Which preposition should be used with combinations in the following sentence?     We should consider all possible combinations between/of different   versions of A on the one hand, and different versions of B on the other.   Generally, "combination of" seems to be way more popular.        But in the above context, we have different versions on each side of the combination, which is why I think maybe "combination between" is a better choice. So, notice please that the expressions on the one hand and on the other are crucial to the context and thus to my question.
I have read both two uses:   The combinations of a, b and c, taken two at a time, are ab, bc, ac. We can not help but noticing a harmonious combination between classic and contemporary.   Are both uses interchangeable? Perhaps substitute combinations among for combinations of in the first sentence? Any idea is appreciated.
Skewness, kurtosis and how many standard deviations values are from the mean As is well known for the normal distribution, 68% of the probability mass is within one standard deviation of the mean, 95% within two standard deviations and 99.7% within 3 standard deviations.   However, I have some empirical distributions that are leptokurtic and negatively skewed. In such circumstances is there a formula based on their higher-order moments to calculate how much of the probability mass is within so many standard deviations of the mean?   I have a measurement and would like to give some sense of how far it is from the midpoint (mean or some other measure of central tendency).  Can this be done? I need to use an alias in the WHERE clause, but It keeps telling me that its an unknown column. Is there any way to get around this issue? I need to select records that have a rating higher than x. Rating is calculated as the following alias:  sum(reviews.rev_rating)/count(reviews.rev_id) as avg_rating Proving this relation is transitive Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?  I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation? I'm trying to find a story for my father, from the 70s or earlier  Earth is at war with aliens (possibly Martians) and was unable to effectively fight back as the aliens had planetary (or at least city-protecting) force fields through which no bombardment was possible.  Earth eventually wins through invention of a device that allows them to teleport weapons (nukes?) under the force fields.  originally believed to be one of Harry Harrison's 'One step from earth' shorts, but this is not the case.  Thanks in advance for any help here! Render every Nth frame of an animation I want to render every Nth frame of an animation, in this case every 12th frame. This way I can play back the animation at 2fps and see if everything is going correctly.  I'm a real novice with Blender Python and cobbled this together:  import bpy  for f in range(startframe,endframe,NthFrame):     bpy.context.scene.frame_current = f     bpy.ops.render.render()     bpy.data.images['Render Result'].save_render(filepath="" + str(f) + '.png')    This kind of works, the right images are rendered and saved. However it completely freezes up Blender, and after about an hour it crashes on windows due to being unresponsive I guess. How can I fix this? Preferably I'd like it to work like rendering an animation normally works, i.e. being able to still use Blender and seeing the progress of each rendered frame. When induction proofs need a choice function Reading the comment to  the user makes the argument that one can pick a dense subset of a separable vector spaces without relying the axiom of countable choices. Since the comment got a lot of up votes, I guess it's correct, but to me this seems very strange. My rule of thumb was/is the following: We can get by without a choice function if   There is a canonical choice at each iteration. The proof doesn't actually require the iteration process to terminate.   None of these seem to be satisfied in the present case, unless there is a canonical choice for the dense subset, or the set is dense at some finite iteration, so my question is:   When can I tell if an induction arguments relies on the existence of a countable choice function? The number of (non-equal) forests on the vertex set V = {1, 2, ...,n} that contains exactly 2 connected components is given by The number of (non-equal) forests on the vertex set V = {1, 2, ...,n} that contains exactly 2 connected components is given by $\sum_{k=1}^{n-1} {n-1 \choose k-1} k^{k-2} (n-k)^{n-k-2}$.  I am unsure how to approach this question. Could someone guide me through the proof? If G is a spanning tree and it is disconnected then I know it is a spanning forest. Also, I know by Cayley's thm that there are $n^{n-2}$ nonequal tree on a vertex set V = {1, 2, ...,n} which is where i'm assuming the $k^{k-2}$ comes from but I do not understand where the rest of it is coming from. How to make terminal transparent while still having its own background image? Is there a way to give terminal specific background image and make it 50% transparent at the same time? How can I take browser screenshots at a higher resolution than my browser supports? Is there a way to forget node names between TikZ pictures? I have a document with several TikZ pictures in it.  I generally use simple names for the nodes such as:  \node (A) at .....  \node (B) at .....   and so forth.  Because of this I have name clashes between the various pictures.  That's generally not a problem.  However I find myself in a position where I may or may not have defined a certain node within a given picture.  So I use the following snippet to test whether the node has been defined:  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother   Unfortunately this fails if I've already created a node by the same name in another picture.  Is there a way to scope the definitions of nodes or is there a way to undefine nodes once you no longer need them?  P.S.  I'm aware that I could embed each picture in a standalone file and import it using \includegraphics.  Unfortunately, this approach will not work for the document I'm creating.  MWE  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc}  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother  \setlength{\parindent}{0pt} \pagestyle{empty} \begin{document}   There's no node \textbf{D} here, so I draw a triangle:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}     There is a node \textbf{D} here, so I draw a square:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1);   \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}      There isn't supposed to be a node \textbf{D} here.  I'm expecting a triangle, but get a square.\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}  \end{document}   produces: How does rooting work? Or, how do I root my device without a rooter?  Rooting instructions for Android devices that I have seen are typically of the following form:   Download Gingerbreak.apk Run Gingerbreak, click “root device” Observe that Superuser.apk has been magically installed, possibly along with busybox and other things, and enjoy your rooty goodness.   What I am wanting to know is this: how does the rooting actually happen? What does Gingerbreak do?  I understand the end result of rooting:   su is present and working Superuser.apk is installed and controls access to su Various utilities are present, provided by BusyBox (optional?)   I also think I understand the basic idea of how Gingerbreak gets started — it exploits some local privilege escalation vulnerability and uses its new-found powers to deploy Superuser and whatever else is necessary.  So, unless I am wrong in some of the understanding I laid out above, the core of thing I want to know is what the root installer does to deploy and activate Superuser. It seems that just putting Superuser.apk in the right place wouldn't be enough — how does it get allowed root access? How does it get hooked up to moderate that access for other applications?  What does Gingerbreak do post-exploit to enable and secure root access? Change default shell for Terminator I want to change the default shell of Terminator to zsh. In /etc/passwd/ I've changed the default shell to /bin/zsh. This settings changes the default shell in LXTerminal (the default terminal on Lubuntu), but not in Terminator. So I tried chsh -s /bin/zsh, but that doesn't work either. The Terminator  says:  How do I change my default shell? Terminator will always start the shell specified by the SHELL environment variable. (It will fall back to /bin/sh if the SHELL environment variable isn't set.) You could simply influence Terminator's environment, but the best solution is to tell your operating system which shell you'd like, and it will ensure that SHELL is set correctly for all applications, not just Terminator. See the documentation for your operating system for details on how to change your default shell.  So I expected my attempts to work, but they don't. How can I change the default shell in Terminator. Endpoint of a line knowing slope, start and distance In a Cartesian system, I've got the slope, start point and distance of a line segment.  What's the formula to find the endpoint? Most MTP () devices show up in Windows File Explorer under their device name or a GUID, but they don't have a drive letter assigned.  How can I access the files on such devices from the command line? Using either cmd.exe or PowerShell. How to snap and align 2 objects Hi! I'm a beginner learner and I'd like to know how to snap these three objects. I'd like to snap the two cylinders with the different faces of the cube. The short one on the left and the highest cylinder on the top of the cube. I'd like to have the center of the cylinder aligned with the center of the cube (the base of the cylinder inscribed into the face of the cube)  I'm sorry for bad explanation I hope someone can help me  Thanks!   I add a screenshoot in which you can see what I have to do   Thanks again if someone can help me with this! Can a child enter the UK on an accompanied visa and then leave the country alone? My daughter who is just under 18 has a child accompanied visa to the UK. I am traveling with her to the UK for a few days. We return from UK on the same day except that she returns to India and I am heading to San Francisco an hour later on the same day.   Is that okay or will we have an issue at immigration into the United Kingdom? I'm trying to solve an exercise about first order stochastic dominance, and there is this one question that I'm not able to answer.  Here are the hypothesis of the exercise:  Let Y ∼ F and Y' ∼ G, that means Y (resp. Y') has a probability law given by the distribution function F (resp. G).  Assume that Y ≥ 0 and Y' ≥ 0 and E(Y) + E(Y')&lt; ∞.  Here comes the question:  -Show that if the survival function of Y first order dominates over the survival function of Y', then if Y and Y' are independent, we have P(Y ≥ Y') ≥ 1/2.  I don't see how to use that Y and Y' are in this case independant.  Maybe I should set X=Y-Y' and then use the convolution?  Any help, would be appreciated, thanks !  Alex Intuitive interface for Composing Boolean Logic? I'm interested to know how people have, or would construct an interface which simplifies constructing logical boolean conditions? (Yes, it's a database query builder) In my mind a bad interface has them typing lots of AND, NOT IN, OR keywords etc. Or else has a million combo boxes.  I'm wondering if anyone has a good example I could glean ideas from? Or some suggestions?  At the moment I'm considering a split screen with a grid for composing the query and a canvas to display a visual representation of the sets of conditions and how they overlap etc. Like a Venn Diagram. This still gives me a problem with the usability/intuitiveness of the controls on the grid.  EDIT:  I'm very interested in ideas which simplify it for less technical users. However, ideas for a version of the UI for advanced users are very helpful as well. Is there an operation similar to inset that creates faces in the opposite direction? That is, instead of creating faces inwards, they create faces outwards? A normal inset operation:  What I would like: Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.