Pigeonhole Principle and Inclusion-Exclusion

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# Pigeonhole Principle and Inclusion-Exclusion ## 2.1 Pigeonhole Principle **Simple form:** If $n$ items are placed into $m$ boxes and $n > m$, then at least one box contains at least 2 items. **Generalized form:** If $n$ items are placed into $m$ boxes, then at least one box contains at least $\lceil n/m \rceil$ it...

Pigeonhole Principle and Inclusion-Exclusion

2.1 Pigeonhole Principle

Simple form: If

n

items are placed into

m

boxes and

n > m

, then at least one box contains at least 2 items. Generalized form: If

n

items are placed into

m

boxes, then at least one box contains at least

\lceil n/m \rceil

items.

Applications

Example 1: Among any 13 people, at least 2 share the same birth month. (

13 > 12

) Example 2: In any set of 5 points in a

2 \times 2

square, some pair are within

\sqrt{2}

distance. (Divide into 4 unit squares; by pigeonhole, 2 points in same square, max distance =

\sqrt{2}

.) Example 3: Any subset of size 6 from

\{1,2,\dots,9\}

contains two numbers summing to 10. (Pairs:

\{1,9\},\{2,8\},\{3,7\},\{4,6\},\{5\}

; 5 pigeonholes, 6 pigeons.)

2.2 Inclusion-Exclusion Principle

For two sets:

|A \cup B| = |A| + |B| - |A \cap B|

For three sets:

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

General Formula

\left|\bigcup_{i=1}^n A_i\right| = \sum_{i} |A_i| - \sum_{i<j} |A_i \cap A_j| + \sum_{i<j<k} |A_i \cap A_j \cap A_k| - \cdots

Example: Derangements

Number of permutations of

n

elements with no fixed point:

D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!}

For

n=4

D_4 = 4!(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 24(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}) = 9

✅ Practice Questions

Q1: How many integers from 1 to 100 are divisible by 2 or 3?

Solution

$|A| = \lfloor 100/2 \rfloor = 50$ (divisible by 2) $|B| = \lfloor 100/3 \rfloor = 33$ (divisible by 3) $|A \cap B| = \lfloor 100/6 \rfloor = 16$ (divisible by 6) $|A \cup B| = 50 + 33 - 16 = 67$ Q2: Prove that among any $n+1$ integers from $\{1,\dots,2n\}$ , one divides another. Solution

Write each number as $2^k \times m$ where $m$ is odd. There are only $n$ odd numbers in $\{1,\dots,2n\}$ . By pigeonhole, two numbers share the same odd part; the smaller divides the larger. Join Discord PreviousAdvanced PermutationsNextRecurrence Relations

What this note is really saying

Pigeonhole Principle and Inclusion-Exclusion

2.1 Pigeonhole Principle

Applications

2.2 Inclusion-Exclusion Principle

General Formula

Example: Derangements

✅ Practice Questions

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