### Including and Excluding

Hi, I am Emon and I am back this time to provide you with some ideas regarding the ''Inclusion and Exclusion Principle", which is a really important topic that we should learn for olympiads.

The Inclusion-Exclusion Principle is a counting technique which generalizes the familiar method of obtaining the number of elements in the union of two finite sets, which is given by the formula

$$\left|A\cup B\right| = \left|A\right|+\left|B\right| - \left|A\cap B\right|.$$

where $A$ and $B$ are two finite sets and $|X|:=$ cardinality of set $X$, $i.e.$ the number of elements present in the set $X$.

The proof of this formula can be understood directly from the above venn diagram.

Additional Note (though a bit out of context). Let $\mathcal{S}$ be a discrete sample space. Then, if $A, B\subset \mathcal{S}$ are two events, then we may write

$$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B),$$ which is equivalent to the previous formula $$|A\cup B|=|A|+|B|-|A\cap B|.$$

Similarly, for $3$ finite sets, we will have,

$$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|.$$

The proof of this is also direct from the venn diagram.

Now, let's state the original, generalised version of the Inclusion-Exclusion Principle.

Formula. If $A_1, A_2, \ldots, A_n$ be $n$ finite sets. Then, we can write
$$|A_1\cup A_2\cup \cdots \cup A_n|=\sum_{i=1}^n|A_i| - \sum_{i<j}|A_i\cap A_j|+\cdots+(-1)^{n+1}|A_1\cap A_2\cap \cdots \cap A_n|.$$

$Proof.$ You may look into the proofs here or here

Derangements
Consider the following 2 examples.

Example 1. A postman has to deliver $n$ letters in $n$ houses. Suppose the letter $\ell_i$ should be delivered at house $h_i$. But, the postman, being a naughty one, decides to deliver the letters in such a way that $\ell_i$ is not delivered to $h_i$ for all $i$. In how many ways can he do so?
Example 2. Suppose that a professor gave a test to $n$ students $\mathcal{S}_1, \mathcal{S}_2, \ldots, \mathcal{S}_n$ and wants to let them grade each other's tests. Of course, no student should grade their own test. How many ways could the professor hand the tests back to the students for grading, such that no student received their own test back?

The above problems are the typical examples of derangements. A derangement can be defined as the permuatation(s) of the elements of a set such that no element appear in it's original position, or in other words, have no fixed points.

Formula (Derangement). The number of derangements of an $n$ element set, $D_n$ or $!n$, is given by the formula
$$D_n = n!\sum_{k=0}^n\frac{(-1)^k}{k!}.$$
$Proof.$ We can prove this using the PIE. So, at first, we consider the all possible arrangements of the whole sequence, which is given by $n!$. Next, we must remove all the arrangements in which each element appears in its original position, $i.e.$ we have to subtract $\binom{n}{1}(n-1)!$. Again, we have to add back the permutations in which each set of two elements stay in their original positions, as we subtracted them twice, so we must add $\binom{n}{2}(n-2)!$ and so on. Thus, we have,
$$D_n=n!-\binom{n}{1}(n-1)!+\binom{n}{2}(n-2)!-\cdots + (-1)^n\binom{n}{n}(n-n)!.$$
But, since $\displaystyle \binom{n}{k}(n-k)!=\frac{n!}{(n-k)!k!}(n-k)!=\frac{n!}{k!}$, we have,
\begin{aligned} D_n &=n!-\frac{n!}{1!}+\frac{n!}{2!}-\cdots+(-1)^n\left(\frac{n!}{n!}\right) \\ &= n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\cdots+(-1)^n\left(\frac{1}{n!}\right)\right) \\ &= n!\sum_{k=0}^n\frac{(-1)^k}{k!},\end{aligned} as desired.
Note. The probability of a derangement is $\displaystyle\frac{1}{e}.$

Problem $1$ (IMO $1987/1$). Let $p_n(k)$ be the number of permutations of the set $\{1,2,3,\ldots,n\}$ which have exactly $k$ fixed points. Prove that $$\sum_{k=0}^nk p_n(k)=n!.$$

Solution. Firstly, observe that $\displaystyle p_n(k)=\binom{n}{k}(n-k)!\sum_{i=0}^{n-k} \frac{(-1)^i}{i!}$, $i.e.$ $p_n(k)=D_{n-k}$ where $D_i:=$ the number of derangements of $\{1, 2, \ldots, i\}$. Now, observe that, \begin{aligned} \frac{p_n(k)}{p_{n-1}(k-1)} &= \frac{n}{k}\\ \iff p_n(k) &= \frac{n}{k}p_{n-1}(k-1)\\ \implies \sum_{k=0}^nkp_n(k) &= n\sum_{k=1}^np_{n-1}(k-1)\\ &= n(n-1)! = n!,\text{ as desired.} \end{aligned}

That's all for today. If you want to try some problems, consider looking at this problem set made by me. Also, if you are interested to learn something more about PIE, then maybe go through thisthisthisthis and this. Hope to see you soon.
Emon.

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