Inclusion–exclusion principle

In combinatorial mathematics, the inclusion–exclusion principle (also known as the sieve principle) states that if A₁, ..., A_n are finite sets, then

${\begin{aligned}{\biggl |}\bigcup _{i=1}^{n}A_{i}{\biggr |}&{}=\sum _{i=1}^{n}\left|A_{i}\right|-\sum _{i,j\,:\,1\leq i<j\leq n}\left|A_{i}\cap A_{j}\right|\\&{}\qquad +\sum _{i,j,k\,:\,1\leq i<j<k\leq n}\left|A_{i}\cap A_{j}\cap A_{k}\right|-\ \cdots \ +\left(-1\right)^{n-1}\left|A_{1}\cap \cdots \cap A_{n}\right|\end{aligned}}$

where |A| denotes the cardinality of the set A. For example, taking n = 2, we get a special case of double counting; in words: we can count the size of the union of sets A and B by adding |A| and |B| and then subtracting the size of their intersection. The name comes from the idea that the principle is based on over-generous inclusion, followed by compensating exclusion. When n > 2 the exclusion of the pairwise intersections is (possibly) too severe, and the correct formula is as shown with alternating signs.

This formula is attributed to Abraham de Moivre; it is sometimes also named for Daniel da Silva, Joseph Sylvester or Henri Poincaré.

Inclusion–exclusion illustrated for three sets

Counts of each region with progressively more terms used for n = 4

For the case of three sets A, B, C the inclusion–exclusion principle is illustrated in the graphic on the right.

Proof

Let A denote the union $\scriptstyle \cup _{i=1}^{n}A_{i}$ of the sets A₁, ..., A_n. To prove the inclusion–exclusion principle in general, we first have to verify the identity

$1_{A}=\sum _{k=1}^{n}(-1)^{k-1}\sum _{\scriptstyle I\subset \{1,\ldots ,n\} \atop \scriptstyle |I|=k}1_{A_{I}}\qquad (*)$

for indicator functions, where

$A_{I}=\bigcap _{i\in I}A_{i}.$

There are at least two ways to do this:

First possibility: It suffices to do this for every x in the union of A₁, ..., A_n. Suppose x belongs to exactly m sets with 1 ≤ m ≤ n, for simplicity of notation say A₁, ..., A_m. Then the identity at x reduces to

$1=\sum _{k=1}^{m}(-1)^{k-1}\sum _{\scriptstyle I\subset \{1,\ldots ,m\} \atop \scriptstyle |I|=k}1.$

The number of subsets of cardinality k of an m-element set is the combinatorical interpretation of the binomial coefficient $\textstyle {\binom {m}{k}}$ . Since $\textstyle 1={\binom {m}{0}}$ , we have

${\binom {m}{0}}=\sum _{k=1}^{m}(-1)^{k-1}{\binom {m}{k}}.$

Putting all terms to the left-hand side of the equation, we obtain the expansion for (1 – 1)^m given by the binomial theorem, hence we see that (*) is true for x.

Second possibility: The following function is identically zero

$(1_{A}-1_{A_{1}})(1_{A}-1_{A_{2}})\cdots (1_{A}-1_{A_{n}})\,=\,0\,,$

because: if x is not in A, then all factors are 0 - 0 = 0; and otherwise, if x does belong to some A_m, then the corresponding mth factor is 1 - 1 = 0. By expanding the product on the right-hand side, equation (*) follows.

Use of (*): To prove the inclusion–exclusion principle for the cardinality of sets, sum the equation (*) over all x in the union of A₁, ..., A_n. To derive the version used in probability, take the expectation in (*). In general, integrate the equation (*) with respect to μ. Always use linearity.

Example

Suppose there is a deck of n cards, each card is numbered from 1 to n. Suppose a card numbered m is in the correct position if it is the mth card in the deck. How many ways, W, can the cards be shuffled with at least 1 card being in the correct position?

Begin by defining set A_m, which is all of the orderings of cards with the mth card correct. Then the number of orders, W, with at least one card being in the correct position, m, is

$W={\biggl |}\bigcup _{m=1}^{n}A_{m}{\biggr |}.$

Apply the principle of inclusion-exclusion,

${\begin{aligned}W&=\sum _{m_{1}=1}^{n}\left|A_{m_{1}}\right|\\&-\sum _{m_{1},m_{2}\,:\,1\leq m_{1}<m_{2}\leq n}\left|A_{m_{1}}\cap A_{m_{2}}\right|\\&+\sum _{m_{1},m_{2},m_{3}\,:\,1\leq m_{1}<m_{2}<m_{3}\leq n}\left|A_{m_{1}}\cap A_{m_{2}}\cap A_{m_{3}}\right|\\&-\cdots \\&+(-1)^{p-1}\sum _{m_{1},\ldots ,m_{p}\,:\,1\leq m_{1}<\cdots <m_{p}\leq n}\left|A_{m_{1}}\cap \cdots \cap A_{m_{p}}\right|\\&\cdots .\\\end{aligned}}$

Each value $A_{m_{1}}\cap \cdots \cap A_{m_{p}}$ represents the set of shuffles having p values m₁, ..., m_p in the correct position. Note that the number of shuffles with p values correct only depends on p, not on the particular values of $m$. For example, the number of shuffles having the 1st, 3rd, and 17th cards in the correct position is the same as the number of shuffles having the 2nd, 5th, and 13th cards in the correct positions. It only matters that of the n cards, 3 were chosen to be in the correct position. Thus there are ${n \choose p}$ terms in each summation (see combination).

${\begin{aligned}W&={n \choose 1}\left|A_{1}\right|\\&-{n \choose 2}\left|A_{1}\cap A_{2}\right|\\&+{n \choose 3}\left|A_{1}\cap A_{2}\cap A_{3}\right|\\&-\cdots \\&+(-1)^{p-1}{n \choose p}\left|A_{1}\cap \cdots \cap A_{p}\right|\\&\cdots .\\\end{aligned}}$

$\left|A_{1}\cap \cdots \cap A_{p}\right|$ is the number of orderings having p elements in the correct position, which is equal to the number of ways of ordering the remaining n − p elements, or (n − p)!. Thus we finally get:

${\begin{aligned}W&={n \choose 1}(n-1)!\\&-{n \choose 2}(n-2)!\\&+{n \choose 3}(n-3)!\\&-\cdots \\&+(-1)^{p-1}{n \choose p}(n-p)!\\&\cdots \\W&=\sum _{p=1}^{n}(-1)^{p-1}{n \choose p}(n-p)!.\\\end{aligned}}$

Noting that ${n \choose p}={\frac {n!}{p!(n-p)!}}$, this reduces to

${\begin{aligned}W&=\sum _{p=1}^{n}(-1)^{p-1}\,{\frac {n!}{p!}}.\end{aligned}}$

A permutation where no card is in the correct position is called a derangement. Taking n! to be the total number of permutations, the probability Q that a random shuffle produces a derangement is given by

$Q=1-{\frac {W}{n!}}=\sum _{p=0}^{n}{\frac {(-1)^{p}}{p!}},$

the Taylor expansion of e⁻¹. Thus the probability of guessing an order for a shuffled deck of cards and being incorrect about every card is approximately 1/e or 36%.

In probability

In probability, for events A₁, ..., A_n in a probability space $\scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )$, the inclusion–exclusion principle becomes for n = 2

$\mathbb {P} (A_{1}\cup A_{2})=\mathbb {P} (A_{1})+\mathbb {P} (A_{2})-\mathbb {P} (A_{1}\cap A_{2}),$

for n = 3

${\begin{aligned}\mathbb {P} (A_{1}\cup A_{2}\cup A_{3})&=\mathbb {P} (A_{1})+\mathbb {P} (A_{2})+\mathbb {P} (A_{3})\\&\qquad -\mathbb {P} (A_{1}\cap A_{2})-\mathbb {P} (A_{1}\cap A_{3})-\mathbb {P} (A_{2}\cap A_{3})\\&\qquad +\mathbb {P} (A_{1}\cap A_{2}\cap A_{3})\end{aligned}}$

and in general

${\begin{aligned}\mathbb {P} {\biggl (}\bigcup _{i=1}^{n}A_{i}{\biggr )}&{}=\sum _{i=1}^{n}\mathbb {P} (A_{i})-\sum _{i,j\,:\,i<j}\mathbb {P} (A_{i}\cap A_{j})\\&\qquad +\sum _{i,j,k\,:\,i<j<k}\mathbb {P} (A_{i}\cap A_{j}\cap A_{k})-\ \cdots \ +(-1)^{n-1}\,\mathbb {P} {\biggl (}\bigcap _{i=1}^{n}A_{i}{\biggr )},\end{aligned}}$

which can be written in closed form as

$\mathbb {P} {\biggl (}\bigcup _{i=1}^{n}A_{i}{\biggr )}=\sum _{k=1}^{n}(-1)^{k-1}\sum _{\scriptstyle I\subset \{1,\ldots ,n\} \atop \scriptstyle |I|=k}\mathbb {P} (A_{I}),$

where the last sum runs over all subsets I of the indices 1, ..., n which contain exactly k elements, and

$A_{I}:=\bigcap _{i\in I}A_{i}$

denotes the intersection of all those A_i with index in I.

According to the Bonferroni inequalities, the sum of the first terms in the formula is alternately an upper bound and a lower bound for the LHS. This can be used in cases where the full formula is too cumbersome.

For a general measure space (S,Σ,μ) and measurable subsets A₁, ..., A_n of finite measure, the above identities also hold when the probability measure $\mathbb {P}$ is replaced by the measure μ.

Special case

If, in the probabilistic version of the inclusion–exclusion principle, the probability of the intersection A_I only depends on the cardinality of I, meaning that for every k in {1, ..., n} there is an a_k such that

$a_{k}=\mathbb {P} (A_{I})\quad {\text{for every}}\quad I\subset \{1,\ldots ,n\}\quad {\text{with}}\quad |I|=k,$

then the above formula simplifies to

$\mathbb {P} {\biggl (}\bigcup _{i=1}^{n}A_{i}{\biggr )}=\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}a_{k}$

due to the combinatorial interpretation of the binomial coefficient $\scriptstyle {\binom {n}{k}}$.

Similarly, when the cardinality of the union of finite sets A₁, ..., A_n is of interest, and these sets are a family with regular intersections, meaning that for every k in {1, ..., n} the intersection

$A_{I}:=\bigcap _{i\in I}A_{i}$

has the same cardinality, say a_k = |A_I|, irrespective of the k-element subset I of {1, ..., n}, then

${\biggl |}\bigcup _{i=1}^{n}A_{i}{\biggr |}=\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}a_{k}\,.$

An analogous simplification is possible in the case of a general measure space (S,Σ,μ) and measurable subsets A₁, ..., A_n of finite measure.

Other forms

The principle is sometimes stated in the form that says that if

$g(A)=\sum _{S\,:\,S\subseteq A}f(S)$

then

$f(A)=\sum _{S\,:\,S\subseteq A}(-1)^{\left|A\right|-\left|S\right|}g(S)\qquad (**)$

We show now that the combinatorial and the probabilistic version of the inclusion-exclusion principle are instances of (**). Take ${\underline {m}}=\{1,2,\ldots ,m\}$, $f({\underline {m}})=0$, and

$f(S)={\bigg |}\bigcap _{i\in {\underline {m}}\backslash S}A_{i}{\bigg \backslash }\bigcup _{i\in S}A_{i}{\bigg |}\qquad {\text{and}}\qquad f(S)=\mathbb {P} {\bigg (}\bigcap _{i\in {\underline {m}}\backslash S}A_{i}{\bigg \backslash }\bigcup _{i\in S}A_{i}{\bigg )}$

respectively for all sets $S$ with $S\subsetneq {\underline {m}}$. Then we obtain

$g(A)={\bigg |}\bigcap _{i\in {\underline {m}}\backslash A}A_{i}{\bigg |},~~g({\underline {m}})={\bigg |}\bigcup _{i\in {\underline {m}}}A_{i}{\bigg |}\qquad {\text{and}}\qquad g(A)=\mathbb {P} {\bigg (}\bigcap _{i\in {\underline {m}}\backslash A}A_{i}{\bigg )},~~g({\underline {m}})=\mathbb {P} {\bigg (}\bigcup _{i\in {\underline {m}}}A_{i}{\bigg )}$

respectively for all sets $A$ with $A\subsetneq {\underline {m}}$. This is because elements $a$ of $\cap _{i\in {\underline {m}}\backslash A}A_{i}$ can be contained in other $A_{i}$'s ($A_{i}$'s with $i\in A$) as well, and the $\cap \backslash \!\!\cup \!{\text{-}}$formula runs exactly through all possible extensions of the sets $\{A_{i}\mid i\in {\underline {m}}\backslash A\}$ with other $A_{i}$'s, counting $a$ only for the set that matches the membership behavior of $a$, if $S$ runs through all subsets of $A$ (as in the definition of $g(A)$).

Since $f({\underline {m}})=0$, we obtain from (**) with $A={\underline {m}}$ that

$\sum _{T\,:\,{\underline {m}}\supseteq T\supsetneq \varnothing }(-1)^{\left|T\right|-1}g({\underline {m}}\backslash T)=\sum _{S\,:\,\varnothing \subseteq S\subsetneq {\underline {m}}}(-1)^{m-\left|S\right|-1}g(S)=g({\underline {m}})$

and by interchanging sides, the combinatorial and the probabilistic version of the inclusion-exclusion principle follow.

If one sees a number $n$ as a set of its prime factors, then (**) is a generalization of Möbius inversion formula for square-free natural numbers. Therefore, (**) is seen as the Möbius inversion formula for the incidence algebra of the partially ordered set of all subsets of A.

For a generalization of the full version of Möbius inversion formula, (**) must be generalized to multisets. For multisets instead of sets, (**) becomes

$f(A)=\sum _{S\,:\,S\subseteq A}\mu (A-S)g(S)\qquad (***)$

where $A-S$ is the multiset for which $(A-S)\uplus S=A$, and

• μ(S) = 1 if S is a set (i.e. a multiset without double elements) of even cardinality.
• μ(S) = -1 if S is a set (i.e. a multiset without double elements) of odd cardinality.
• μ(S) = 0 if S is a proper multiset (i.e. S has double elements).

Notice that $\mu$$(A-S)$ is just the $(-1)^{\left|A\right|-\left|S\right|}$ of (**) in case $A-S$ is a set.

Proof of (***): Substitute

$g(S)=\sum _{T\,:\,T\subseteq S}f(T)$

on the right hand side of (***). Notice that $f(A)$ appears once on both sides of (***). So we must show that for all $T$ with $T\subsetneq A$, the terms $f(T)$ cancel out on the right hand side of (***). For that purpose, take a fixed $T$ such that $T\subsetneq A$ and take an arbitrary fixed $a\in A$ such that $a\not \in T$.

Notice that $A-S$ must be a set for each positive or negative appearance of $f(T)$ on the right hand side of (***) that is obtained by way of the multiset $S$ such that $T\subseteq S\subseteq A$. Now each appearance of $f(T)$ on the right hand side of (***) that is obtained by way of $S$ such that $A-S$ is a set that contains $a$ cancels out with the one that is obtained by way of the corresponding $S$ such that $A-S$ is a set that does not contain $a$. This gives the desired result.

Applications

In many cases where the principle could give an exact formula (in particular, counting prime numbers using the sieve of Eratosthenes), the formula arising doesn't offer useful content because the number of terms in it is excessive. If each term individually can be estimated accurately, the accumulation of errors may imply that the inclusion–exclusion formula isn't directly applicable. In number theory, this difficulty was addressed by Viggo Brun. After a slow start, his ideas were taken up by others, and a large variety of sieve methods developed. These for example may try to find upper bounds for the "sieved" sets, rather than an exact formula.

Counting derangements

Main article: Derangement

A well-known application of the inclusion–exclusion principle is to the combinatorial problem of counting all derangements of a finite set. A derangement of a set A is a bijection from A into itself that has no fixed points. Via the inclusion–exclusion principle one can show that if the cardinality of A is n, then the number of derangements is [n! / e] where [x] denotes the nearest integer to x; a detailed proof is available here.

This is also known as the subfactorial of n, written !n. It follows that if all bijections are assigned the same probability then the probability that a random bijection is a derangement quickly approaches 1/e as n grows.

Counting intersections

The principle of inclusion–exclusion, combined with de Morgan's theorem, can be used to count the intersection of sets as well. Let $\scriptstyle {\overline {A}}_{k}$ represent the complement of A_k with respect to some universal set A such that $\scriptstyle A_{k}\,\subseteq \,A$ for each k. Then we have

$\bigcap _{i=1}^{n}A_{i}={\overline {\bigcup _{i=1}^{n}{\overline {A}}_{i}}}$

thereby turning the problem of finding an intersection into the problem of finding a union.

See also

• Combinatorial principles
• Boole's inequality
• Necklace problem
• Schuette–Nesbitt formula
• Maximum-minimums identity

References

principle of inclusion-exclusion at PlanetMath.