Problem:     $50$ people go to teather. The price of the ticket is $5$\$.   $25$ of these people have only a $5$\$ note in their wallet, while the other $25$ only have a $10$\$ banknote. The cashier doesn't have any cash at the beginning of the queue. I would like to know how many queues exist such that the cashier always has the possibility to give the rest.   For instance: if the first person pays with a $10$\$ note, the queue does not work, so we know that the first person who pays, needs to pay with a $5$\$ note, because if the first one comes with $5$\$, then the second one can pay with $10$\$, since in this case the cashier would have the rest.  So, how many different queues are possible?  My thoughts:  I thought i would have found the solution by finding the number of arrangements of $50$ elements by using only $\{0,1\}$, where $0$ stands for a person who pays with a $5$\$ note, while $1$ stands for a person who pays with a $10$\$ note. In this way I know that the number of $0$s must always be bigger or equal to the number of $1$s, if you read the arrangement from left to right.  I also found that this number must be smaller than $\dfrac{1}{2}\begin{pmatrix} 50 \\ 25 \end{pmatrix}$, so a half of total number of arrangements.  Can somebody help me?
My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me.     Let $m$ and $n$ be nonnegative integers with $n\geq m$. There are $m+n$ people in line to get into a theater for which admission if $50$ cents. Of the $m+n$ people, $n$ have a $50$-cent piece and $m$ have a $\$ 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is   $$ \frac{n-m+1}{n+1}\binom{m+n}{m}. $$     I first noted that the first person to enter must be one of the $n$ with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the $n$ people with a half-dollar enters, the number of half-dollars in the register increases by $1$, and when one of the $m$ people with a bill enters, the number of half-dollars decreases by $1$ but the number of bills increases by $1$.   I tried to model this by looking at paths in $\mathbb{Z}^2$. The $x$-axis is like the number of half-dollars, and the $y$-axis is the number of bills. You start at $(0,0)$, and you can take steps forward $(1,0)$ or backwards diagonally $(-1,1)$ corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make $m+n$ moves, and I figured maybe the number of such paths is counted by $\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you.
I need to run a program as a different user + make it run with admin privileges (Run as administrator).  I use the '' for running the program as a different user  How can I run the program as admin with that command? (or a different one so will give me my requirements)  EDIT :  I need 2 things   Run as a different user Run as admin - not just as the admin user but also with full permissions   As this image Can I complete the Pokedex with only 6th Generation games? antonym for beneficiary Is the below brief explanation correct and complete (at least in context)? Can it be improved/simplified without getting technical? I am especially concerned about the last sentence: Is that reasoning absolutely clear to a layman?            The problem can be viewed as drawing the above graph without lifting your hand and without retracing a line. Observe that each dot is connected by three lines. Also observe that you have to draw a line to arrive at a dot, and you have to draw a line to leave that dot. Because each dot is connected by three lines, each must be visited twice. But it is impossible to visit each dot twice (even number of times) and end up with each dot connected with three lines (odd number of times). Word/phrase for the activity of hiding mistakes What word/phrase can I use to describe the activity of hiding a stupid mistake/incident, such as quickly cleaning up tea spilled by accident before anyone could notice? Centre of $GL_n(\mathbb R)$ How to prove that center of $GL_n(\mathbb R)$ is scalar matrices?  Again what can you say about the center of set of all upper triangular matrices and center of set of all diagonal matrices. Why isn't persistence working on Lubuntu 12.04 Live-USB? Dwellers return automatically after finding a low amount of items in Wasteland What is the scientific name of this evergreen plant? This plant is found in India. It is an evergreen plant. Would love to know the scientific name. $r$ primitive root of prime $p$, where $p \equiv 1 \mod 4$: prove $-r$ is also a primitive root question is as follows:  Let $p$ be a prime with $p \equiv  1 \mod 4$, and $r$ be a primitive root of $p$.  Prove that $-r$ is also a primitive root of $p$.  I have shown that $-r^{\phi(p)} \equiv 1 \mod p$.  What I am having trouble showing, however, is that the order of $-r$ modulo $p$ is not some number (dividing $\phi(p)$ that is LESS than ($\phi(p)$).  By way of contradiction, I've shown that the order of $-r$ cannot be an EVEN number less than $\phi(p)$.  But my methodology does not work for the hypothetical possibility of an ODD order that is less than $\phi(p)$.  Any and all help appreciated.  Happy to show methodology for any of the parts I have managed to do, if requested. When adding image from Wikimedia Commons, automatically add author/license/link Many Stack Exchange questions and answers embed images from Wikimedia Commons. It is probably the most used image source here. Good news Thanks to their license, all images from Wikimedia Commons can be legally used inside Stack Exchange. Bad news The only requirement is to cite the author/license (), an easy task, but we are totally failing at it, only maybe 1% of Commons images used here are properly cited. That makes our use of these images illegal, and could expose Stack Exchange to various legal threats. Not my problem, you could say, but this is also important because , bringing Stack Exchange closer to closed-data sites like Yahoo! Answers or Experts-Exchange. Proposed solution When using the image addition widget, the widget should detect whether the image's URL has a *.wikimedia.org domain, and if yes:  (optional) Transfer the image to imgur, for perennity reasons. Retrieve the author name and license name from Commons (there is a very reliable API that gives that). Concatenate into a proper citation, maybe something like &quot;*Credit: [&quot; + author + &quot;](&quot; + commonsPageURL + &quot;) , [&quot; + license + &quot;](&quot; + licenseURL + &quot;)*&quot; Insert the citation at the same time as the image's MarkDown.  Benefits:  Remove a large part of Stack Exchange's copyright violations. Makes us a good citizen in the open data world. Remember what happens to sites who disregard ethics (Experts Exchange). Consistent citations. Zero added burden for the users. Numbering equations If I am writing a LaTeX document on TexMaker and I want to number my equation but most probably I will be adding details later on and I do not want to fall into retyping it all over again. So if for example I have an equation and currently called equation (1.2) that I want to refer later as in the following sentence "As you can see in (1.2)" but then I had to edit my document in a way that it becomes (1.3) now instead of (1.2) what should I do in order to avoid editing large number of equations and my reference to them? Fundamental question about dimensional analysis Why a tesselation of the plane by a convex polygon of 7 or more sides is not possible? I read in several places, including Wikipedia, that a tessellation of the plane by a single, convex, $n-$sided polygon is not possible for $n\geq7$. I was not able to locate a proof, or a paper that discusses this. Maybe it is too trivial, but I am not able to figure it out myself.   So I would like to ask you for help. Where can I find a proof for this?  Also, I am interested in the mathematical background behind plane tiling (especially periodic), If you can suggest papers on the topic it would really help. Uncertainty principle and multiple observers vs. . Which to use? How to embed a view in a specific node? I try to embed a view in a specific node. I found tutorials for doing this with Drupal 7 and PHP-Templates:  &lt;?php print views_embed_view('view_name'); ?&gt;   But I look for a way in Drupal 8 with twig-Templates. I tried to call the views_embed_view() function in the hook_node_view() and assign it to the node so that I can print it out in the twig-Template. But it seems that this hook never gets called.  Any hints/solutions how to do this in the Drupal 8 way? I have the following recurrence that I've been pounding on:  $$ a(0)=1\\ a(1)=1\\ a(2)=2\\ a(2n)=a(n)+a(n+1)+n\ \ \ (\forall n&gt;1)\\ a(2n+1)=a(n)+a(n-1)+1\ \ \ (\forall n\ge 1) $$  I don't have much background in solving these things, so I've been googling around looking for different techniques. Unfortunately, I haven't been able to correctly apply them yet. So far I've tried:   Characteristic equations Generating functions Plugging it into Wolfram Alpha Telescoping Observation   I'm sure that one or more of these is the way to go, but my biggest problem right now is figuring out how to deal with the idea that there are different equations for odd and even values of $n$. I'm not necessarily looking for a solution (though it would be gratefully accepted), but any nudges in the right direction would be much appreciated.    To follow up, it turned out that the speed problem described in my comment, below was related to the Decimal implementation in Python 2.7. Switching to Python 3.3 (and Java) made everything many orders of magnitude better. I had created the xml document with xml version="1.0".  In that document I need to use the greater than symbol &gt; and less than symbol &lt;.  How should I include those symbols? It's not working.  &amp;gt; and &amp;lt; are not working for me.  Is there any special encoder for this? I built a model of multiple linear regression to predict the revenues at the box office of a movie: the variable $Y$ is revenue while the variables $X$ are two: Activity (number of tweets per day) and sentiment (connotation of these tweets).  The problem is that the intercept is not significant. I thought of a possible explanation: the intercept in practice is the value of the variable $Y$ when the variables $X$ assume values of 0; but because of the very strong development of social networks, it is virtually impossible not to have even a tweet (and then a wave of activity) about a movie that just came out, I thought that the significance or not the intercept could be neglected because the assumed value has no meaning in reality (a tweet of 0 would never be satisfied, given the market today). In this way, I consider the intercept a mere constant for the calculation of the line, without worrying about its significance (because in some cases would take negative values, which is impossible for revenues).  The interpretation may be correct?