Prove by induction of binomial coefficients Prove using induction that the binomial coefficient $\binom nr$ is always a natural number for all $n$ and for $0≤ r ≤ n$   I tried proving it but i failed .what should i do?
Proof that a Combination is an integer From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.  This is clearly an integer, however I was curious as to why the expression  $$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.  So far I figured:  $n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.
How do I switch sides in the Civil War? I have a blender file where a sphere is orbiting a force field. I want it to be more red when it's closer to the center, and more blue when it's further. I already have the math figured out, I just need to know how to get x/y coordinates in nodes.  Here's the file: How do I change the angle of the axis without changing the angle of the object? Hey guys I am currently working on a scene with multiple objects.  There is one object (a wheel) that I need to rotate the origin for without rotating the wheel itself.  I have the exact angle that I need so a manual solution would be the best.  Any help would be much appreciated. I am used to parametric modeling and Blender is a little out of my depth. Total of attribute field of point layer per polygon layer I have a polygon shapefile of many lakes, I also have a point shapefile of locations on those lakes where fish were caught.  The point shapefile contains the attributes: Fisherman_iD, Fish_Caught, Lat, Long.  The polygon shapefile contains the attributes Lake_name, Points_per_polygon(PNTCNT)  Each point obviously has differing entries under the fish_caught field.   What I want to get is the average fish caught per lake (polygon).  So I need to have a total number of fish caught, per individual polygon, and then I can divide that by the PNTCNT file I create using the Points in a Polygon tool.  Ideas? I am relatively new to QGIS. Problem with pstricks and htlatex/tex4ht Consider the following MWE:  \documentclass{article} \usepackage{pst-all}  \begin{document}   \begin{pspicture}(0,-3)(12,5)       \psline[linewidth=2pt]{-&gt;}(1.5,0)(0,0)   \end{pspicture} \end{document}   This compiles fine with latex (not with pdflatex) but when run with htlatex it leads to the ghostscript error:  System call: dvips -E -q -Ppdf -f q.idv -pp 2 &gt; zzq.ps System return: 0 System call: gs -sDEVICE=pngalpha -sOutputFile=q0x.png -r110x110 -dEPSCrop -dBackgroundColor=16#ffffff -dTextAlphaBits=2 -dGraphicsAlphaBits=2 -q -dbatch -dNOPAUSE zzq.ps -c quit Error: /configurationerror in --setpagedevice-- Additional information: [/PageSize [0 0]] Operand stack:    --nostringval--   (%%BoundingBox: 72 72 72 72)   --dict:1/2(ro)(G)--   --nostringval--   --nostringval--   (%%BoundingBox: 72 72 72 72)   1   72   72   false   --dict:1/1(L)-- Execution stack:    %interp_exit   .runexec2   --nostringval--   --nostringval--   --nostringval--   2   %stopped_push   --nostringval--   --nostringval--   --nostringval--   false   1   %stopped_push   1951   1   3   %oparray_pop   1950   1   3   %oparray_pop   --nostringval--   1934   1   3   %oparray_pop   1820   1   3   %oparray_pop   --nostringval--   %errorexec_pop   .runexec2   --nostringval--   --nostringval--   --nostringval--   2   %stopped_push   --nostringval--   %ztokenexec_continue   (scanner state)   --nostringval--   --nostringval--   --nostringval--   --nostringval--   --nostringval--   --nostringval--   1913   10   3   %oparray_pop   --nostringval--   --nostringval--   --dict:1/10(L)--   --nostringval--   15   %dict_continue Dictionary stack:    --dict:1190/1684(ro)(G)--   --dict:0/20(G)--   --dict:82/200(L)-- Current allocation mode is local Last OS error: Invalid argument Current file position is 169 GPL Ghostscript 9.16: Unrecoverable error, exit code 1 --- Warning --- System return: 256 Entering q.css Entering q.tmp   This code used to work fine. From other examples I suspect that the problem has something to do with dvips having trouble determining the bounding box for the image (compare with  -- although the solution given here, of using separate files for each image, is not a viable solution for me), but I have not been able to work out what the problem is.  Does anyone have any ideas? Can creatures without hands (and as a consequence fingers) use rings? Pretty much what it says on a label. I have a monster for my next adventure that has a pretty interesting ring in his treasure. Since it is intelligent, and would be able to use the ring, I don't see why the ring should be lying there in a chest waiting to be looted.  Monster would use the ring, if it only had hands or even fingers. Hold a second, are fingers even necessary in 5e?  For reference, assume it only has grasping / manipulating tentacles like a  (although it is not one); Said tentacles are very flexible, capable of fine manipulation, and have varying thickness. Showing that $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$ I need prove the following: ($\zeta(s)$ is the Riemann zeta function)  $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$  I really don't know, I have tried hard, but have obtained nothing for now. Show that $\mu (E \setminus A)>0$ if $A$ is a nonmeasurable subset of $X$ and $A \subseteq E$, $E$ is measurable. This question was asked , but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.  Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.  For clarity, $\mu$ is an outer measure.  A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.  E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.  This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).  So we can attempt a proof (or the start of one) by contradiction.  Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.  So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. I assume I'm supposed to show a contradiction using that A is nonmeasurable. I'm unsure how.  Another way I think could work would be to use: $\mu (E \setminus A)= \mu (E \bigcap A^c) = 0$ and arrive at a contradiction given that A is nonmeasurable, but again, I'm unsure how.  I just need some direction from here. Thank you in advance for any advice. What is the Big-O time complexity of the following nested loops: for (int i = 0; i &lt; N; i++) {     for (int j = i + 1; j &lt; N; j++) {         System.out.println(&quot;i = &quot; + i + &quot; j = &quot; + j);     } }  Would it be O(N^2) still? How do you concatenate characters in java? Concatenating strings would only require a + between the strings, but concatenating chars using + will change the value of the char into ascii and hence giving a numerical output. I want to do System.out.println(char1+char2+char3... and create a String word like this.  I could do   System.out.print(char1); System.out.print(char2); System.out.print(char3);   But, this will only get me the characters in 1 line. I need it as a string. Any help would be appreciated.  Thanks I am writing a report in a 2 column page format. Now I want to insert a rather large picture that requires to use full width of the page and cut the columns horizontally. How can I achiveve this? Please see the images below for what I mean : Which graphics cards are supported by the new AMDGPU driver in Ubuntu 16.04? The new open source AMDGPU driver pre-installed in Ubuntu 16.04 should support all the "Volcanic Islands" cards out-of-the-box. This means that AMD Radeon R9 285, R9 380/380X and R9 Fury/FuryX graphics cards will be supported out-of-the-box if Ubuntu 16.04 LTS is installed.  However I read that the open source AMDGPU driver has an experimental support for the "Sea Islands" graphics cards (HD7790, R7 260, R9 290, R7 360, R9 390). Is this support enabled by default on Ubuntu 16.04, or will these graphics cards use the Radeon driver?  Thanks How can I prove these statements? If I have NM cable running into a junction box, can I strip the sheath of the NM cable starting inside the junction box and run the conductors without a sheath through conduit that is connected to the junction box? I understand that the wiring you'd typically use through conduit is THHN or NM (if you want to use the conduit solely as protection).  But I'd like to be able to take this incoming NM cable and have it's conductors only continue on through conduit along with some other THHN that's coming from another source. The individual conductors of the NM cable while not labeled can still be traced back to where the sheathing ends, so technically the conductors are still identifiable as to what they are. If need be, I could splice the NM conductors to THHN wire in the junction box, but I don't see a reason to add a splice where I don't think it's needed. My reasoning for why I think this should be OK is that if you terminate NM cable in a panel, you typically have a foot or so of the conductors without the sheath and there's nothing identifying those other than tracing it back to the sheath it came from. I need to view the saved WiFi passwords on my non-rooted Android phone I have an LG Optimus 7 (not-rooted), and all my WiFi passwords are there. I want to view the saved WiFi passwords to transfer thm to my new Samsung Galaxy S3, but I don't know how.  I tried to search on the Internet and Google play for an application that would allow me to view this information, but I can't find one. Is there any way for me to accomplish this? If $A|B$ and $B|A$ then prove $A=\pm B$   So far I have $A|B \iff AX=B$ and $B|A \iff BY=A$ with $X,Y \in \mathbb{Z}$  Not sure how to finish, any help. I wonder how a researcher send an unsolicited email to another one asking for feedback to their work. I guess it will be something like this?  Hello Dr. X To introduce myself, I am Y, in institute Z. I'm sorry if this email is not convenient at this time, but my work about A is highly relevant to your work in B, so I think it will be interesting to you too. If possible, can you take a look at it and tell me what you think? Here is the link:  Thank you for your reading, hope you enjoy it.  I wonder if the summary/abstract should be included in the email too, since the principle for  is to show everything you know about it. Maybe it's not necessary, since the title of the link should prove its interestingness nevertheless? I'd like to have your confirm. I'm also interested in the case where Y has no Z, and the link is just a collection of observations posted in a blog, not a full paper with proper citation and literary review. It is possible that the observations may have been covered in the field, but Y isn't aware of that yet. I also think that if the link is interesting enough, then putting a tracking method in there is fine too? Even if they finds out, they wouldn't feel insulted either, because they find that the link is indeed interesting.  Related: •  •  • texstudio: show line numbers in editor How do I show line numbers in the editor? According to the picture in the doc (e.g. about bookmarks) there should appear by default.  I'm using Texstudio version 2.7, win8.1. Extract float/double value How do I extract a double value from a string using regex.  import re  pattr = re.compile(???)     x = pattr.match("4.5") This is not a duplicate question because I am looking for an explanation directed to a general audience as to the mistakes (if any) in  (reproduced below). (Numberphile is a YouTube channel devoted to pop-math and this particular video has garnered over 3m views.)   By general audience, I mean the same sort of audience as the millions who watch Numberphile. (Which would mean, ideally, making little or no mention of things that a general audience will never have heard of - e.g. Riemann zeta functions, analytic continuations, Casimir forces; and avoiding tactics like appealing to the fact that physicists and other clever people use it in string theory, so therefore it must be correct.)  Numberphile's Proof.  The proof proceeds by evaluating each of the following:  $S_1 = 1 - 1 + 1 - 1 + 1 - 1 + \ldots$   $S_2 = 1 - 2 + 3 - 4 + \ldots $   $S = 1 + 2 + 3 + 4 + \ldots $   "Now the first one is really easy to evaluate ... You stop this at any point. If you stop it at an odd point, you're going to get the answer $1$. If you stop it at an even point, you get the answer $0$. Clearly, that's obvious, right? ... So what number are we going to attach to this infinite sum? Do we stop at an odd or an even point? We don't know, so we take the average of the two. So the answer's a half."   Next:  $S_2 \ \ = 1 - 2 + 3 - 4 + \cdots$  $S_2 \ \ = \ \ \ \ \ \ \ 1 - 2 + 3 - 4 + \cdots$  Adding the above two lines, we get:  $2S_2 = 1 - 1 + 1 - 1 + \cdots$  Therefore, $2S_2=S_1=\frac{1}{2}$ and so $S_2=\frac{1}{4}$.  Finally, take  \begin{align} S - S_2 &amp; = 1 + 2 + 3 + 4 + \cdots           \\ &amp; - (1 - 2 + 3 - 4 + \cdots)          \\ &amp; =   0 + 4 + 0 + 8 + \cdots           \\ &amp; = 4 + 8 + 12 + \cdots           \\ &amp; = 4S \end{align}  Hence $-S_2=3S$ or $-\frac{1}{4}=3S$.  And so $S=-\frac{1}{12}$. $\blacksquare$