Help on proof of $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$ The proof is required to be made through the binomial theorem. I will expose the demonstration I was tought, and forward my questions after exposing it. You'll see question marks like this one (?-n) on points I don't quite understand, where $n$ is the numeration of the mark. This are the doubts I have about the demonstration, the which I hope someone can clarify.  Prove that $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$.  We will use the following equality, and call it $P$:   $(1+x)^n(1+x)^n=(1+x)^{2n}$    (?-1)  The result will be proved finding the $x^n$ coefficient of both terms of this equality (?-2).  According to the binomial theorem, the left-hand side of this equation is the product of two factors, both equal to  $\binom{n}{0}1+\binom{n}{1}x+...+\binom{n}{r}x^r+...+\binom{n}{n}x^n$  When both factors multiply, a term on $x^n$ is obtained when a term of the first factor has some $x^i$ and the term of the second factor has some $x^{n-i}$. Therefor the coefficients of $x^n$ are  $\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+...\binom{n}{n}\binom{n}{0}$ .  Since $\binom{n}{n-r}=\binom{n}{r}$, the previous summation is equal to $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2$. So the left hand side of the equation we are asked to proove is a coefficient of $x^n$. When we expand the right-hand side of the equation $P$, we find that $\binom{2n}{n}$ is a coefficient of $x^n$. Therefore (?-3) the left-hand side of the equation we were asked to prove is in deed equal to $\binom{2n}{n}$. In conclussion,  $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$.  This was all the demonstration. My doubt one (?-1) goes about where the heck does this equation come from? How would I know what equation to come up with if requested to prove a different equality?  Doubt two (?-2) goes about why would the solution of the first equation would have anything to do with finding the $x^n$ coefficients of the one I just made up (see doubt one).  Doubt three (?-3) goes about why demonstrating that $a$ is a coefficient of $x^n$ on the left hand side of the equation I made up, and that $b$ is a coefficient of $x^n$ on the right-hand side of this equation as well, would prove my original equation, the one I was supposed to prove on the first place?  I know there are many doubts here, I hope you guys can help me. Sorry for the long post, it's a long demonstration.
Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$ I was hoping to find a more "mathematical" proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$.  I already know the logical Proof:  $${n \choose k}^2 = {n \choose k}{ n \choose n-k}$$  Hence summation can be expressed as:  $$\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n}\binom{n}{0}$$  One can think of it as choosing $n$ people from a group of $2n$  (imagine dividing a group of $2n$ into $2$ groups of $n$ people each. I can get $k$ people from group $1$ and another $n-k$ people from group $2$. We do this from $k = 0$ to $n$.
I am trying to copy specific attribute table rows from one shapefile to a new shapefile that will contain only the specified rows in its attribute table, that is, if a field GDP_TOP10 = 1.   I am using this at the moment:  #Grab rows using Search Cursor with WHERE clause rows = arcpy.SearchCursor("C:/CGDP.shp", "GDP_TOP10 = 1") for row in rows:     arcpy.CopyFeatures_management("CGDP.shp", "New.shp") del row del rows   However, this just copies the entire shapefile, not the rows I want to select with the cursor. If I execute a print statement under my for loop instead, it works and prints only the relevant fields, but the copy to new shapefile command does not seem to copy specifically.   Is there a different ArcPy command to only copy rows that match a query? It is stated as a problem in Spivak's Calculus and I can't wrap my head around it. Check if a services is installed using C My new 1,5 TB ext4 Hard Disk makes mysterious noises and fails to read and write. But not always. Sometimes it just works. But then I get those following errors. Does anyone know a solution for that or is my new WD Caviar Black 1.5 TB hard disk broken?  [  551.162286] ata7: SATA link down (SStatus 0 SControl 310) [  551.162294] ata7.00: link offline, clearing class 1 to NONE [  551.178223] ata7: EH complete [  551.190802] ata7: exception Emask 0x10 SAct 0x0 SErr 0x4040000 action 0xe frozen [  551.190807] ata7: irq_stat 0x00000040, connection status changed [  551.190811] ata7: SError: { CommWake DevExch } [  551.190818] ata7: limiting SATA link speed to 1.5 Gbps [  551.190822] ata7: hard resetting link   Update Mainboard P7P55D LE: The escape velocity of earth is roughly $11 kms^{-1}$. However, what if a long ladder was built extending out of Earth's atmosphere and considerably more. Then if something was to climb up at much less than the escape velocity, what would happen when it reached the end?  And what if the object that climbed the "ladder" then fired some kind of thruster/rocket and was going fast enough so that it orbited the Earth. It would mean less energy required to get into orbit? In my recent study of Artinian modules I have been looking at Prüfer $p$-groups (which I will denote by $\mathbb Z_{p^\infty}$ from now on).   I am attempting to prove that $\mathbb Z_{p^\infty}$ is an Artinian $\mathbb{Z}$-module which is not finitely generated.      To do this I am trying to show that every proper non-trivial submodule of $\mathbb Z_{p^\infty}$ is equal to $\left(\frac{1}{p^n}\mathbb{Z_{}}\right)/\mathbb{Z}$ for some $n$.   I am using the following definition of $\mathbb Z_{p^\infty}$: $$\mathbb Z_{p^\infty}=\left\{x\in \mathbb{Q}/\mathbb{Z}\mid \exists n:p^nx=0 \right\}$$ I have read in several places, Atiyah &amp; MacDonald Example 6.3 and Wikipedia to name a few, that the subgroups of $\mathbb Z_{p^\infty}$ are of this form. This would demonstrate that $\mathbb Z_{p^\infty}$ is Artinian. However, I can't seem to find anywhere a proof or example showing why all the subgroups are of this form. I would like to understand this further, as it is not immediate to me that this would be the case, but my attempts to figure out why have so far been futile. I may be thinking too much and missing a simple explanation. I have looked at starting with an arbitrary subgroup $H$ such that $H\neq 0$ and $H\neq\mathbb Z_{p^\infty}$ and seeing if I can deduce anything from that. I got this far, but it doesn't seem to be going anywhere. I have also looked at Sylow's theorems, but I'm not sure if they will be any use to me as $|\mathbb Z_{p^\infty}|=\infty$ (unless I'm wrong about this!).  I would be extremely grateful if anybody could offer a few hints here as I am slowly losing my sanity trying to figure it out!  Thanks,  Andy. The annual student intake of Hogwarts seems about 100. Since in book 7 Hogwarts is declared mandatory for all children, this means that the total magical population of Britain cannot be much more than 10,000.  How do these people make a living?  All jobs mentioned in the books are non-productive occupations (government, stores, sports &amp;c) - not farmers or engineers. There are exceptions - makers of wands/robes/broomsticks - but where do they get the raw materials from?  Who and how "creates goods" in the wizarding world?  Note that if you say "wizards exchange their gold for Muggle money and buy stuff from Muggles (through intermediaries, if necessary)" then you would have to explain where they get their gold: they would either have to mine it, or produce something of value to the Muggles (in which case the separate financial system of galleons/sickles/knuts does not make sense).  PS. A simple issue of maintaining a steam locomotive for Hogwarts express is pretty much insurmountable for 10k people - without Magic, that is. So, I can imagine that the skilled tradesmen can transfigure dirt into raw materials for their trade - except for food, which "is the first of the five Principal Exceptions to Gamp's Law of Elemental Transfiguration". Norms Induced by Inner Products and the Parallelogram Law What does the $\ni$ (backwards element of) symbol mean?  It doesn't appear in the , and a Google search for "backwards element of" or "backwards epsilon" turns up contradictory (or unreliable) information.  It seems it can mean both "such that", or "contains as an element".  Is this correct, and if so, which is the more common usage? What is the meaning of "build off of"? What is the meaning of "build off of"? I Googled it, but can't find any definition!  I encountered it in the sentence in a technical book. The sentence is:     For the application we will build, the todos example is the closest   fit, so we'll go ahead and build off of that example.   There is a link related to it I found by Google, But it didn't give the definition! There is some conflict or error I don't understand. I can skip it and nothing visible happens. The output is okay and everything seems to be fine. Still I would like to get rid of the error message. If I run the following minimal example:   \documentclass[]{scrbook}  \usepackage[applemac, grafiken]{tahbiblatex} \usepackage[ngerman]{babel}   \begin{document}  Testing äüöaüä.  \end{document}   I get the error message:      ! Missing = inserted for \ifnum. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel} ?  ! Missing number, treated as zero. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel}   Here is my .sty file:  \NeedsTeXFormat{LaTeX2e}\relax \ProvidesPackage{tahbiblatex}[2012, by tah modified by sm]  \RequirePackage{ifthen}  \DeclareOption{latin1}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{applemac}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{cp850}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{utf8}{\PassOptionsToPackage{\CurrentOption}{inputenc}}  \DeclareOption{dvi}{% \newif\ifpdf \ifx\pdfoutput\undefined     \pdffalse \else     \pdfoutput=0     \pdftrue \fi}  \newcounter{Grafiken}\setcounter{Grafiken}{0} \DeclareOption{grafiken}{\setcounter{Grafiken}{1}}  \newcounter{Randausgleich}\setcounter{Randausgleich}{0} \DeclareOption{randausgleich}{\setcounter{Randausgleich}{1}}  \newcounter{Tabellen}\setcounter{Tabellen}{0} \DeclareOption{tabellen}{\setcounter{Tabellen}{1}}  \newcounter{Sync}\setcounter{Sync}{0} \DeclareOption{sync}{\setcounter{Sync}{1}}  \ProcessOptions  \RequirePackage{fixltx2e} \RequirePackage{inputenc} \RequirePackage[T1]{fontenc} \RequirePackage[T1]{url} \RequirePackage{setspace} %\RequirePackage[german]{varioref} \RequirePackage{textcomp}  \newcommand{\urlangabe}[2]{% $\langle$\url{#1}$\rangle$ -- Zugriff am #2}  \def\hierlockern {\begingroup \def\par{\endgraf\endgroup}%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% }  \newcommand{\alleslockern}{%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% } \ifthenelse{\value{Grafiken}=1}   {   \ifx\pdfoutput\undefined \newcount\pdfoutput \fi   \ifcase\pdfoutput       \RequirePackage[dvips]{graphicx}       \DeclareGraphicsExtensions{.eps}       \typeout{Paket tah Meldung: Die Option `grafiken' fuer DVI wurde gewaehlt.}% \else     \RequirePackage[pdftex]{graphicx}     \DeclareGraphicsExtensions{.jpg,.tif,.png,.pdf}     \typeout{Paket tah Meldung: Die Option `grafiken' fuer PDF wurde gewaehlt.}%   \fi}   {}  \ifthenelse{\value{Sync}=1}   {     \ifx\pdfoutput\undefined \newcount\pdfoutput \fi     \ifcase\pdfoutput     \RequirePackage{srcltx}     \typeout{Paket tah Meldung: Die Option `sync' fuer DVI wurde gewaehlt.}%     \else     \RequirePackage{pdfsync}     \typeout{Paket tah Meldung: Die Option `sync' fuer PDF wurde gewaehlt.}%     \fi }   {}   \ifthenelse{\value{Randausgleich}=1}   {\typeout{Paket tah Meldung: Die Option `randausgleich' wurde gewaehlt.}%   \RequirePackage[activate]{pdfcprot}}   {}  \ifthenelse{\value{Tabellen}=1}   {\typeout{Paket tah Meldung: Die Option `tabellen' wurde gewaehlt.}%   \RequirePackage{tabularx}   \RequirePackage{ragged2e}   \RequirePackage{dcolumn}   \RequirePackage{booktabs}   \newcolumntype{L}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}l}   \newcolumntype{Y}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}X}   \newcolumntype{d}[1]{&gt;{\DC@{.}{,}{#1}}l&lt;{\DC@end}}   }   {}    \endinput   I also don't really need all the features. So it would be nice to reduce it to minimal features, which are needed.. What are Java command line options to set to allow JVM to be remotely debugged? I know there's some JAVA_OPTS to set to remotely debug a Java program.  What are they and what do they mean ? I just graduated, and I'm starting a new job soon. My employer has pretty good policy for matching 401(k) contributions, so I want to invest as much as possible, especially during this first year.  My question is: should I invest in a traditional 401(k), a Roth 401(k), or some combination of the two? On the one hand, tax rates may be much higher when I retire, and I might be in a higher tax bracket, so a Roth 401(k) would be good. On the other hand, I might be in a lower tax bracket when I retire, so a traditional 401(k) might be better. It's hard to know where I will stand in 30 years.  Is this the same trade-off as with a traditional / Roth IRA, or is there any other tax difference I should know about? I need to install more applications - can I use my external HD (samsung T5) and add an additional applications folder? I have loads of space on it and it the speeds are great. Do lenses have focal planes or focal spheres/ellipsoids? I'd like to use a different + symbol in addition to + in math mode, preferably one which is just bolder than the standard one.  I'm currently using \pmb{+} from the amsbsy package, but it's a bit too lean for my taste, and it looks a bit pointy to the ends.   I am aware of \ding{58} from the pifont package, as well as \Plus from bbding, both of which seem to be non-math mode symbols, and look really bad next to math symbols.  Am I missing some other option to produce a bold +, apart from drawing the symbol myself? Formula for expected value of continuous random variable without using density I am using Blender to mix the video from my camera with the audio from my external recording setup.  It works great mostly and I am very pleased with it.  But the problem is that I can't start the audio recorder an even number of thirtieths of a second from when I started the camera.  And Blender seems to only let me move the audio track in the VSE by whole frames (1/30 of a second).  Here are a couple screenshots:   The track on top is the external audio and the one on the bottom is the camera's audio.  Between the two screenshots I have only moved the audio track by one frame.  But as you can see, I need to move it by about 1/2 frame to line up the audio properly with the video.  So can I move a track in the VSE by fractions of a frame? Did any Quidditch games ever end by agreement of the Captains? The game can only end when the Snitch has been caught, or by mutual agreement of the two teams' Captains; the latter is very rare, however, as one team would have to lose. (source: HP Wikia on Golden Snitch)   Is there an example from canon of this resolution to the game ever happening? Does IBIS reduce image resolution? How does it compare to lens based IS?