\sum_{x\le z\le y}\mu(x,z)-\sum_{x\le z< y}\mu(x,z)</math>, which gives the above inductive definition of Möbius function.
=== Computing Möbius functions===
We consider the simple poset <math>P=[n]</math>, where <math>\le</math> is the total order. It follows directly from the recursive definition of Möbius function that
:<math>\mu(i,j)=\begin{cases}1 & \text{if }i=j,\\
-1 & \text{if }i+1=j,\\
0 & \text{otherwise.}
\end{cases}
</math>
Usually for general posets, it is difficult to directly compute the Möbius function from its definition. We introduce a rule helping us compute the Möbius function by decomposing the poset into posets with simple structures.
{{Theorem|Theorem (the product rule)|
: Let <math>P</math> and <math>Q</math> be two finite posets, and <math>P\times Q</math> be the poset resulted from Cartesian product of <math>P</math> and <math>Q</math>, where for all <math>(x,y), (x',y')\in P\times Q</math>, <math>(x,y)\le (x',y')</math> if and only if <math>x\le x'</math> and <math>y\le y'</math>. Then
::<math>\mu_{P\times Q}((x,y),(x',y'))=\mu_P(x,x')\mu_Q(y,y')</math>.
}}
{{Proof|
We use the recursive definition
:<math>\mu(x,y)=\begin{cases}
-\sum_{x\le z< y}\mu(x,z)&\text{if }x<y,\\
1&\text{if }x=y,\\
0&\text{if }x\not\le y.
\end{cases}
</math>
to prove the equation in the theorem.
If <math>(x,y)=(x',y')</math>, then <math>x=x'</math> and <math>y=y'</math>. It is easy to see that both sides of the equation are 1. If <math>(x,y)\not\le(x',y')</math>, then either <math>x\not\le x'</math> or <math>y\not\le y'</math>. It is also easy to see that both sides are 0.
The only remaining case is that <math>(x,y)<(x',y')</math>, in which case either <math>x<x'</math> or <math>y<y'</math>.
:<math>
\begin{align}
\sum_{(x,y)\le (u,v)\le (x',y')}\mu_P(x,u)\mu_Q(y,v)
&=\left(\sum_{x\le u\le x'}\mu_P(x,u)\right)\left(\sum_{y\le v\le y'}\mu_Q(y,v)\right)=I(x,x')I(y,y')=0,
\end{align}
</math>
where the last two equations are due to the proposition for <math>\mu</math>. Thus
:<math>\mu_P(x,x')\mu_Q(y,y')=-\sum_{(x,y)\le (u,v)< (x',y')}\mu_P(x,u)\mu_Q(y,v)</math>.
By induction, assume that the equation <math>\mu_{P\times Q}((x,y),(u,v))=\mu_P(x,u)\mu_Q(y,v)</math> is true for all <math>(u,v)< (x',y')</math>. Then
:<math>
\begin{align}
\mu_{P\times Q}((x,y),(x',y'))
&=-\sum_{(x,y)\le (u,v)< (x',y')}\mu_{P\times Q}((x,y),(u,v))\\
&=-\sum_{(x,y)\le (u,v)< (x',y')}\mu_P(x,u)\mu_Q(y,v)\\
&=\mu_P(x,x')\mu_Q(y,y'),
\end{align}
</math>
which complete the proof.
}}
;Poset of subsets
:Consider the poset defined by all subsets of a finite universe <math>U</math>, that is <math>P=2^U</math>, and for <math>S,T\subseteq U</math>, <math>S\le_P T</math> if and only if <math>S\subseteq T</math>.
{{Theorem| Möbius function for subsets|
:The Möbius function for the above defined poset <math>P</math> is that for <math>S,T\subseteq U</math>,
::<math>\mu(S,T)=
\begin{cases}
(-1)^{|T|-|S|} & \text{if }S\subseteq T,\\
0 &\text{otherwise.}
\end{cases}
</math>
}}
{{Proof|
We can equivalently represent each <math>S\subseteq U</math> by a boolean string <math>S\in\{0,1\}^U</math>, where <math>S(x)=1</math> if and only if <math>x\in S</math>.
For each element <math>x\in U</math>, we can define a poset <math>P_x=\{0, 1\}</math> with <math>0\le 1</math>. By definition of Möbius function, the Möbius function of this elementary poset is given by <math>\mu_x(0,0)=\mu_x(1,1)=1</math>, <math>\mu_x(0,1)=-1</math> and <math>\mu(1,0)=0</math>.
The poset <math>P</math> of all subsets of <math>U</math> is the Cartesian product of all <math>P_x</math>, <math>x\in U</math>. By the product rule,
:<math>\mu(S,T)=\prod_{x\in U}\mu_x(S(x), T(x))=\prod_{x\in S\atop x\in T}1\prod_{x\not\in S\atop x\not\in T}1\prod_{x\in S\atop x\not\in T}0\prod_{x\not\in S\atop x\in T}(-1)=\begin{cases}
(-1)^{|T|-|S|} & \text{if }S\subseteq T,\\
0 &\text{otherwise.}
\end{cases}</math>
}}
:Note that the poset <math>P</math> is actually the [http://en.wikipedia.org/wiki/Boolean_algebra_(structure) Boolean algebra] of rank <math>|U|</math>. The proof relies only on that the fact that the poset is a Boolean algebra, thus the theorem holds for Boolean algebra posets.
;Posets of divisors
:Consider the poset defined by all devisors of a positive integer <math>n</math>, that is <math>P=\{a>0\mid a|n\}</math>, and for <math>a,b\in P</math>, <math>a\le_P b</math> if and only if <math>a|b\,</math>.
{{Theorem|Möbius function for divisors|
:The Möbius function for the above defined poset <math>P</math> is that for <math>a,b>0</math> that <math>a|n</math> and <math>b|n</math>,
::<math>\mu(a,b)=
\begin{cases}
(-1)^{r} & \text{if }\frac{b}{a}\text{ is the product of }r\text{ distinct primes},\\
0 &\text{otherwise, i.e. if }a\not|b\text{ or }\frac{b}{a}\text{ is not squarefree.}
\end{cases}
</math>
}}
{{Proof|
Denote <math>n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k}</math>. Represent <math>n</math> by a tuple <math>(n_1,n_2,\ldots,n_k)</math>. Every <math>a\in P</math> corresponds in this way to a tuple <math>(a_1,a_2,\ldots,a_k)</math> with <math>a_i\le n_i</math> for all <math>1\le i\le k</math>.
Let <math>P_i=\{1,2,\ldots,n_i\}</math> be the poset with <math>\le</math> being the total order. The poset <math>P</math> of divisors of <math>n</math> is thus isomorphic to the poset constructed by the Cartesian product of all <math>P_i</math>, <math>1\le i\le k</math>. Then
:<math>
\begin{align}
\mu(a,b)
&=\prod_{1\le i\le k}\mu(a_i,b_i)=\prod_{1\le i\le k\atop a_i=b_i}1\prod_{1\le i\le k\atop b_i-a_i=1}(-1)\prod_{1\le i\le k\atop b_i-a_i\not\in\{0,1\}}0
=\begin{cases}
(-1)^{\sum_{i}(b_i-a_i)} & \text{if all }b_i-a_i\in\{0,1\},\\
0 &\text{otherwise.}
\end{cases}\\
&=\begin{cases}
(-1)^{r} & \text{if }\frac{b}{a}\text{ is the product of }r\text{ distinct primes},\\
0 &\text{otherwise.}
\end{cases}
\end{align}
</math>
}}
=== Principle of Möbius inversion ===
We now introduce the the famous Möbius inversion formula.
{{Theorem|Möbius inversion formula|
:Let <math>P</math> be a finite poset and <math>\mu</math> its Möbius function. Let <math>f,g:P\rightarrow \mathbb{R}</math>. Then
::<math>\forall x\in P,\,\, g(x)=\sum_{y\le x} f(y)</math>,
:if and only if
::<math>\forall x\in P,\,\, f(x)=\sum_{y\le x}g(y)\mu(y,x)</math>.
}}
The functions <math>f,g:P\rightarrow\mathbb{R}</math> are vectors. Evaluate the matrix multiplications <math>f\zeta</math> and <math>g\mu</math> as follows:
:<math>(f\zeta)(x)=\sum_{y\in P}f(y)\zeta(y,x)=\sum_{y\le x}f(y)</math>,
and
:<math>(g\mu)(x)=\sum_{y\in P}g(y)\mu(y,x)=\sum_{y\le x}g(y)\mu(y,x)</math>.
The Möbius inversion formula is nothing but the following statement
:<math>f\zeta=g\Leftrightarrow f=g\mu</math>,
which is trivially true due to <math>\mu\zeta=I</math> by basic linear algebra.
The following dual form of the inversion formula is also useful.
{{Theorem|Möbius inversion formula, dual form|
:Let <math>P</math> be a finite poset and <math>\mu</math> its Möbius function. Let <math>f,g:P\rightarrow \mathbb{R}</math>. Then
::<math>\forall x\in P, \,\, g(x)=\sum_{y{\color{red}\ge} x} f(y)</math>,
: if and only if
::<math>\forall x\in P, \,\, f(x)=\sum_{y{\color{red}\ge} x}\mu(x,y)g(y)</math>.
}}
To prove the dual form, we only need to evaluate the matrix multiplications on left:
:<math>\zeta f=g\Leftrightarrow f=\mu g</math>.
;Principle of Inclusion-Exclusion
:Let <math>A_1,A_2,\ldots,A_n\subseteq U</math>. For any <math>J\subseteq\{1,2,\ldots,n\}</math>,
:*let <math>f(J)</math> be the number of elements that belongs to ''exactly'' the sets <math>A_i, i\in J</math> and to no others, i.e.
:::<math>f(J)=\left|\left(\bigcap_{i\in J}A_i\right)\setminus\left(\bigcup_{i\not\in J}A_i\right)\right|</math>;
:*let <math>g(J)=\left|\bigcap_{i\in J}A_i\right|</math>.
:For any <math>J\subseteq\{1,2,\ldots,n\}</math>, the following relation holds for the above defined <math>f</math> and <math>g</math>:
::<math>g(J)=\sum_{I\supseteq J}f(I)</math>.
:Applying the dual form of the Möbius inversion formula, we have that for any <math>J\subseteq\{1,2,\ldots,n\}</math>,
::<math>f(J)=\sum_{I\supseteq J}\mu(J,I)g(I)=\sum_{I\supseteq J}\mu(J,I)\left|\bigcap_{i\in I}A_i\right|</math>,
:where the Möbius function is for the poset of all subsets of <math>\{1,2,\ldots,n\}</math>, ordered by <math>\subseteq</math>, thus it holds that <math>\mu(J,I)=(-1)^{|I|-|J|}\,</math> for <math>J\subseteq I</math>. Therefore,
::<math>f(J)=\sum_{I\supseteq J}(-1)^{|I|-|J|}\left|\bigcap_{i\in I}A_i\right|</math>.
:We have a formula for the number of elements with exactly those properties <math>A_i, i\in J</math> for any <math>J\subseteq\{1,2,\ldots,n\}</math>. For the special case that <math>J=\emptyset</math>, <math>f(\emptyset)</math> is the number of elements satisfying no property of <math>A_1,A_2,\ldots,A_n</math>, and
::<math>f(\emptyset)=\left|U\setminus\bigcup_iA_i\right|=\sum_{I\subseteq \{1,\ldots,n\}}(-1)^{|I|}\left|\bigcap_{i\in I}A_i\right|</math>
:which gives precisely the Principle of Inclusion-Exclusion.
;Möbius inversion formula for number theory
:The number-theoretical Möbius inversion formula is stated as such: Let <math>N</math> be a positive integer,
::<math>g(n)=\sum_{d|n}f(d)\,</math> for all <math>n|N</math>
:if and only if
::<math>f(n)=\sum_{d|n}g(d)\mu\left(\frac{n}{d}\right)\,</math> for all <math>n|N</math>,
:where <math>\mu</math> is the [http://en.wikipedia.org/wiki/M%C3%B6bius_function number-theoretical Möbius function], defined as
::<math>\mu(n)=\begin{cases}1 & \text{if }n\text{ is product of an even number of distinct primes,}\\
-1 &\text{if }n\text{ is product of an odd number of distinct primes,}\\
0 &\text{otherwise.}\end{cases}</math>
:The number-theoretical Möbius inversion formula is just a special case of the Möbius inversion formula for posets, when the poset is the set of divisors of <math>N</math>, and for any <math>a,b\in P</math>, <math>a\le_P b</math> if <math>a|b</math>.


== Reference ==
== Reference ==
* ''Stanley,'' Enumerative Combinatorics, Volume 1, Chapter 2.
* ''Stanley,'' Enumerative Combinatorics, Volume 1, Chapter 2.
* ''van Lin and Wilson'', A course in combinatorics, Chapter 10, 25.
* ''van Lin and Wilson'', A course in combinatorics, Chapter 10, 25.

Latest revision as of 06:18, 20 March 2013

Principle of Inclusion-Exclusion

Let [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] be two finite sets. The cardinality of their union is

[math]\displaystyle{ |A\cup B|=|A|+|B|-{\color{Blue}|A\cap B|} }[/math].

For three sets [math]\displaystyle{ A }[/math], [math]\displaystyle{ B }[/math], and [math]\displaystyle{ C }[/math], the cardinality of the union of these three sets is computed as

[math]\displaystyle{ |A\cup B\cup C|=|A|+|B|+|C|-{\color{Blue}|A\cap B|}-{\color{Blue}|A\cap C|}-{\color{Blue}|B\cap C|}+{\color{Red}|A\cap B\cap C|} }[/math].

This is illustrated by the following figure.

Generally, the Principle of Inclusion-Exclusion states the rule for computing the union of [math]\displaystyle{ n }[/math] finite sets [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math], such that

[math]\displaystyle{ \begin{align} \left|\bigcup_{i=1}^nA_i\right| &= \sum_{I\subseteq\{1,\ldots,n\}}(-1)^{|I|-1}\left|\bigcap_{i\in I}A_i\right|. \end{align} }[/math]

In combinatorial enumeration, the Principle of Inclusion-Exclusion is usually applied in its complement form.

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n\subseteq U }[/math] be subsets of some finite set [math]\displaystyle{ U }[/math]. Here [math]\displaystyle{ U }[/math] is some universe of combinatorial objects, whose cardinality is easy to calculate (e.g. all strings, tuples, permutations), and each [math]\displaystyle{ A_i }[/math] contains the objects with some specific property (e.g. a "pattern") which we want to avoid. The problem is to count the number of objects without any of the [math]\displaystyle{ n }[/math] properties. We write [math]\displaystyle{ \bar{A_i}=U-A_i }[/math]. The number of objects without any of the properties [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] is

[math]\displaystyle{ \begin{align} \left|\bar{A_1}\cap\bar{A_2}\cap\cdots\cap\bar{A_n}\right|=\left|U-\bigcup_{i=1}^nA_i\right| &= |U|+\sum_{I\subseteq\{1,\ldots,n\}}(-1)^{|I|}\left|\bigcap_{i\in I}A_i\right|. \end{align} }[/math]

For an [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math], we denote

[math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math]

with the convention that [math]\displaystyle{ A_\emptyset=U }[/math]. The above equation is stated as:

Principle of Inclusion-Exclusion
Let [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be a family of subsets of [math]\displaystyle{ U }[/math]. Then the number of elements of [math]\displaystyle{ U }[/math] which lie in none of the subsets [math]\displaystyle{ A_i }[/math] is
[math]\displaystyle{ \sum_{I\subseteq\{1,\ldots, n\}}(-1)^{|I|}|A_I| }[/math].

Let [math]\displaystyle{ S_k=\sum_{|I|=k}|A_I|\, }[/math]. Conventionally, [math]\displaystyle{ S_0=|A_\emptyset|=|U| }[/math]. The principle of inclusion-exclusion can be expressed as

[math]\displaystyle{ \left|\bar{A_1}\cap\bar{A_2}\cap\cdots\cap\bar{A_n}\right|= S_0-S_1+S_2+\cdots+(-1)^nS_n. }[/math]

Surjections

In the twelvefold way, we discuss the counting problems incurred by the mappings [math]\displaystyle{ f:N\rightarrow M }[/math]. The basic case is that elements from both [math]\displaystyle{ N }[/math] and [math]\displaystyle{ M }[/math] are distinguishable. In this case, it is easy to count the number of arbitrary mappings (which is [math]\displaystyle{ m^n }[/math]) and the number of injective (one-to-one) mappings (which is [math]\displaystyle{ (m)_n }[/math]), but the number of surjective is difficult. Here we apply the principle of inclusion-exclusion to count the number of surjective (onto) mappings.

Theorem
The number of surjective mappings from an [math]\displaystyle{ n }[/math]-set to an [math]\displaystyle{ m }[/math]-set is given by
[math]\displaystyle{ \sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n }[/math].
Proof.

Let [math]\displaystyle{ U=\{f:[n]\rightarrow[m]\} }[/math] be the set of mappings from [math]\displaystyle{ [n] }[/math] to [math]\displaystyle{ [m] }[/math]. Then [math]\displaystyle{ |U|=m^n }[/math].

For [math]\displaystyle{ i\in[m] }[/math], let [math]\displaystyle{ A_i }[/math] be the set of mappings [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] that none of [math]\displaystyle{ j\in[n] }[/math] is mapped to [math]\displaystyle{ i }[/math], i.e. [math]\displaystyle{ A_i=\{f:[n]\rightarrow[m]\setminus\{i\}\} }[/math], thus [math]\displaystyle{ |A_i|=(m-1)^n }[/math].

More generally, for [math]\displaystyle{ I\subseteq [m] }[/math], [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math] contains the mappings [math]\displaystyle{ f:[n]\rightarrow[m]\setminus I }[/math]. And [math]\displaystyle{ |A_I|=(m-|I|)^n\, }[/math].

A mapping [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is surjective if [math]\displaystyle{ f }[/math] lies in none of [math]\displaystyle{ A_i }[/math]. By the principle of inclusion-exclusion, the number of surjective [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is

[math]\displaystyle{ \sum_{I\subseteq[m]}(-1)^{|I|}\left|A_I\right|=\sum_{I\subseteq[m]}(-1)^{|I|}(m-|I|)^n=\sum_{j=0}^m(-1)^j{m\choose j}(m-j)^n }[/math].

Let [math]\displaystyle{ k=m-j }[/math]. The theorem is proved.

[math]\displaystyle{ \square }[/math]

Recall that, in the twelvefold way, we establish a relation between surjections and partitions.

  • Surjection to ordered partition:
For a surjective [math]\displaystyle{ f:[n]\rightarrow[m] }[/math], [math]\displaystyle{ (f^{-1}(0),f^{-1}(1),\ldots,f^{-1}(m-1)) }[/math] is an ordered partition of [math]\displaystyle{ [n] }[/math].
  • Ordered partition to surjection:
For an ordered [math]\displaystyle{ m }[/math]-partition [math]\displaystyle{ (B_0,B_1,\ldots, B_{m-1}) }[/math] of [math]\displaystyle{ [n] }[/math], we can define a function [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] by letting [math]\displaystyle{ f(i)=j }[/math] if and only if [math]\displaystyle{ i\in B_j }[/math]. [math]\displaystyle{ f }[/math] is surjective since as a partition, none of [math]\displaystyle{ B_i }[/math] is empty.

Therefore, we have a one-to-one correspondence between surjective mappings from an [math]\displaystyle{ n }[/math]-set to an [math]\displaystyle{ m }[/math]-set and the ordered [math]\displaystyle{ m }[/math]-partitions of an [math]\displaystyle{ n }[/math]-set.

The Stirling number of the second kind [math]\displaystyle{ \left\{{n\atop m}\right\} }[/math] is the number of [math]\displaystyle{ m }[/math]-partitions of an [math]\displaystyle{ n }[/math]-set. There are [math]\displaystyle{ m! }[/math] ways to order an [math]\displaystyle{ m }[/math]-partition, thus the number of surjective mappings [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is [math]\displaystyle{ m! \left\{{n\atop m}\right\} }[/math]. Combining with what we have proved for surjections, we give the following result for the Stirling number of the second kind.

Proposition
[math]\displaystyle{ \left\{{n\atop m}\right\}=\frac{1}{m!}\sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n }[/math].

Derangements

We now count the number of bijections from a set to itself with no fixed points. This is the derangement problem.

For a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math], a fixed point is such an [math]\displaystyle{ i\in\{1,2,\ldots,n\} }[/math] that [math]\displaystyle{ \pi(i)=i }[/math]. A derangement of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] is a permutation of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] that has no fixed points.

Theorem
The number of derangements of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] given by
[math]\displaystyle{ n!\sum_{k=0}^n\frac{(-1)^k}{k!}\approx \frac{n!}{\mathrm{e}} }[/math].
Proof.

Let [math]\displaystyle{ U }[/math] be the set of all permutations of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math]. So [math]\displaystyle{ |U|=n! }[/math].

Let [math]\displaystyle{ A_i }[/math] be the set of permutations with fixed point [math]\displaystyle{ i }[/math]; so [math]\displaystyle{ |A_i|=(n-1)! }[/math]. More generally, for any [math]\displaystyle{ I\subseteq \{1,2,\ldots,n\} }[/math], [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math], and [math]\displaystyle{ |A_I|=(n-|I|)! }[/math], since permutations in [math]\displaystyle{ A_I }[/math] fix every point in [math]\displaystyle{ I }[/math] and permute the remaining points arbitrarily. A permutation is a derangement if and only if it lies in none of the sets [math]\displaystyle{ A_i }[/math]. So the number of derangements is

[math]\displaystyle{ \sum_{I\subseteq\{1,2,\ldots,n\}}(-1)^{|I|}(n-|I|)!=\sum_{k=0}^n(-1)^k{n\choose k}(n-k)!=n!\sum_{k=0}^n\frac{(-1)^k}{k!}. }[/math]

By Taylor's series,

[math]\displaystyle{ \frac{1}{\mathrm{e}}=\sum_{k=0}^\infty\frac{(-1)^k}{k!}=\sum_{k=0}^n\frac{(-1)^k}{k!}\pm o\left(\frac{1}{n!}\right) }[/math].

It is not hard to see that [math]\displaystyle{ n!\sum_{k=0}^n\frac{(-1)^k}{k!} }[/math] is the closest integer to [math]\displaystyle{ \frac{n!}{\mathrm{e}} }[/math].

[math]\displaystyle{ \square }[/math]

Therefore, there are about [math]\displaystyle{ \frac{1}{\mathrm{e}} }[/math] fraction of all permutations with no fixed points.

Permutations with restricted positions

We introduce a general theory of counting permutations with restricted positions. In the derangement problem, we count the number of permutations that [math]\displaystyle{ \pi(i)\neq i }[/math]. We now generalize to the problem of counting permutations which avoid a set of arbitrarily specified positions.

It is traditionally described using terminology from the game of chess. Let [math]\displaystyle{ B\subseteq \{1,\ldots,n\}\times \{1,\ldots,n\} }[/math], called a board. As illustrated below, we can think of [math]\displaystyle{ B }[/math] as a chess board, with the positions in [math]\displaystyle{ B }[/math] marked by "[math]\displaystyle{ \times }[/math]".

a b c d e f g h
8 a8 __ b8 cross c8 cross d8 __ e8 cross f8 __ g8 __ h8 cross 8
7 a7 cross b7 __ c7 __ d7 cross e7 __ f7 __ g7 cross h7 __ 7
6 a6 cross b6 __ c6 cross d6 cross e6 __ f6 cross g6 cross h6 __ 6
5 a5 __ b5 cross c5 __ d5 __ e5 cross f5 __ g5 cross h5 __ 5
4 a4 cross b4 __ c4 __ d4 __ e4 cross f4 cross g4 cross h4 __ 4
3 a3 __ b3 cross c3 __ d3 cross e3 __ f3 __ g3 __ h3 cross 3
2 a2 __ b2 __ c2 cross d2 __ e2 cross f2 __ g2 __ h2 cross 2
1 a1 cross b1 __ c1 __ d1 cross e1 __ f1 cross g1 __ h1 __ 1
a b c d e f g h

For a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ \{1,\ldots,n\} }[/math], define the graph [math]\displaystyle{ G_\pi(V,E) }[/math] as

[math]\displaystyle{ \begin{align} G_\pi &= \{(i,\pi(i))\mid i\in \{1,2,\ldots,n\}\}. \end{align} }[/math]

This can also be viewed as a set of marked positions on a chess board. Each row and each column has only one marked position, because [math]\displaystyle{ \pi }[/math] is a permutation. Thus, we can identify each [math]\displaystyle{ G_\pi }[/math] as a placement of [math]\displaystyle{ n }[/math] rooks (“城堡”，规则同中国象棋里的“车”) without attacking each other.

For example, the following is the [math]\displaystyle{ G_\pi }[/math] of such [math]\displaystyle{ \pi }[/math] that [math]\displaystyle{ \pi(i)=i }[/math].

a b c d e f g h
8 a8 white rook b8 __ c8 __ d8 __ e8 __ f8 __ g8 __ h8 __ 8
7 a7 __ b7 white rook c7 __ d7 __ e7 __ f7 __ g7 __ h7 __ 7
6 a6 __ b6 __ c6 white rook d6 __ e6 __ f6 __ g6 __ h6 __ 6
5 a5 __ b5 __ c5 __ d5 white rook e5 __ f5 __ g5 __ h5 __ 5
4 a4 __ b4 __ c4 __ d4 __ e4 white rook f4 __ g4 __ h4 __ 4
3 a3 __ b3 __ c3 __ d3 __ e3 __ f3 white rook g3 __ h3 __ 3
2 a2 __ b2 __ c2 __ d2 __ e2 __ f2 __ g2 white rook h2 __ 2
1 a1 __ b1 __ c1 __ d1 __ e1 __ f1 __ g1 __ h1 white rook 1
a b c d e f g h

Now define

[math]\displaystyle{ \begin{align} N_0 &= \left|\left\{\pi\mid B\cap G_\pi=\emptyset\right\}\right|\\ r_k &= \mbox{number of }k\mbox{-subsets of }B\mbox{ such that no two elements have a common coordinate}\\ &=\left|\left\{S\in{B\choose k} \,\bigg|\, \forall (i_1,j_1),(i_2,j_2)\in S, i_1\neq i_2, j_1\neq j_2 \right\}\right| \end{align} }[/math]

Interpreted in chess game,

  • [math]\displaystyle{ B }[/math]: a set of marked positions in an [math]\displaystyle{ [n]\times [n] }[/math] chess board.
  • [math]\displaystyle{ N_0 }[/math]: the number of ways of placing [math]\displaystyle{ n }[/math] non-attacking rooks on the chess board such that none of these rooks lie in [math]\displaystyle{ B }[/math].
  • [math]\displaystyle{ r_k }[/math]: number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math].

Our goal is to count [math]\displaystyle{ N_0 }[/math] in terms of [math]\displaystyle{ r_k }[/math]. This gives the number of permutations avoid all positions in a [math]\displaystyle{ B }[/math].

Theorem
[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)! }[/math].
Proof.

For each [math]\displaystyle{ i\in[n] }[/math], let [math]\displaystyle{ A_i=\{\pi\mid (i,\pi(i))\in B\} }[/math] be the set of permutations [math]\displaystyle{ \pi }[/math] whose [math]\displaystyle{ i }[/math]-th position is in [math]\displaystyle{ B }[/math].

[math]\displaystyle{ N_0 }[/math] is the number of permutations avoid all positions in [math]\displaystyle{ B }[/math]. Thus, our goal is to count the number of permutations [math]\displaystyle{ \pi }[/math] in none of [math]\displaystyle{ A_i }[/math] for [math]\displaystyle{ i\in [n] }[/math].

For each [math]\displaystyle{ I\subseteq [n] }[/math], let [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math], which is the set of permutations [math]\displaystyle{ \pi }[/math] such that [math]\displaystyle{ (i,\pi(i))\in B }[/math] for all [math]\displaystyle{ i\in I }[/math]. Due to the principle of inclusion-exclusion,

[math]\displaystyle{ N_0=\sum_{I\subseteq [n]} (-1)^{|I|}|A_I|=\sum_{k=0}^n(-1)^k\sum_{I\in{[n]\choose k}}|A_I| }[/math].

The next observation is that

[math]\displaystyle{ \sum_{I\in{[n]\choose k}}|A_I|=r_k(n-k)! }[/math],

because we can count both sides by first placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math] and placing [math]\displaystyle{ n-k }[/math] additional non-attacking rooks on [math]\displaystyle{ [n]\times [n] }[/math] in [math]\displaystyle{ (n-k)! }[/math] ways.

Therefore,

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)! }[/math].
[math]\displaystyle{ \square }[/math]

Derangement problem

We use the above general method to solve the derange problem again.

Take [math]\displaystyle{ B=\{(1,1),(2,2),\ldots,(n,n)\} }[/math] as the chess board. A derangement [math]\displaystyle{ \pi }[/math] is a placement of [math]\displaystyle{ n }[/math] non-attacking rooks such that none of them is in [math]\displaystyle{ B }[/math].

a b c d e f g h
8 a8 cross b8 __ c8 __ d8 __ e8 __ f8 __ g8 __ h8 __ 8
7 a7 __ b7 cross c7 __ d7 __ e7 __ f7 __ g7 __ h7 __ 7
6 a6 __ b6 __ c6 cross d6 __ e6 __ f6 __ g6 __ h6 __ 6
5 a5 __ b5 __ c5 __ d5 cross e5 __ f5 __ g5 __ h5 __ 5
4 a4 __ b4 __ c4 __ d4 __ e4 cross f4 __ g4 __ h4 __ 4
3 a3 __ b3 __ c3 __ d3 __ e3 __ f3 cross g3 __ h3 __ 3
2 a2 __ b2 __ c2 __ d2 __ e2 __ f2 __ g2 cross h2 __ 2
1 a1 __ b1 __ c1 __ d1 __ e1 __ f1 __ g1 __ h1 cross 1
a b c d e f g h

Clearly, the number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math] is [math]\displaystyle{ r_k={n\choose k} }[/math]. We want to count [math]\displaystyle{ N_0 }[/math], which gives the number of ways of placing [math]\displaystyle{ n }[/math] non-attacking rooks such that none of these rooks lie in [math]\displaystyle{ B }[/math].

By the above theorem

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)!=\sum_{k=0}^n(-1)^k{n\choose k}(n-k)!=\sum_{k=0}^n(-1)^k\frac{n!}{k!}=n!\sum_{k=0}^n(-1)^k\frac{1}{k!}\approx\frac{n!}{e}. }[/math]

Problème des ménages

Suppose that in a banquet, we want to seat [math]\displaystyle{ n }[/math] couples at a circular table, satisfying the following constraints:

  • Men and women are in alternate places.
  • No one sits next to his/her spouse.

In how many ways can this be done?

(For convenience, we assume that every seat at the table marked differently so that rotating the seats clockwise or anti-clockwise will end up with a different solution.)

First, let the [math]\displaystyle{ n }[/math] ladies find their seats. They may either sit at the odd numbered seats or even numbered seats, in either case, there are [math]\displaystyle{ n! }[/math] different orders. Thus, there are [math]\displaystyle{ 2(n!) }[/math] ways to seat the [math]\displaystyle{ n }[/math] ladies.

After sitting the wives, we label the remaining [math]\displaystyle{ n }[/math] places clockwise as [math]\displaystyle{ 0,1,\ldots, n-1 }[/math]. And a seating of the [math]\displaystyle{ n }[/math] husbands is given by a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ [n] }[/math] defined as follows. Let [math]\displaystyle{ \pi(i) }[/math] be the seat of the husband of he lady sitting at the [math]\displaystyle{ i }[/math]-th place.

It is easy to see that [math]\displaystyle{ \pi }[/math] satisfies that [math]\displaystyle{ \pi(i)\neq i }[/math] and [math]\displaystyle{ \pi(i)\not\equiv i+1\pmod n }[/math], and every permutation [math]\displaystyle{ \pi }[/math] with these properties gives a feasible seating of the [math]\displaystyle{ n }[/math] husbands. Thus, we only need to count the number of permutations [math]\displaystyle{ \pi }[/math] such that [math]\displaystyle{ \pi(i)\not\equiv i, i+1\pmod n }[/math].

Take [math]\displaystyle{ B=\{(0,0),(1,1),\ldots,(n-1,n-1), (0,1),(1,2),\ldots,(n-2,n-1),(n-1,0)\} }[/math] as the chess board. A permutation [math]\displaystyle{ \pi }[/math] which defines a way of seating the husbands, is a placement of [math]\displaystyle{ n }[/math] non-attacking rooks such that none of them is in [math]\displaystyle{ B }[/math].

a b c d e f g h
8 a8 cross b8 cross c8 __ d8 __ e8 __ f8 __ g8 __ h8 __ 8
7 a7 __ b7 cross c7 cross d7 __ e7 __ f7 __ g7 __ h7 __ 7
6 a6 __ b6 __ c6 cross d6 cross e6 __ f6 __ g6 __ h6 __ 6
5 a5 __ b5 __ c5 __ d5 cross e5 cross f5 __ g5 __ h5 __ 5
4 a4 __ b4 __ c4 __ d4 __ e4 cross f4 cross g4 __ h4 __ 4
3 a3 __ b3 __ c3 __ d3 __ e3 __ f3 cross g3 cross h3 __ 3
2 a2 __ b2 __ c2 __ d2 __ e2 __ f2 __ g2 cross h2 cross 2
1 a1 cross b1 __ c1 __ d1 __ e1 __ f1 __ g1 __ h1 cross 1
a b c d e f g h

We need to compute [math]\displaystyle{ r_k }[/math], the number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math]. For our choice of [math]\displaystyle{ B }[/math], [math]\displaystyle{ r_k }[/math] is the number of ways of choosing [math]\displaystyle{ k }[/math] points, no two consecutive, from a collection of [math]\displaystyle{ 2n }[/math] points arranged in a circle.

We first see how to do this in a line.

Lemma
The number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive objects from a collection of [math]\displaystyle{ m }[/math] objects arranged in a line, is [math]\displaystyle{ {m-k+1\choose k} }[/math].
Proof.

We draw a line of [math]\displaystyle{ m-k }[/math] black points, and then insert [math]\displaystyle{ k }[/math] red points into the [math]\displaystyle{ m-k+1 }[/math] spaces between the black points (including the beginning and end).

[math]\displaystyle{ \begin{align} &\sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \\ &\qquad\qquad\qquad\quad\Downarrow\\ &\sqcup \, \bullet \,\, {\color{Red}\bullet} \, \bullet \,\, {\color{Red}\bullet} \, \bullet \, \sqcup \, \bullet \,\, {\color{Red}\bullet}\, \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \,\, {\color{Red}\bullet} \end{align} }[/math]

This gives us a line of [math]\displaystyle{ m }[/math] points, and the red points specifies the chosen objects, which are non-consecutive. The mapping is 1-1 correspondence. There are [math]\displaystyle{ {m-k+1\choose k} }[/math] ways of placing [math]\displaystyle{ k }[/math] red points into [math]\displaystyle{ m-k+1 }[/math] spaces.

[math]\displaystyle{ \square }[/math]

The problem of choosing non-consecutive objects in a circle can be reduced to the case that the objects are in a line.

Lemma
The number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive objects from a collection of [math]\displaystyle{ m }[/math] objects arranged in a circle, is [math]\displaystyle{ \frac{m}{m-k}{m-k\choose k} }[/math].
Proof.

Let [math]\displaystyle{ f(m,k) }[/math] be the desired number; and let [math]\displaystyle{ g(m,k) }[/math] be the number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive points from [math]\displaystyle{ m }[/math] points arranged in a circle, next coloring the [math]\displaystyle{ k }[/math] points red, and then coloring one of the uncolored point blue.

Clearly, [math]\displaystyle{ g(m,k)=(m-k)f(m,k) }[/math].

But we can also compute [math]\displaystyle{ g(m,k) }[/math] as follows:

  • Choose one of the [math]\displaystyle{ m }[/math] points and color it blue. This gives us [math]\displaystyle{ m }[/math] ways.
  • Cut the circle to make a line of [math]\displaystyle{ m-1 }[/math] points by removing the blue point.
  • Choose [math]\displaystyle{ k }[/math] non-consecutive points from the line of [math]\displaystyle{ m-1 }[/math] points and color them red. This gives [math]\displaystyle{ {m-k\choose k} }[/math] ways due to the previous lemma.

Thus, [math]\displaystyle{ g(m,k)=m{m-k\choose k} }[/math]. Therefore we have the desired number [math]\displaystyle{ f(m,k)=\frac{m}{m-k}{m-k\choose k} }[/math].

[math]\displaystyle{ \square }[/math]

By the above lemma, we have that [math]\displaystyle{ r_k=\frac{2n}{2n-k}{2n-k\choose k} }[/math]. Then apply the theorem of counting permutations with restricted positions,

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)!=\sum_{k=0}^n(-1)^k\frac{2n}{2n-k}{2n-k\choose k}(n-k)!. }[/math]

This gives the number of ways of seating the [math]\displaystyle{ n }[/math] husbands after the ladies are seated. Recall that there are [math]\displaystyle{ 2n! }[/math] ways of seating the [math]\displaystyle{ n }[/math] ladies. Thus, the total number of ways of seating [math]\displaystyle{ n }[/math] couples as required by problème des ménages is

[math]\displaystyle{ 2n!\sum_{k=0}^n(-1)^k\frac{2n}{2n-k}{2n-k\choose k}(n-k)!. }[/math]

The Euler totient function

Two integers [math]\displaystyle{ m, n }[/math] are said to be relatively prime if their greatest common diviser [math]\displaystyle{ \mathrm{gcd}(m,n)=1 }[/math]. For a positive integer [math]\displaystyle{ n }[/math], let [math]\displaystyle{ \phi(n) }[/math] be the number of positive integers from [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] that are relative prime to [math]\displaystyle{ n }[/math]. This function, called the Euler [math]\displaystyle{ \phi }[/math] function or the Euler totient function, is fundamental in number theory.

We now derive a formula for this function by using the principle of inclusion-exclusion.

Theorem (The Euler totient function)

Suppose [math]\displaystyle{ n }[/math] is divisible by precisely [math]\displaystyle{ r }[/math] different primes, denoted [math]\displaystyle{ p_1,\ldots,p_r }[/math]. Then

[math]\displaystyle{ \phi(n)=n\prod_{i=1}^r\left(1-\frac{1}{p_i}\right) }[/math].
Proof.

Let [math]\displaystyle{ U=\{1,2,\ldots,n\} }[/math] be the universe. The number of positive integers from [math]\displaystyle{ U }[/math] which is divisible by some [math]\displaystyle{ p_{i_1},p_{i_2},\ldots,p_{i_s}\in\{p_1,\ldots,p_r\} }[/math], is [math]\displaystyle{ \frac{n}{p_{i_1}p_{i_2}\cdots p_{i_s}} }[/math].

[math]\displaystyle{ \phi(n) }[/math] is the number of integers from [math]\displaystyle{ U }[/math] which is not divisible by any [math]\displaystyle{ p_1,\ldots,p_r }[/math]. By principle of inclusion-exclusion,

[math]\displaystyle{ \begin{align} \phi(n) &=n+\sum_{k=1}^r(-1)^k\sum_{1\le i_1\lt i_2\lt \cdots \lt i_k\le n}\frac{n}{p_{i_1}p_{i_2}\cdots p_{i_k}}\\ &=n-\sum_{1\le i\le n}\frac{n}{p_i}+\sum_{1\le i\lt j\le n}\frac{n}{p_i p_j}-\sum_{1\le i\lt j\lt k\le n}\frac{n}{p_{i} p_{j} p_{k}}+\cdots + (-1)^r\frac{n}{p_{1}p_{2}\cdots p_{r}}\\ &=n\left(1-\sum_{1\le i\le n}\frac{1}{p_i}+\sum_{1\le i\lt j\le n}\frac{1}{p_i p_j}-\sum_{1\le i\lt j\lt k\le n}\frac{1}{p_{i} p_{j} p_{k}}+\cdots + (-1)^r\frac{1}{p_{1}p_{2}\cdots p_{r}}\right)\\ &=n\prod_{i=1}^r\left(1-\frac{1}{p_i}\right). \end{align} }[/math]
[math]\displaystyle{ \square }[/math]

Reference

  • Stanley, Enumerative Combinatorics, Volume 1, Chapter 2.
  • van Lin and Wilson, A course in combinatorics, Chapter 10, 25.
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