Lecture 027

Counting in Two Ways Proofs

Template:

• define S you want to count clearly

• count in two ways of the set S

• conclude

Pascal's Formula: ${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}$ TODO Proof: by partition

Summation Identity: $\sum_{i = 0}^n {i \choose k} = {n+1 \choose k+1}$

Theory: k {n \choose k} = n {n-1 \choose k-1}

• Proof: by Committees and Leaders

Theory: $n \times 2^{n-1} = \sum_{k=1}^n {n \choose k}$

Principle of Inclusion / Exclusion

$|\bar{A_1}\cap \bar{A_2}\cap ... \cap \bar{A_m}| = |S|-\sum_{i\in [m]} |A_i| + \sum_{i<j\in[m]}|A_i \cap A_j|-...+(-1)^m|\bigcap_{i\in[m]}A_i|=\sum_{X\subseteq [m]}(-1)^{|X|}|\bigcap_{i\in X}A_i|$ with $\bigcap_{i = \emptyset}A_i = S$

Corollary to Principle of Inclusion / Exclusion

Let $A_1, A_2, ..., A_n \subseteq S$, and for $X \subseteq [n]$, $|\bigcap_{i \in X} A_i| \text{depends only on |X|}$. Then $|\bar{A_1} \cap \bar{A_2} \cap ... \cap \bar{A_n}| = \sum_{k=0}^n (-1)^k {n \choose k} |\bigcap_{i \in [k]}A_i|$

TODO: understand it