Let A be a finite set. \{A_i\}_{i \in \mathbb{I}} is called a partition iff U_{i\in \mathbb{I}} = A \land A_i \cap A_j = \emptyset \text{for} i\neq j
For finite partition of A \mathbb{F}=\{A_1, ..., A_m\}, then |A|=\sum_{i=1}^m {|A_i|} = |A_1|+ ...+|A_m|
For finite partition of A A = A_1 \times ... \times A_n, then |A|=\prod_{i=1}^m {|A_i|} = |A_1|\times ...\times |A_m|
About making choice
Story: If a lot of pigeons fly into not too many pigeonholes, then at least 1 pigeonhole will be occupied by multiple pigeons.
Let A be a finite set partitioned into A_1, ... A_k. |A|=n \implies (\exists i \in [k])(|A_1|\geq \lceil \frac{n}{k} \rceil).
Proof: give an answer, suppose not, then contradictory.
Principle: if we count the size of a set in two different ways, then both expressions for the cardinality of the set must be equal.
The Handshake Lemma: \frac{(n(n-1))}{2} = \sum_{i=1}^{n-1}i
Combinatorial Methods:
arrangements without repetition
arrangements with repetition
selection without repetition
selection with repetition
k-arrangements of X: a (number of possible) injection f: [k] \hookrightarrow X with k \leq |X|
we view positions(k) as domain, choice as codomain
Theorem: |X|\times(|X|-1)\times,..., \times(|X|-k+1) = \sum_{i=0}^{k-1}(n-i)=\frac{n!}{(n-k)!}
Example: possible arrangements for 1st, 2nd, 3rd prizes TODO: proof
permutation of X: a (number of possible) bijection f: [|X|] \to X.
Example: arrangements of seats for classmates can be viewed as a (set of) ordered |X|-tuple: (a, b, c), (b, a, c), (c, b, a)
a special type of arrangements when k=|X|.
Corollary: There are |X|! many permutations on X
k-arrangements with repetition: a (number of possible) functions f:[k]\to X or an ordered k-tuple from X.
selections: arrangements without order k-selection: a (number of possible) subset of X of size k with k \leq |X|.
binomial coefficient (n choose k) {n\choose k} \text{or} \binom{n}{k}
k < 0 \lor k > n \implies {n\choose k} = 0
otherwise: {n\choose k} = \frac{n!}{k!(n-k)!}.
because there are k! ways to order k-many elements, |P(n, k)| = k! {n \choose k}
note: {n\choose k} = {n\choose n-k}
Example: 5 cards dealt from 52 cards. Since order doesn't matter {52 \choose 5} distinct hands.
The Binomial theorem: Let x, y \in \mathbb{R} \land n \in \mathbb{N}, then (x+y)^n = \sum_{k=0}^n {n \choose k} x^ky^{n-k}
Poker: 52 cards, 4 suits, 13 ranks, Ace, Jack, King, Queen.
3-of-a-kind: 5 card with 3 cards of one rank, and others are different ranks. (三带两单张)
one-pair: (二带三单张)
2^n = \sum_{i=0}^{n} {n \choose i}.
| OO | OOO | O | | Flavor 1 | Flavor 2 | Flavor 3 |
Can be viewed as (00100010) and asking how many repeated way to buy 6 donuts with 3 flavors is the same as how to arrange binary strings with exactly 6 0's and 2 1's.
There are {n+k-1 \choose k-1} = {n+k-1 \choose n} ways to select n-many objects from k-many types with repetition.
Example: anagrams of LIMITING ROP Procedure:
choose 3 of 8 positions for I's -> {8 \choose 3}
remaining 5 positions, choose a permutation of 5 letters -> 5!
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