Webdiscrete math counting cheat sheet.pdf - | Course Hero University of California, Los Angeles MATH MATH 61 discrete math counting cheat sheet.pdf - discrete math >> endobj To guarantee that a graph with n vertices is connected, minimum no. Discrete mathematics cheat sheet WebTrig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p <= n; 0 otherwise4. \newcommand{\Q}{\mathbb Q} %PDF-1.5 o[rgQ *q$E$Y:CQJ.|epOd&\AT"y@$X 14 0 obj (c) Express P(k + 1). E(aX+bY+c) =aE(X) +bE(Y) +c If two Random Variables have the same distribution, even when theyare dependent by theproperty of Symmetrytheir expected Discrete Math Cheat Sheet by Dois - Cheatography If the outcome of the experiment is contained in $E$, then we say that $E$ has occurred. Sum of degree of all vertices is equal to twice the number of edges.4. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. x3T0 BCKs=S\.t;!THcYYX endstream This number is also called a binomial coefficient since it occurs as a coefficient in the expansion of powers of binomial expressions.Let and be variables and be a non-negative integer. /SA true endobj In general, use the form It is determined as follows: Characteristic function A characteristic function $\psi(\omega)$ is derived from a probability density function $f(x)$ and is defined as: Euler's formula For $\theta \in \mathbb{R}$, the Euler formula is the name given to the identity: Revisiting the $k^{th}$ moment The $k^{th}$ moment can also be computed with the characteristic function as follows: Transformation of random variables Let the variables $X$ and $Y$ be linked by some function. There are two very important equivalences involving quantifiers. Necessary condition for bijective function |A| = |B|5. For example A = {1, 3, 9, 7} and B = {3, 1, 7, 9} are equal sets. Pascal's Identity. WebBefore tackling questions like these, let's look at the basics of counting. /Parent 22 0 R Note that zero is an even number, so a string. of functions from A to B = nm2. Counting 69 5.1. stream The number of ways to choose 3 men from 6 men is $^6C_{3}$ and the number of ways to choose 2 women from 5 women is $^5C_{2}$, Hence, the total number of ways is $^6C_{3} \times ^5C_{2} = 20 \times 10 = 200$. Reference Sheet for Discrete Maths - GitHub Pages <> Prove the following using a proof by contrapositive: Let x be a rational number. From his home X he has to first reach Y and then Y to Z. Hence, there are (n-2) ways to fill up the third place. No. /Length 58 Size of a SetSize of a set can be finite or infinite. + \frac{ (n-1)! } How many ways can you choose 3 distinct groups of 3 students from total 9 students? The no. \). After filling the first place (n-1) number of elements is left. We have: Chebyshev's inequality Let $X$ be a random variable with expected value $\mu$. \newcommand{\imp}{\rightarrow} Counting problems may be hard, and easy solutions are not obvious Approach: simplify the solution by decomposing the problem Two basic decomposition rules: Product rule A count decomposes into a sequence of dependent counts (each element in the first count is associated with all elements of the second count) Sum rule 2 0 obj << Proof : Assume that m and n are both squares. Ten men are in a room and they are taking part in handshakes. stream By using this website, you agree with our Cookies Policy. If n pigeons are put into m pigeonholes where n > m, there's a hole with more than one pigeon. | x |. Complemented Lattice : Every element has complement17. Let G be a connected planar simple graph with n vertices and m edges, and no triangles. /Type /ObjStm WebBefore tackling questions like these, let's look at the basics of counting. The Pigeonhole Principle 77 Chapter 6. Toomey.org Tutoring Resources Simple is harder to achieve. }}\], \[\boxed{P(A|B)=\frac{P(B|A)P(A)}{P(B)}}\], \[\boxed{\forall i\neq j, A_i\cap A_j=\emptyset\quad\textrm{ and }\quad\bigcup_{i=1}^nA_i=S}\], \[\boxed{P(A_k|B)=\frac{P(B|A_k)P(A_k)}{\displaystyle\sum_{i=1}^nP(B|A_i)P(A_i)}}\], \[\boxed{F(x)=\sum_{x_i\leqslant x}P(X=x_i)}\quad\textrm{and}\quad\boxed{f(x_j)=P(X=x_j)}\], \[\boxed{0\leqslant f(x_j)\leqslant1}\quad\textrm{and}\quad\boxed{\sum_{j}f(x_j)=1}\], \[\boxed{F(x)=\int_{-\infty}^xf(y)dy}\quad\textrm{and}\quad\boxed{f(x)=\frac{dF}{dx}}\], \[\boxed{f(x)\geqslant0}\quad\textrm{and}\quad\boxed{\int_{-\infty}^{+\infty}f(x)dx=1}\], \[\textrm{(D)}\quad\boxed{E[X]=\sum_{i=1}^nx_if(x_i)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{E[X]=\int_{-\infty}^{+\infty}xf(x)dx}\], \[\textrm{(D)}\quad\boxed{E[g(X)]=\sum_{i=1}^ng(x_i)f(x_i)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{E[g(X)]=\int_{-\infty}^{+\infty}g(x)f(x)dx}\], \[\textrm{(D)}\quad\boxed{E[X^k]=\sum_{i=1}^nx_i^kf(x_i)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{E[X^k]=\int_{-\infty}^{+\infty}x^kf(x)dx}\], \[\boxed{\textrm{Var}(X)=E[(X-E[X])^2]=E[X^2]-E[X]^2}\], \[\boxed{\sigma=\sqrt{\textrm{Var}(X)}}\], \[\textrm{(D)}\quad\boxed{\psi(\omega)=\sum_{i=1}^nf(x_i)e^{i\omega x_i}}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{\psi(\omega)=\int_{-\infty}^{+\infty}f(x)e^{i\omega x}dx}\], \[\boxed{e^{i\theta}=\cos(\theta)+i\sin(\theta)}\], \[\boxed{E[X^k]=\frac{1}{i^k}\left[\frac{\partial^k\psi}{\partial\omega^k}\right]_{\omega=0}}\], \[\boxed{f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|}\], \[\boxed{\frac{\partial}{\partial c}\left(\int_a^bg(x)dx\right)=\frac{\partial b}{\partial c}\cdot g(b)-\frac{\partial a}{\partial c}\cdot g(a)+\int_a^b\frac{\partial g}{\partial c}(x)dx}\], \[\boxed{P(|X-\mu|\geqslant k\sigma)\leqslant\frac{1}{k^2}}\], \[\textrm{(D)}\quad\boxed{f_{XY}(x_i,y_j)=P(X=x_i\textrm{ and }Y=y_j)}\], \[\textrm{(C)}\quad\boxed{f_{XY}(x,y)\Delta x\Delta y=P(x\leqslant X\leqslant x+\Delta x\textrm{ and }y\leqslant Y\leqslant y+\Delta y)}\], \[\textrm{(D)}\quad\boxed{f_X(x_i)=\sum_{j}f_{XY}(x_i,y_j)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{f_X(x)=\int_{-\infty}^{+\infty}f_{XY}(x,y)dy}\], \[\textrm{(D)}\quad\boxed{F_{XY}(x,y)=\sum_{x_i\leqslant x}\sum_{y_j\leqslant y}f_{XY}(x_i,y_j)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{F_{XY}(x,y)=\int_{-\infty}^x\int_{-\infty}^yf_{XY}(x',y')dx'dy'}\], \[\boxed{f_{X|Y}(x)=\frac{f_{XY}(x,y)}{f_Y(y)}}\], \[\textrm{(D)}\quad\boxed{E[X^pY^q]=\sum_{i}\sum_{j}x_i^py_j^qf(x_i,y_j)}\quad\quad\textrm{and}\quad\textrm{(C)}\quad\boxed{E[X^pY^q]=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}x^py^qf(x,y)dydx}\], \[\boxed{\psi_Y(\omega)=\prod_{k=1}^n\psi_{X_k}(\omega)}\], \[\boxed{\textrm{Cov}(X,Y)\triangleq\sigma_{XY}^2=E[(X-\mu_X)(Y-\mu_Y)]=E[XY]-\mu_X\mu_Y}\], \[\boxed{\rho_{XY}=\frac{\sigma_{XY}^2}{\sigma_X\sigma_Y}}\], Distribution of a sum of independent random variables, CME 106 - Introduction to Probability and Statistics for Engineers, $\displaystyle\frac{e^{i\omega b}-e^{i\omega a}}{(b-a)i\omega}$, $\displaystyle \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$, $e^{i\omega\mu-\frac{1}{2}\omega^2\sigma^2}$, $\displaystyle\frac{1}{1-\frac{i\omega}{\lambda}}$.
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