$\text{Square}$

$\text{About}$: $(a+b)^r$

$$(a+b)^r=a^r+(C_r^1)a^{n-1}b^1+(C_r^2)a^{n-2}b^2+(C_r^3)a^{n-3}b^{3}+\dots+(C_r^{n-2})a^2b^{n-2}+(C_r^{n-1})a^1b^{n-1} $$

$\text {Set}$

$8$ $\texttt{bit}$ $\text{location}$：

$\text {Condition}$ $\text A$ : "$1$" $\text {start}$ $2^7$ 个

$\text {Condition}$ $\text B$ : "$00$" $\text {last}$ $2^6$ 个

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

$\text{Ring multiplication principle}$

$\text {At a round table, there are four people seated, and if in both seating arrangements, everyone's left and right neighbors are the same, how many ways are there to think the same?}$

$$\texttt{Configurages}=\frac {4!}{4}$$

$\text {Binomial theorem}$

$$\sum_{i=0}^{n}{2^i}\cdot(C_n^i)=(2+1)^n=3^n$$

$\text{Examples of complexity}$：

#include <iostream>
#include <bitset>

using namespace std;

int main() {
    int s = 9;
    for (int i = s; ; i = (i - 1) & s) { // s is a set.
        cout << bitset<8>(i);
        if (i == 0) {
            break;
        }
    }
}
/*
a subset of 9:
1001
0001
1000
0000 *
*/

$\text{The time complexity is}$：

$$\sum_{i=0}^{n}2^i\cdot$$

$\text{Plus find the complexity}$：

#include <iostream>
#include <bitset>

using namespace std;

int main()
{
    int sum = 0;
    int n = 5;
    for (int s = 0; s < (1 << n); s++) {
        for (int i = s; ; i = (i - 1) & s) {
            cout << bitset<8>(i) << '\n';
            sum++;
            if (i == 0) {
                break;
            }
        }
    }
    cout << sum << '\n';
    return 0;
}

$\text{Yang Hui triangle}$

1 1
1 2 1
1 3 3 1
1 4 6 4 1
...

$$C_n^m=C_{n-1}^{m-1}+C_{n-1}^m$$

$\text{Yang Hui triangle diagram}$

$\text{Combinatorics}$

$\text {VANDERMONDE'S IDENTITY}$ $\text {Let m, n, and r, be nonnegative integers with r not ex-ceeding either m or b. Then}$

$$C_{m+n}^r=\sum_{k=0}^{r}C_m^{r-k}C_n^k$$

$\text {COROLLARY If n is a nonnegative integer. Then}$:

$$C_{2n}^n=\sum_{k=0}^{n}(C_n^k)^2$$

$\text {THEOREM Let n and r be nonnegative integer with } r\leqslant n \text {. Then}$

$$C_{n+1}^{r+1}=\sum_{j=r}^{n}C_j^i$$

$\text {The principle of repulsion}$

$\text{A probabilistic form of the repulsion principle}$:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B) P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B \cap C) $$

$\text{Probability}$

$\text {Fundamental theorem:}$

$$P(A)=1-P(\overline A) \\ P(\phi)=0 \\ \text {If } A \subset B, \text {So} P(A)\leqslant P(B) \\ \text {To any event: } 0\leqslant P(A)\leqslant 1 $$

$\text{Discrete probability}$：

$\text{Formula}$

$$\text {Probability} (\%)=\frac {\text{Number of valid scenarios}}{\text{Total number of scenarios}} $$

$\text {Expectation}$:

$\text {Defined as: a weighted average weighted by probability (or density).}$

$$\text E [X]=\sum_{x\in X} xP(x)$$

$\text{Classical generalization}$

$\text{Example:}$

$\text{There are 4 red balls and 6 black balls in the bag, which are now placed in the bag Return to the ground to take the ball 3 times in turn, and find the probability of getting the black ball for the first 2 times and the red ball for the 3rd time.}$

$\text{Theorem:}$

• $\text {If the random event is satisfied:}$

1. $\text {Limited sample space: The number of sample points is limited.}$
2. $\text{And other possibilities,}$ $\text{Each basic event has the same probability of occurring.}$
   • $\text{Ideally, the probability of any one side facing up is equal when a die is rolled.}$

• $\text {The test is a classical generalization}$.
• $\text {Event A's Probability}$ $P(A)=\frac {|A|}{|S|}$

$\text{Geometric generalizations}$

$\text{Example:}$

$\text{At 0 moments, two people are waiting for the bus, car A arrives with a probability of arriving at the first 60 minutes, and car B arrives with an equal probability of arriving at the first 120 minutes. Ask about the probability that car A will arrive before car B.}$

$\text{Math Expectations}$

$\text {Let the random variable } X \text {have the pmf}$

$$f(x)=\frac{1}{3}\\ x\in S_x$$

$\text{where } S_x={-1,0,1}\text{ Let } u(X)=X^2\text{. Then}$

$$E(X^2)=\sum_{x\in S_x}x^2f(x)=(-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{3})+(1)^2(\frac{1}{3})=\frac{2}{3} $$

$\text{Segma } \sum \text{'s Quality}$

$\text{Plus quality:}$

$$\sum{(a_i+b_i)}=\sum{a_i}+\sum{b_i} \\ \sum{(a_i-b_i)}=\sum{a_i}-\sum{b_i} $$

$\text{Frac quality:}$

$$\sum{\frac{a}{b}}=\frac{1}{b}\sum{a}$$

$\text{Cycle's segma quality:}$

$$\sum_{cyc}{a}=\sum_{cyc}{b}=\sum_{cyc}{c}=a+b+c \\ \sum_{cyc}ab=\sum_{cyc}bc=\sum_{cyc}{ca}=ab+bc+ca \\ \sum_{cyc}a^2b=\sum_{cyc}b^2c=\sum_{cyc}c^2b=a^2b+b^2c+c^2a $$