Let $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c\}$  How many surjective functions are there from $A \to B$?  My first thought is 3*3*2*1, but I've already written 19 out by hand.
It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.  The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.  Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.  Is this anything like correct or have I made a major mistake here?
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For me, I never knew whether it was acceptable grammar. However, what I did learn was that it was a logic distractor, could lead to confusion, and therefore should be reworded to avoid this. Feature selection with Random Forests I've recently gotten into The Flash TV show, and so I've been reading about his powers. What I don't understand is why his typical rogues pose a threat - it seems like there should be very little limit to his powers.  He can go back in time, so if he loses a battle or the bad guys do something, he can go back and preempt it. In the new 52, he can all but predict the immediate future, so he can see bad things that are going to happen and stop them. He can phase matter through matter, or let matter pass through him. This means he is all but immune to weapons fire.  I haven't read enough of the old comics so I don't the know challenges he's had. For a guy who can't be attacked, can time-travel and can race around the globe several times in a minute, I can't imagine there is much that could pose a threat. dpkg: error: parsing file '/var/lib/dpkg/updates/0045' near line 0: Alternative to LibreOffice Base for Ubuntu Diagnostics for logistic regression? Why are C# interface methods not declared abstract or virtual? SSL_read failing with SSL_ERROR_SYSCALL error Is there an easy way to disable this laptop's touchpad? Specifically, I'm on a . I'd like to disable the  so that I can re-enable if my mouse were to run out of batteries (so physically breaking the connection to the motherboard or uninstalling the driver will not be the best options). In the Control Panel, when I go to "Mouse" and find the touch pad, there is no disable option. Any ideas?  Edit: I am on Windows 7. There is a driver on the Dell site that might allow an option to disable it, but I cannot install it on Windows 7. Different mechanics of \hyperlink vs. \hyperref I am fiddling around with the hyperref package to build a hypertext structure into my pdf. At first glance, hyperref provides two similar macros to put links between parts of my document: \hyperlink{target}{text} and \hyperref[target]{text}. You might notice their slightly different syntax. What really interests me is their different working mechanics. I will illustrate this by the following example: \documentclass{article}  \usepackage{pdfpages} \usepackage[]{hyperref} \usepackage{mwe}  \begin{document}  \section{Section Heading} \label{sectionheading}  \includepdf[link=true, linkname=pdfpage]{example-image}   Link to section works with \hyperref[sectionheading]{hyperref} and with \hyperlink{sectionheading}{hyperlink} (but with slightly different targets on the page).  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Thanks for helping me :) Bijection between an infinite set and its union of a countably infinite set I have $A$ as an infinite set and $S$ as a countably infinite set, (so that means there exists a one-to-one correspondence between $S$ and $\mathbb{N}$).  How do I show that there always exists a bijection between $A$ and $A\cup S$? Can it be done by showing that there is a bijection from $A$ to $S$ or from $A$ to itself? I am lost on this one.   Oh, and can it be possible that there is no bijection if it is between $S$ and $A\cup S$? What about a map that maps $\mathbb{N}\to \mathbb{N}\cup\mathbb{R}$? Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?  I proceeded thus: $\phi(2^n)=2^{n-1}$ for all $n\ge1$ $$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$ $$3^{1024}-1\equiv0\pmod {2^{11}}$$ Since $\phi(2^{11})=2^{10}$.  So, maximum power of $2$ must be $11$. But the answer says it is $12$. Where am I wrong and how to solve it correctly? I have gotten as far as to create the bootable usb, booting into that bootable usb and getting to the install options however i quit because I had some concerns.  I currently have a Samsung 850 evo 1tb that has win 10ninstalled. I want to dual boot with Ubuntu 18.04 by installing it on a brand new unformatted uninitialized Samsung 860bevo 250gb. Should i format the brand new ssd as a new volume in win 10 and then proceed with the Ubuntu install on this volume. What else should i do? Im a first time Ubuntu user and installer.  Any and all help is appreciated. I am posting this question just to get a better understanding of the questions that can be asked on Stack Overflow.  I find that  can several answers based on users opinion and cause a discussion.  Why is this a good question when several other requests for books and tutorials are not acceptable?