MAS334 Combinatorics 4 The pigeonhole principle 6 Matching problems

5. The inclusion-exclusion principle

Problem 5.1.

Consider a sports club in which people can play tennis or squash. Some members play both sports, some play only one, and some just drink at the bar. Suppose that $10$ members play tennis (and maybe squash as well), $12$ members play squash (and maybe tennis as well), and $3$ members play both sports. How many members play at least one sport?

Solution.

Interactive demo

Let $M_{t}$ be the set of members who play tennis, and let $M_{s}$ be the set of members who play squash. We are given that $|M_{t}|=10$ and $|M_{s}|=12$ and $|M_{s}\cap M_{t}|=3$ , and we need to find $|M_{s}\cup M_{t}|$ . For this, we need to fill in the numbers in the following Venn diagram:

Region $2$ is $M_{t}\cap M_{s}$ , which has $3$ elements, so we write $3$ there. Regions $1$ and $2$ together make up $M_{t}$ , which has $10$ elements, so we need $10-3=7$ elements in region $1$ to make the total correct. Regions $2$ and $3$ together make up $M_{s}$ , which has $12$ elements, so we need $12-3=9$ elements in region $3$ to make the total correct. Now $M_{t}\cup M_{s}$ consists of regions $1$ , $2$ and $3$ , so it has $7+3+9=19$ elements in total. In other words, there are $19$ members who play at least one sport.

Another way to describe the solution is as follows: we could just take $|M_{t}|+|M_{s}|$ , but that would count the people who play both sports twice, once as members of $M_{t}$ , and once more as members of $M_{s}$ . To compensate for this, we need to subtract the number of people who play both sports, which is $|M_{t}\cap M_{s}|$ . This gives

|M_{t}\cup M_{s}|=|M_{t}|+|M_{s}|-|M_{t}\cap M_{s}|=10+12-3=19

as before.

Problem 5.2.

Now suppose instead that we have a club that offers tennis (t), squash (s) and badminton (b). We are given the following data:

10 members play $t$	5 members play $t$ and $s$
15 members play $s$	4 members play $t$ and $b$
12 members play $b$	3 members play $s$ and $b$
2 members play $t$ , $s$ and $b$	There are 40 members altogether.

How many members play no sport at all?

Solution.

Interactive demo

This time we will give a more algebraic explanation. We write $M$ for the set of all members, and $m=|M|$ . We write $M_{t}$ for the set of members who play $t$ , and $m_{t}=|M_{t}|$ for the number of players who play $t$ , and similarly for $M_{s}$ , $M_{tb}$ and so on. The initial data is then as follows:

$\displaystyle m_{t}$	$\displaystyle=10$	$\displaystyle m_{ts}$	$\displaystyle=5$
$\displaystyle m_{s}$	$\displaystyle=15$	$\displaystyle m_{tb}$	$\displaystyle=4$
$\displaystyle m_{b}$	$\displaystyle=12$	$\displaystyle m_{sb}$	$\displaystyle=3$
$\displaystyle m_{tsb}$	$\displaystyle=2$	$\displaystyle m$	$\displaystyle=40.$

Now let $M_{t}^{*}$ be the set of members who play $t$ and nothing else, and let $M_{ts}^{*}$ be the set of members who play $t$ and $s$ and nothing else, and so on. To complete the pattern, we put $M^{*}_{tsb}=M_{tsb}$ , and we write $M^{*}$ for the set of members who play no sport at all, so our problem is to find $|M^{*}|$ . The people who play $t$ can be divided into four groups:

•

Those that play $t$ and nothing else
•

Those that play $t$ and $s$ but not $b$
•

Those that play $t$ and $b$ but not $s$
•

Those that play $t$ and $s$ and $b$ .

From this we get $M_{s}=M_{s}^{*}\cup M_{ts}^{*}\cup M_{tb}^{*}\cup M_{tsb}^{*}$ , and these sets do not overlap, so we get $m_{s}=m_{s}^{*}+m_{ts}^{*}+m_{tb}^{*}+m_{tsb}^{*}$ . By a similar analysis, we get the following equations:

	$\displaystyle m_{tsb}^{*}$	$\displaystyle=m_{tsb}=2$
	$\displaystyle m_{ts}^{}+m_{tsb}^{}$	$\displaystyle=m_{ts}=5$
	$\displaystyle m_{tb}^{}+m_{tsb}^{}$	$\displaystyle=m_{tb}=4$
	$\displaystyle m_{sb}^{}+m_{tsb}^{}$	$\displaystyle=m_{sb}=3$
	$\displaystyle m_{t}^{}+m_{st}^{}+m_{tb}^{}+m_{tsb}^{}$	$\displaystyle=m_{t}=10$
	$\displaystyle m_{s}^{}+m_{st}^{}+m_{sb}^{}+m_{tsb}^{}$	$\displaystyle=m_{s}=15$
	$\displaystyle m_{b}^{}+m_{tb}^{}+m_{sb}^{}+m_{tsb}^{}$	$\displaystyle=m_{b}=12$
	$\displaystyle m^{}+m_{t}^{}+m_{s}^{}+m_{b}^{}+m_{ts}^{}+m_{tb}^{}+m_{sb}% ^{}+m_{tsb}^{}$	$\displaystyle=m=40.$

These equations are easily solved to give

	$\displaystyle m_{tsb}^{*}$	$\displaystyle=m_{tsb}=2$
	$\displaystyle m_{ts}^{*}$	$\displaystyle=m_{ts}-m_{tsb}=5-2=3$
	$\displaystyle m_{tb}^{*}$	$\displaystyle=m_{tb}-m_{tsb}=4-2=2$
	$\displaystyle m_{sb}^{*}$	$\displaystyle=m_{sb}-m_{tsb}=3-2=1$
	$\displaystyle m_{t}^{*}$	$\displaystyle=m_{t}-m_{ts}-m_{tb}+m_{tsb}=10-5-4+2=3$
	$\displaystyle m_{s}^{*}$	$\displaystyle=m_{s}-m_{ts}-m_{sb}+m_{tsb}=15-5-3+2=9$
	$\displaystyle m_{b}^{*}$	$\displaystyle=m_{b}-m_{tb}-m_{sb}+m_{tsb}=12-4-3+2=7$
	$\displaystyle m^{*}$	$\displaystyle=m-m_{t}-m_{s}-m_{b}+m_{ts}+m_{tb}+m_{sb}-m_{tsb}=40-10-15-12+5+4% +3-2=13.$

In particular, we have $m^{*}=13$ , so there are $13$ members who play no sport; this answers the original question.

We now want to discuss the Inclusion-Exclusion Principle (IEP), which generalises the last two problems. Suppose we have a finite set $B$ , together with a family of subsets $B_{a}\subseteq B$ for each $a$ in some set $A$ of labels. (For example, in Problem 5.2 we have a set $B=M$ , together with subsets $M_{s}$ , $M_{t}$ and $M_{b}$ , indexed by the set $A=\{s,t,b\}$ of available sports.) We put

	$\displaystyle B^{\prime}$	$\displaystyle=\{b\in B\;\|\;b\text{ lies in at least one of the sets }B_{a}\}$
	$\displaystyle B^{*}$	$\displaystyle=B\setminus B^{\prime}=\{b\in B\;\|\;b\text{ lies in none of the % sets }B_{a}\}.$

In the common case where $A=\{1,\dotsc,n\}$ , this can be written as

	$\displaystyle B^{\prime}$	$\displaystyle=B_{1}\cup B_{2}\cup\dotsb\cup B_{n}$
	$\displaystyle B^{*}$	$\displaystyle=B\setminus(B_{1}\cup B_{2}\cup\dotsb\cup B_{n}).$

The IEP tells us about $|B^{\prime}|$ and $|B^{*}|$ . To formulate it, we use the following notation. Given a subset $I\subseteq A$ , we put $B_{I}=\bigcap_{i\in I}B_{i}$ . This means that

	$\displaystyle B_{\{i\}}$	$\displaystyle=B_{i}$
	$\displaystyle B_{\{i,j\}}$	$\displaystyle=B_{i}\cap B_{j}$
	$\displaystyle B_{\{i,j,k\}}$	$\displaystyle=B_{i}\cap B_{j}\cap B_{k}$

and so on. For the case $I=\emptyset$ , we interpret this as $B_{\emptyset}=B$ . We will often abbreviate the notation, by writing $B_{ijk}$ for $B_{\{i,j,k\}}$ and so on.

Theorem 5.3.

For $B$ and $B_{i}$ as above, we have

	$\displaystyle\|B^{*}\|$	$\displaystyle=\sum_{I\subseteq A}(-1)^{\|I\|}\|B_{I}\|$
	$\displaystyle\|B^{\prime}\|$	$\displaystyle=\sum_{I\neq\emptyset}(-1)^{\|I\|+1}\|B_{I}\|.$

In the case where $A=\{1,\dotsc,n\}$ , this can be written as

	$\displaystyle\|B^{*}\|$	$\displaystyle=\|B\|-\|B_{1}\|-\dotsb-\|B_{n}\|+\|B_{12}\|+\dotsb+\|B_{n-1,n}\|-\dotsb\pm% \|B_{12\dotsb n}\|$
	$\displaystyle\|B^{\prime}\|$	$\displaystyle=\|B_{1}\|+\dotsb+\|B_{n}\|-\|B_{12}\|-\dotsb-\|B_{n-1,n}\|+\dotsb\mp\|B_{% 12\dotsb n}\|$

There is a single video covering the statement and proof of the IEP, together with two lemmas required for the proof:

Video

The equation for $|B^{*}|$ is called the negative form of the IEP, and the equation for $|B^{\prime}|$ is called the positive form. Because $B^{*}=B\setminus B^{\prime}$ we have $|B^{*}|=|B|-|B^{\prime}|$ , which makes it easy to see that the two forms are equivalent. For $n=2$ and $n=3$ the equations are as follows:

	$\displaystyle\|B_{1}\cup B_{2}\|$	$\displaystyle=\|B_{1}\|+\|B_{2}\|-\|B_{12}\|$
	$\displaystyle\|B\setminus(B_{1}\cup B_{2})\|$	$\displaystyle=\|B\|-\|B_{1}\|-\|B_{2}\|+\|B_{12}\|$
	$\displaystyle\|B_{1}\cup B_{2}\cup B_{3}\|$	$\displaystyle=\|B_{1}\|+\|B_{2}\|+\|B_{3}\|-\|B_{12}\|-\|B_{13}\|-\|B_{23}\|+\|B_{123}\|$
	$\displaystyle\|B\setminus(B_{1}\cup B_{2})\|$	$\displaystyle=\|B\|-\|B_{1}\|-\|B_{2}\|-\|B_{3}\|+\|B_{12}\|+\|B_{13}\|+\|B_{23}\|-\|B_{123}\|.$

Problem 5.1 is just an example of the positive IEP with $n=2$ . Problem 5.2 is an example of the negative IEP with $n=3$ .

We will prove the IEP after some preliminary discussion.

Lemma 5.4.

Let $I$ be a finite set, and consider the sum $s=\sum_{J\subseteq I}(-1)^{|J|}$ . Then $s=1$ if $I$ is empty, and $s=0$ if $I$ is not empty.

Proof.

If $I$ is empty, then the only term in the sum is for $J=\emptyset$ , and that term is $(-1)^{0}=1$ , so $s=1$ . Suppose instead that $I\neq\emptyset$ , and put $n=|I|>0$ . Then there are $\binom{n}{k}$ possible choices of $J$ with $|J|=k$ , and this gives $s=\sum_{k}\binom{n}{k}(-1)^{k}$ . This is just the binomial expansion of $(1-1)^{n}=0^{n}=0$ , so $s=0$ . Alternatively, we can choose an element $a\in I$ , and put $I^{\prime}=I\setminus\{a\}$ . For every $J^{\prime}\subseteq I^{\prime}$ we have a term $(-1)^{|J^{\prime}|}$ in $s$ for $J=J^{\prime}$ , and another term $(-1)^{1+|J^{\prime}|}$ for $J=J^{\prime}\cup\{a\}$ , and these terms cancel out. All the terms cancel in pairs in this way, so we are left with $s=0$ . ∎

Definition 5.5.

In the context of the IEP, for an element $b\in B$ , we put

A\langle b\rangle=\{a\in A\;|\;b\in B_{a}\}\subseteq A.

For example, consider a member $x$ of the club in Problem 5.2. Then $A\langle x\rangle$ is just the set of sports that $x$ plays. For example, if $x$ plays tennis and badminton but not squash, then $A\langle x\rangle=\{t,b\}$ .

•

Consider the set $I=\{t,s,b\}$ and the corresponding set $M_{I}=M_{tsb}=M_{t}\cap M_{s}\cap M_{b}$ . Member $x$ does not lie in this set $M_{I}$ , because $x$ does not play s. Here $I\not\subseteq A\langle x\rangle=\{t,b\}$ .
•

Consider instead the set $I=\{t,s\}$ and the corresponding set $M_{I}=M_{ts}=M_{t}\cap M_{s}$ . Member $x$ does not lie in this set $M_{I}$ , because $x$ does not play s. Here $I\not\subseteq A\langle x\rangle=\{t,b\}$ .
•

Now consider the set $I=\{t,b\}$ and the corresponding set $M_{I}=M_{tb}=M_{t}\cap M_{b}$ . Member $x$ does lie in this set $M_{I}$ , because $x$ does not play both $t$ and b. Here $I\subseteq A\langle x\rangle=\{t,b\}$ .
•

Similarly, if $I=\{b\}$ then $x$ does lie in the set $M_{I}=M_{b}$ , and again $I\subseteq A\langle x\rangle$ .
•

For an arbitrary subset $I\subseteq A=\{t,s,b\}$ we find that $x\in M_{I}$ iff $x$ plays all the sports in $I$ iff $I\subseteq K\langle x\rangle$ .

We record the obvious generalisation as a lemma:

Lemma 5.6.

Suppose we have a family of subsets $(B_{a})_{a\in A}$ as before, and an element $b\in B$ , and a subset $I\subseteq A$ . Then $b\in B_{I}$ iff $I\subseteq A\langle b\rangle$ .

Proof.

By definition $B_{I}=\bigcap_{i\in I}B_{i}$ , so $b\in B_{I}$ iff $b\in B_{i}$ for all $i\in I$ . However, we have $b\in B_{i}$ iff $i\in A\langle b\rangle$ , by the definition of $A\langle b\rangle$ . Thus, we can say that $b\in B_{I}$ iff for all $i\in I$ , we have $i\in A\langle b\rangle$ . This is clearly equivalent to the condition $I\subseteq A\langle b\rangle$ . ∎

Lemma 5.7.

$A\langle b\rangle$ is empty iff $b\in B^{*}$ .

Proof.

We have $i\in A\langle b\rangle$ iff $b\in B_{i}$ . Thus $A\langle b\rangle$ is empty iff the condition $b\in B_{i}$ is false for all $i$ , which means that $b$ lies in none of the sets $B_{i}$ , which means that $b\in B^{*}$ . ∎

Proof of Theorem 5.3.

Put $u=\sum_{I\subseteq A}(-1)^{|I|}|B_{I}|$ . We need to prove that this is the same as $|B^{*}|$ . We have $|B_{I}|=\sum_{b\in B_{I}}1$ , so we can rewrite the definition of $u$ as

u=\sum_{I\subseteq A}\sum_{b\in B_{I}}(-1)^{|I|}.

Lemma 5.6 tells us that $b\in B_{I}$ iff $I\subseteq A\langle b\rangle$ , so we can regroup this sum as

u=\sum_{b\in B}\sum_{I\subseteq A\langle b\rangle}(-1)^{|I|}.

Now Lemma 5.4 tells us that $\sum_{I\subseteq A\langle b\rangle}(-1)^{|I|}$ is zero if $A\langle b\rangle\neq\emptyset$ , but is $1$ if $A\langle b\rangle=\emptyset$ . Using Lemma 5.7, we therefore see that $\sum_{I\subseteq A\langle b\rangle}(-1)^{|I|}$ is zero if $b\not\in B^{*}$ , but is $1$ if $b\in B^{*}$ . We now have

u=\sum_{b\in B^{*}}1=|B^{*}|,

as required. This proves the negative form of the IEP. For the positive form, we note that

|B^{\prime}|=|B|-|B^{*}|=|B|-\sum_{I\subseteq A}(-1)^{|I|}|B_{I}|.

The term for $I=\emptyset$ in the sum cancels out the extra term of $|B|$ outside the sum. We can also bring the minus sign inside the sum to get

|B^{\prime}|=\sum_{I\neq\emptyset}(-1)^{|I|+1}|B_{I}|.

∎

Definition 5.8.

Let $S_{n}$ be the set of all permutations of the set $N=\{1,\dotsc,n\}$ , so $|S_{n}|=n!$ . A derangement of $\{1,\dotsc,n\}$ is a permutation $\sigma\in S_{n}$ with the property that for all $i$ we have $\sigma(i)\neq i$ . We write $D_{n}$ for the set of derangements, so $D_{n}\subseteq S_{n}$ . We also write $p_{n}=|D_{n}|/|S_{n}|$ (which is the probability that a randomly chosen permutation is a derangement).

Example 5.9.

This picture lists all $24$ possible permutations of the set $N=\{1,2,3,4\}$ . For example, the top right box contains $1432$ , which refers to the permutation sending $1$ , $2$ $3$ and $4$ to $1$ , $4$ , $3$ and $2$ respectively. (In disjoint cycle notation, this would be $(2\;\;4)$ .) The numbers $1$ and $3$ are sent to themselves, so they are underlined. As some numbers are sent to themselves, this is not a derangement. However, in the bottom right box we have the permutation $4321$ . This does not send anything to itself, so no numbers are underlined, and we have a derangement. All the derangements are circled; there are $9$ of them. Thus, the fraction of derangements is $p_{4}=9/24=3/8=0.375$ .

Example 5.10.

Suppose that $n$ people arrive at a party, each wearing a hat. At the end of the party, no one can remember which hat they brought, so they pick one up at random. This means that guest $i$ picks up the hat belonging to guest $\sigma(i)$ , for some randomly chosen permutation $\sigma$ . This permutation is a derangement iff no one gets the right hat. Thus, the probability that no one gets the right hat is $p_{n}$ . How does this change as $n$ increases? For any given guest, there are more hats to choose from, so the probability of getting the right hat goes down. On the other hand, as there are more guests, there are more chances for at least one guest to get the right hat. It is not obvious how these competing effects balance out, but the answer is given by our next result.

Proposition 5.11.

$p_{n}=\sum_{k=0}^{n}(-1)^{k}/k!$ , and this converges to $e^{-1}\simeq 0.368$ as $n\to\infty$ .

Proof.

Interactive demo

Put

	$\displaystyle N$	$\displaystyle=\{1,\dotsc,n\}$
	$\displaystyle P$	$\displaystyle=S_{n}=\{\text{ all permutations of }N\}$
	$\displaystyle P_{i}$	$\displaystyle=\{\sigma\in P\;\|\;\sigma(i)=i\}=\{\text{ permutations that fix }% i\}.$

Note that a permutation is a derangement iff it lies in none of the sets $P_{i}$ , so $D_{n}=P^{*}$ in the usual notation of the IEP. We therefore have

	$\displaystyle\|D_{n}\|$	$\displaystyle=\|P^{*}\|=\sum_{I\subseteq N}(-1)^{\|I\|}\|P_{I}\|$
	$\displaystyle p_{n}$	$\displaystyle=n!^{-1}\|D_{n}\|=\sum_{I\subseteq N}(-1)^{\|I\|}\|P_{I}\|/n!.$

Here $P_{I}$ is the set of permutations $\sigma$ that fix all the elements of $I$ , but are free to permute the remaining elements of $N\setminus I$ in any way. If $|I|=k$ we have $|N\setminus I|=n-k$ so there are $(n-k)!$ possible permutations of $N\setminus I$ . This means that $|P_{I}|=(n-k)!$ . On the other hand, there are $\binom{n}{k}$ possible choices of $I$ with $|I|=k$ . Putting this together, we get

p_{n}=\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}\frac{(n-k)!}{n!}.

However, we also have

\binom{n}{k}\frac{(n-k)!}{n!}=\frac{n!}{k!(n-k)!}\frac{(n-k)!}{n!}=\frac{1}{k!},

so our previous expression simplifies to $p_{n}=\sum_{k=0}^{n}(-1)^{k}/k!$ as claimed. As $n$ tends to infinity, this converges to $\sum_{k=0}^{\infty}(-1)^{k}/k!$ , which is $e^{-1}$ by the standard Taylor series for $e^{x}$ . ∎

Problem 5.12.

Of the numbers $0,1,\dotsc,41$ , how many are coprime with $42$ ?

Solution.

Interactive demo

Put $D=\{0,\dotsc,41\}$ , and let $U$ be the subset of numbers that are coprime with $42$ . We need to find $|U|$ . Put $P=\{2,3,7\}$ , which is the set of primes that divide $42$ . For any $p\in P$ , put

D_{p}=\{i\in D\;|\;i\text{ is divisible by }p\}.

In the standard notation for the IEP, we have $U=D^{*}$ , and so $|U|=\sum_{I\subseteq P}(-1)^{|I|}|D_{I}|$ . We therefore need to understand $|D_{I}|$ . Let $q_{I}$ be the product of the primes in $I$ (to be interpreted as $q_{I}=1$ in the case $I=\emptyset$ ). We note that $q_{I}$ divides $42$ , and $D_{I}$ is the set of multiples of $q_{I}$ in $D$ , so $D_{I}=\{k\,q_{I}\;|\;0\leq k<42/q_{I}\}$ and $|D_{I}|=42/q_{I}$ . We now have

|U|=42\sum_{I\subseteq P}\frac{(-1)^{I}}{q_{I}}=42\left(1-\frac{1}{2}-\frac{1}% {3}-\frac{1}{7}+\frac{1}{2\times 3}+\frac{1}{2\times 7}+\frac{1}{3\times 7}-% \frac{1}{2\times 3\times 7}\right).

It is not hard to see that this factors as

|U|=42\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{7}% \right)=12.

Alternatively, we could say that the proportion of coprime numbers is $|U|/|D|=(1-\tfrac{1}{2})(1-\tfrac{1}{3})(1-\tfrac{1}{7})=2/7$ .

The following more general statement can be proved in the same way:

Proposition 5.13.

Consider an integer $m>1$ , and let $P$ be the set of primes that divide $m$ . Put $D=\{0,1,\dotsc,m-1\}$ , and let $x$ be the proportion of numbers in $D$ that are coprime with $m$ . Then

x=\prod_{p\in P}(1-p^{-1}).\qed

	$\displaystyle\|B^{*}\|$	$\displaystyle=\sum_{I\subseteq A}(-1)^{\|I\|}\|B_{I}\|$
	$\displaystyle\|B^{\prime}\|$	$\displaystyle=\sum_{I\neq\emptyset}(-1)^{\|I\|+1}\|B_{I}\|.$

	$\displaystyle\|B^{*}\|$	$\displaystyle=\|B\|-\|B_{1}\|-\dotsb-\|B_{n}\|+\|B_{12}\|+\dotsb+\|B_{n-1,n}\|-\dotsb\pm% \|B_{12\dotsb n}\|$
	$\displaystyle\|B^{\prime}\|$	$\displaystyle=\|B_{1}\|+\dotsb+\|B_{n}\|-\|B_{12}\|-\dotsb-\|B_{n-1,n}\|+\dotsb\mp\|B_{% 12\dotsb n}\|$

	$\displaystyle\|B_{1}\cup B_{2}\|$	$\displaystyle=\|B_{1}\|+\|B_{2}\|-\|B_{12}\|$
	$\displaystyle\|B\setminus(B_{1}\cup B_{2})\|$	$\displaystyle=\|B\|-\|B_{1}\|-\|B_{2}\|+\|B_{12}\|$
	$\displaystyle\|B_{1}\cup B_{2}\cup B_{3}\|$	$\displaystyle=\|B_{1}\|+\|B_{2}\|+\|B_{3}\|-\|B_{12}\|-\|B_{13}\|-\|B_{23}\|+\|B_{123}\|$
	$\displaystyle\|B\setminus(B_{1}\cup B_{2})\|$	$\displaystyle=\|B\|-\|B_{1}\|-\|B_{2}\|-\|B_{3}\|+\|B_{12}\|+\|B_{13}\|+\|B_{23}\|-\|B_{123}\|.$

	$\displaystyle\|D_{n}\|$	$\displaystyle=\|P^{*}\|=\sum_{I\subseteq N}(-1)^{\|I\|}\|P_{I}\|$
	$\displaystyle p_{n}$	$\displaystyle=n!^{-1}\|D_{n}\|=\sum_{I\subseteq N}(-1)^{\|I\|}\|P_{I}\|/n!.$