Inclusion-Exclusion and its various Applications

Difficulty Level : Easy
Last Updated : 17 Aug, 2021

In the field of Combinatorics, it is a counting method used to compute the cardinality of the union set. According to basic Inclusion-Exclusion principle:

For 2 finite sets $A_1$ and $A_2$ , which are subsets of Universal set, then $(A_1-A_2), (A_2-A_1)$ and $(A_1\bigcap A_2)$ are disjoint sets.

Hence it can be said that,
$\left|(A_1-A_2)\bigcup(A_2-A_1)\bigcup(A_1\bigcap A_2)\right| = \left|A_1\right| - \left|A_1\bigcap A_2\right| + \left|A_2\right| - \left|A_1\bigcap A_2\right| + \left|A_1\bigcap A_2\right|$

$\left|A_1 \bigcup A_2\right| = \left|A_1\right| + \left|A_2\right| -\left|A_1 \bigcap A_2\right|$ .
Similarly for 3 finite sets $A_1$ , $A_2$ and $A_3$ ,
$\left|A_1 \bigcup A_2 \bigcup A_3\right| = \left|A_1\right| + \left|A_2\right| + \left|A_3\right| - \left|A_1 \bigcap A_2\right| - \left|A_2 \bigcap A_3\right| - \left|A_1 \bigcap A_3\right|+ \left|A_1 \bigcap A_2 \bigcap A_3\right|$

Principle :

Inclusion-Exclusion principle says that for any number of finite sets $A_1, A_2, A_3... A_i$ , Union of the sets is given by = Sum of sizes of all single sets – Sum of all 2-set intersections + Sum of all the 3-set intersections – Sum of all 4-set intersections .. + $(-1)^{i+1}$ Sum of all the i-set intersections.

In general it can be said that,

$\left|A_1 \bigcup A_2 \bigcup A_3 .. \bigcup A_i\right| = \sum\limits_{1 \leq k \leq i} \left|A_k\right| + (-1)\sum\limits_{1 \leq k_1 \textless k_2 \leq i} \left|A_{k_1} \bigcap A_{k_2}\right| + (-1)^2\sum\limits_{1 \leq k_1 \textless k_2 \textless k_3 \leq i} \left|A_{k_1} \bigcap A_{k_2} \bigcap A_{k_3}\right| .. + (-1)^{i+1}\sum\limits_{1 \leq k_1 \textless k_2 \textless k_3 \textless .. k_i\leq i} \left|A_{k_1} \bigcap A_{k_2} \bigcap A_{k_3} ..\bigcap A_{k_i}\right|$

Properties :

Computes the total number of elements that satisfy at least one of several properties.
It prevents the problem of double counting.

Example 1:
As shown in the diagram, 3 finite sets A, B and C with their corresponding values are given. Compute $\left|A \bigcup B \bigcup C\right|$ .

Applications :

Derangements
To determine the number of derangements( or permutations) of n objects such that no object is in its original position (like Hat-check problem).
As an example we can consider the derangements of the number in the following cases:
For i = 1, the total number of derangements is 0.
For i = 2, the total number of derangements is 1. This is $2 1$ .
For i = 3, the total number of derangements is 2. These are $2 3 1$ and 3 1 2.

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Inclusion-Exclusion and its various Applications

Principle :

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