I would appreciate if somebody could help me with the following problem  Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!} )$  $$\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times  2^{2n}$$
In  I prove, using generating functions, a statement equivalent to  $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even.  (Clearly the sum is $0$ when $n$ is odd.)  The nice expression on the right-hand side indicates that there should be a pretty combinatorial proof of this statement.  The proof should start by associating objects with even parity and objects with odd parity counted by the left-hand side.  The number of leftover (unassociated) objects should have even parity and should "obviously" be $2^n \binom{n}{n/2}$.  I'm having trouble finding such a proof, though.  So, my question is     Can someone produce a combinatorial proof that, for even $n$, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}?$$   Some thoughts so far:   Combinatorial proofs for $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$ are given by  and by .  The proofs are basically equivalent.  In Phira's argument, both sides count the number of paths of length $2n$ starting from $(0,0)$ using steps of $(1,1)$ and $(1,-1)$.  By conditioning on the largest value of $2k$ for which a particular path returns to the horizontal axis at $(2k,0)$ and using the facts that there are $\binom{2k}{k}$ paths from $(0,0)$ to $(2k,0)$ and $\binom{2n-2k}{n-k}$ paths of length $2n-2k$ that start at the horizontal axis but never return to the axis we obtain the left-hand side.  With these interpretations of the central binomial coefficients $2^n \binom{n}{n/2}$ could count (1) paths that do not return to the horizontal axis by the path's halfway point of $(n,0)$, or (2) paths that touch the point $(n,0)$.  But I haven't been able to construct the association that makes these the leftover paths (nor do all of these paths have even parity anyway).  So perhaps there's some other interpretation of $2^n \binom{n}{n/2}$ as the number of leftover paths.    Update. Some more thoughts:  There's another way to view the identity $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$.  Both sides count the number of lattice paths of length $n$ when north, south, east, and west steps are allowed.  The right side is obvious. The left side has a similar interpretation as before: $\binom{2k}{k}$ counts the number of NSEW lattice paths of length $k$ that end on the line $y=0$, and $\binom{2n-2k}{n-k}$ counts the number of NSEW lattice paths of length $n-k$ that never return to the line $y =0$.  So far, this isn't much different as before.  However, $2^n \binom{n}{n/2}$ has an intriguing interpretation: It counts the number of NSEW lattice paths that end on the diagonal $y = x$ (or, equivalently, $y = -x$).  So maybe there's an involution that leaves these as the leftover paths.  (Proofs of all of these claims can be found on , for those who are interested.)
I know that a similar question has been asked in:        although I didn't really understand the answer and couldn't really make it work.  However, I have tried something different. I decided to read the manual page for ls and it mentions different environment variables that can be set when dealing with the ls command. If one goes to the -G option it says to look at the CLICOLOR environment variables. I did and that one links you to documentation for the LSCOLORS environment variable. I went to it and it mentions:     "The value of this variable describes what color to use for which attribute when colors are enabled with CLICOLOR. This string is a concatenation of pairs of the format fb, where f is the foreground color and b is the background color. "   It also mentions that the default value of LSCOLORS is "exfxcxdxbxegedabagacad" and that each part specifies the color of some specific thing in the ls command. For example, the first pair ex in example specifies the color of directories. the second fx the color of symbolic links etc... e means blue and x means default "foreground".  1) First what does this mean: "when colors are enabled with CLICOLOR"?  2) What does "foreground" mean?  What must environment variables be for the ls command to color the output however I desire?    I also tried:  ls --color   But it threw me the following error:  ls: illegal option -- - usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]   The same error happened when I did ls --color=auto: it alone also throws an error. I have no idea why, and am not sure if it was due to iTerm2 or OS X or why that happened. It seems it works on other systems... I was supposed to find all possible Jordan canonical forms of a $5\times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester.  I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2\times 2$ while the largest Jordan block with eigenvalue 1 is $1\times 1$.  Did I miss any matrices or interrupt the minimal polynomial incorrectly?  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;2  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;2  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;1  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;0  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;1  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix} 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? When one wants to list various cases/classes/categories/types of objects in a string of conjunctions, is it preferable (or even mandatory) to keep on using (the same) preposition in front of each one them?  Moreover, does singular or plural form of the word "case", "class", "category" make any difference in the previous question?  Examples:   In the present article, we study the case of X, of Y and of Z objects. (in this example, as in most of the ones I have in mind, it is the preposition "of" which is relevant to my question). We study the cases/classes of X and of Y objects. vs. We study the cases/classes of X and Y objects.   Although it is rather clear that, in the singular case, the absence of "of" in later conjuncts may lead to a logical confusion (e.g., mistakenly taken as the one class/case of objects which are both X and Y), when it comes to plural of the words "case", "class", etc., this might not be an issue. Still, there might anyway be a rule in the plural setting as well. What are the visa (or any other crucial) requirements of Bhutan for tourists? I heard that access to Bhutan is quite restricted for all people of outside South-Asian countries?  Can I have the details on that? Ubuntu 14.04 doesn' t boot after upgrade from 12.04 installed inside Windows 8.1 I have Ubuntu 12.04 installed like an app on windows 8.1 (Ubuntu 12.04 allow to be installed like an app in Windows 8.1 and it can be removed when you don't need it any more from Control Panel). Usually, to chose what os to boot when you start the laptop, you can choose between windows 8.1 and Ubuntu after windows logo appeared at start up and that was ok until I made this upgrade. Now when I try to choose Ubuntu the laptop try to boot it but, after that full colored screen is showed the screen go black and this messages appear:  mount: mounting /dev/loop0/ on /root failed : Invalid argument mount: mounting /dev on /root/dev failed: No such file or directory mount: mounting /sys on /root/sys failed: No such file or directory mount: mounting /proc on /root/proc failed: No such file or directory Target filesystem doesn' t have requested /sbin/init No init found. Try passing init = bootarg.  BusyBox v1.21.1 (Ubuntu 1:1:21.0-1ubuntu1) built-in shell (ash) Enter 'help' for a list of built-in commands  (initramfs) _   I don' t know what to do after this screen appears. Please help ! Converting Raw HTTP Request into HTTPWebRequest Object How to remove the left part of a string? I have some simple python code that searches files for a string e.g. path=c:\path, where the c:\path part may vary. The current code is:  def find_path(i_file):     lines = open(i_file).readlines()     for line in lines:         if line.startswith("Path="):             return # what to do here in order to get line content after "Path=" ?   What is a simple way to get the text after Path=? I remember after Stargate: Universe was cancelled, that I read, somewhere on the web, either part of an interview or an article by one of the producers about what they were planning if there had been a third season.  (He said they weren't sure whether they'd set it soon after the ending of the second season, or later, or were going to do something else.)  He also wrote that, while he felt okay discussing that, he did not feel like he could reveal how the story about the superstructure the Destiny had found would turn out.  He said that was something the other two producers had worked out and he didn't want to reveal it and would leave it up to them to reveal it if they wanted to.  Was this ever revealed?  Did the other producers ever discuss or reveal anything about what was going to happen with that storyline?    Addition: I'm referring to the superstructure that is the message found in the background radiation throughout the universe, not anything from the planet builders or other easily seen physical structures. How do I skip specific tests when building an rpm? I am building an RPM of git version 2.13.0. The build has gone ok but one of the tests are failing:  Config variable ${LIB_HTTPD_SVN} is not defined Config variable ${LIB_HTTPD_SVN_PATH} is not defined   I cannot find any info about these variables so I was going to just skip the tests. How can I do that? Is there an alternative to Cantor's cardinalities that makes proper subsets smaller than their sets? Cantor defined an infinite set as a set whose subset can be placed in a one-to-one correspondence with its subset. That is, take the set of all natural numbers: {0, 1, 2, 3, 4,...}. From that set, you can form a subset of all even numbers: {0, 2, 4, 6,...}. There is a one-to-one correspondence between both these sets. After all, you can multiply each member in the set of all natural numbers by 2, and you’ll get the set of all even numbers. Because of this one-to-one correspondence, the two sets have the same cardinality, despite the set of even numbers being a subset of the set of natural numbers. This is counter-intuitive to me. How can a subset of a set be as big as the set itself? To make the intuition more concrete, consider the set of multiples of 1,000,000: {0, 1,000,000, 2,000,000,...}. There is a one-to-one correspondence between this set and the set of all natural numbers. So they have the same cardinality. This is really counter-intuitive.  From as far as I can tell, there are basically two axioms at play here:  A1: If all members of set A can be placed in a one-to-one correspondence with all the members of set B, then both sets have the same cardinality.  A2: If set A is a proper subset of set B, then A has a smaller cardinality than set B.  How Cantor defined infinite sets is that he accepted A1 and rejected A2. That is counter-intuitive to me. But mathematicians might say that my intuition is just wrong. So my question is whether we can define infinite sets in another way, i.e. accept A2 but reject A1. That is, the set of even numbers is smaller than the set of all natural numbers, even though both sets have a one-to-one correspondence. If you think this is counter-intuitive because sets with one-to-one correspondence with each other should have the same cardinality, and thus think this is wrong, why can’t I say the same when Cantor accepted A1 and rejected A2? Either way, there is something counter-intuitive and paradoxical.   So, my questions are:   Can we accept A2 and reject A1 for the definition of an infinite set? What arguments, if any, are given for accepting A1 and rejecting A2? My port of entry and exit is Frankfurt, Germany. However i will only be spending a day or two there. I am visiting Spain and Portugal for a week each. So should I be applying to Spain embassy or German embassy? I am an Indian citizen. IllegalArgumentException or NullPointerException for a null parameter? I have a simple setter method for a property and null is not appropriate for this particular property. I have always been torn in this situation: should I throw an , or a ? From the javadocs, both seem appropriate. Is there some kind of an understood standard?  Or is this just one of those things that you should do whatever you prefer and both are really correct? Reference for the subgroup structure of $\rm{PSL}_2(q)$ Class of sets of a given infinite cardinality This question was inspired by the question on .  From the discussion in the comments there, it seems there does not exist a set of all sets of a given cardinality.  To me, it seems easy to see that this is true for any nonzero finite cardinality $n$. Given any set $x$, you can always make a set of size $n$ by adding $n-1$ other sets to $x$ to make a set of size $n$, which exists by repeated use of the pairing axiom.   But how would you do this for infinite cardinalities? For suppose $\kappa$ is some given infinite cardinality. To show that the set of all sets of cardinality $\kappa$ does not exist, it seems you would have to show that for any set $x$, there exists a set of that size with $x$ as an element, and then you could take the union of that set to find the set of all sets, and thus a contradiction. But it doesn't seem reasonable to simply say, for a given set of size $\kappa$ if $x$ is in the set, we have no problem. If not, just take an element out of the set and put $x$ in. Something about that seems like it would not be allowed.  So how would you do this for infinite cardinalities? I used the jQuery Form plugin for asynchronous form submission. For forms that contain files, it copies the form to a hidden iframe, submits it, and copies back the iframe's contents. The problem is that I can't figure out how to find what HTTP status code was returned by the server. For example, if the server returns 404, the data from the iframe will be copied as normal and treated as a regular response.  I've tried poking around in the iframe objects looking for some sort of status_code attribute, but haven't been able to find anything like that.    The $.ajax() function can't be used, because it does not support uploading files. The only way to asynchronously upload files that I know of is using the hidden iframe method. How do I bypass "this connection is not private" warnings in safari? How to demystify the axioms of propositional logic? How might I go about getting some intuition on the typical axiom schemes given for propositional logic? They seem rather mysterious at first glance.  For example, these are taken from:    $\phi \to \phi$ $\phi \to \left( \psi \to \phi \right)$ $\left ( \phi \to ( \psi \rightarrow \xi \right)) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$ $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$ How to change the theme of Snap Apps? I am enjoying the new "snap" applications, but apps installed using this method doesn't seem to be using the system gtk theme I have set.   Some use Adwaita or worse, Raleigh.   How can I have them use the standard Ambience theme I'm using across the whole system? Index and length must refer to a location within the string