My attempt was based on principles of exclusion-inclusion but I'm unsure how to exclude in this case as the arrangements in which 'EEE' occur together are a complete subset of the arrangements in which 'EE' occur together. Could someone recommend a better method/ Elaborate how inclusion-exclusion works in such cases? The answer is supposed to be 960.
I understand that there are $\frac{8!}{3!2!}$=3600 permutations possible without any restrictions in place. I follow this by attaching all like elements as such {1,2,{3,3,3},{4,4},5} where I can account for 5! permutations where I have three 3s and two 4s in consecutive positions. However, when it comes to accounting for the permutations that, for example, will have three consecutives 3s but not so for the 4s, I am at a loss. I wrote some simple code and came up with a set of 960 permutations as the acceptable answer. As the question implies, I am unsure of how to reach that through other means. My apologies if this is considered elementary and I appreciate any help whatsoever.
A water pipe from sea level to beyond the atmosphere If a pipe extended from just above the ocean floor to outside the atmosphere, would water be sucked up it by the vacuum beyond the atmosphere? If a hole was made in the pipe, above sea level, how would that affect the flow of water? Would it stop it completely? Sum of $\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$ Does a closed form exist for      $$\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$$   in terms of $k$ and other functions? The best that I have been able to do is solve the case where $k=1$, since the sum is just the infinite series for $e$. I would guess that any closed form must involve the exponential function, but am at a loss to prove it. I have an expression for a covariance matrix $C$ in terms of the indices $i$ and $j$.  In this way I can analytically calculate the elements of my covariance matrix, however when I try to invert $C$ matlab gives a warning about the matrix being close to singular.  The inversion therefore doesn't work, by which I mean the multiplying $C$ by the inverse given i do not get the identity.  I have tried calculating the pseudo inverse but this also does not work.  I have also tried adding a small constant along the diagonal but again the results do not work.  In general I'm working with matrices of dimension 1200 but I will give low dimensional example matrix that has the same properties, i.e. the matrices are symmetric about the diagonal and the anti-diagonal     19.9939  19.9954  19.9958  19.9951  19.9933  19.9905  19.9865    19.9954  19.9973  19.9981  19.9978  19.9965  19.9940  19.9905    19.9958  19.9981  19.9993  19.9995  19.9985  19.9965  19.9933    19.9951  19.9978  19.9995  20.0000  19.9995  19.9978  19.9951    19.9933  19.9965  19.9985  19.9995  19.9993  19.9981  19.9958    19.9905  19.9940  19.9965  19.9978  19.9981  19.9973  19.9954    19.9865  19.9905  19.9933  19.9951  19.9958  19.9954  19.9939   As mentioned in the title the matrix isn't positive, however the the negative eigenvalues are very small suggesting that the matrix is not positive only due to machine precision.  The negative eigenvalues are $-1.4048e-14$ and $-2.4571e-15$.  How can I go about inverting these matrices? QuickStart basic layout/template for web development? I'm looking for a quick start or template, or something, to get a web site's basic layout done quickly and looking fairly professional, and then let me do the coding.  I don't want a CMS, as this is for a highly customised reporting application, but I need something like frame header and two columns and frame footer.  I can design the pretty reports for the wide right hand column myself, as the left hand column menu. How to make `sudo` preserve $PATH? I have a program that is installed in a custom directory under /opt. To make it easier to run it, I edited my bashrc to add said directory to my path: export PATH=$PATH:/opt/godi/bin:/opt/godi/sbin  This works fine if I want to run the program without sudo. However, if I try to run it with sudo it fails with a &quot;command not found&quot; error. $ sudo godi_console sudo: godi_console: command not found  Inspecting the PATH variable after using sudo reveals that its not including the same PATH I have as a normal user: $ sudo sh # echo $PATH                  /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin  Why is the PATH not the same? Am I doing something wrong? I'm on Debian Jessie, if it makes a difference. One thing I tried was to invoke /opt/godi/sbin/godi_console directly, passing the absolute path to the executable. Unfortunatelly, that didn't help in this particular case because godi_console itself depends on the PATH being correctly set. Existence of a local geodesic frame Let $(M,g)$ be a Riemannian manifold of dimension $n$ with Riemannian connection $\nabla,$ and let $p \in M.$ Show that there exists a neighborhood $U \subset M$ of $p$ and $n$ (smooth) vector fields $E_1,...,E_n \in \chi (U),$ orthonormal at each point of $U,$ s.t. at $p,$ $\nabla_{E_i}E_j(p)=0.$  What I've found so far;  Let $V$ be a normal neighborhood at $p.$ Let $(W,x(x_1,...,x_n))$ be a local coordinate system at $p$ s.t. $x(W) \subset V.$  Let $\{X_i:=\frac{\partial}{\partial x_i}\}$ be the standard basis of $T_pM.$ By Gram-Schmidt, there exists an orthonormal basis $\{E_i\}$ of $T_pM.$ Consider the orthonormal frame $E_i(t),$ by parallel transporting $\{E_i(0)=E_i\}$ along a  a geodesic $\gamma :I \to M$ starting at $\gamma(0)=p$ and ending in $\gamma(1)=q,$ where $q \in V.$ This gives us an orthonormal frame at each point of $V.$ To satisfy the second condition, let $E_i(t)=\sum_{l}a_{li}X_l$ and  $E_j(t)=\sum_{s}b_{sj}(t)X_s.$ Then   $$\nabla_{E_i}E_j=\sum_{k}(\sum_{l,s} a_{li}(t)b_{sj}(t)\Gamma^k_{ls}+E_i(b_{kj}))X_k,$$  where $\Gamma^k_{is}$ are Christoffel symbols.   To have $\nabla_{E_i}E_j(p)=0,$ we must have $\sum_{l,s}a_{li}(0)b_{sj}(0)\Gamma^k_{ls}+E_i(b_{kj})(p)=0$ for each $k.$      How can I conclude the argument? Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)? Let's say we have a cloud of dust which is a lightyear across and someone shoots a beam of light from point A to B , why it is not possible for an observer far far away to see the light while it travels through the cloud at the speed of light? In the Landau theory of phase transitions, is the order parameter a thermodynamic variable of state? In the Landau theory of phase transitions, one typically considers a "free energy" $F$ as a function of the temperature $T$ and the "order parameter" $\psi$: $F(T, \psi)$.  For the sake of clarity, let's consider the liquid-gas transition, in which the order parameter is usually taken to be the difference of densities between the ordered phase and the disordered phase.  In this case, it looks that the order parameter $\psi$ is closely related to the the volume $V$, a global variable of state.  What is the correct interpretation of $\psi$ as a thermodynamic variable?   I consider several logical possibilities, but all interpretations look troublesome to me:    $\psi$ is a variable of state replacing the volume. In this case I have a few related concerns:   Is the "free energy" $F$ the Helmholtz or the Gibbs free energy?    If $F$ is the Gibbs free energy, $F$ should depend on the temperature $T$ and the pressure $p$ (the conjugate variable of the volume), so it shouldn't depend on $\psi$.  If it is the Helmholtz free energy, it's ok that it depends on $\psi$, but the experiments are usually carried out at constant pressure, not constant volume.  Usually, state variables are something one can select at will [for instance, think of the phase diagram $(p, T)$]. However, the order parameter $\psi$ is not selected at will, but it is selected by the system as a result of the minimization process according to the Landau theory.  $\psi$ is yet another variable of state in addition to volume and temperature.    Why the pressure is not usually taken into account as a variable of state, and only the temperature is considered? What is the conjugate variable to $\psi$? Can really $\psi$ and $V$ be taken as independent?  $\psi$ is not a variable of state, is something else.   What else can $\psi$ be from the thermodynamic point of view? Same question as above, why is pressure not usually taken into account in Landau theory? How to install MariaDB? I can't see anything - Teams banner and cookie toaster obscure my whole screen There have been  for a while, but this is ridiculous. It reminds me of , but this time it's network-wide.   The banner must be scrolled past (or dismissed) on every page unless both first-party JavaScript and first-party cookies are enabled. If first-party JavaScript is enabled, and the banner is visible, the rest of the page jumps up and down every so often.  Under certain circumstances (not quite sure what they are, yet – a direct link to an answer, maybe? I'm seeing it ) the entire page jumps up and down a line every few seconds even when the banner is off-screen.   The cookie banner keeps coming back unless both third-party JavaScript  are enabled (and  is solved). The cookie box occludes the “Learn more about Teams” link in the banner (with default Firefox window sizes) and the Job / Teams link in the sidebar.  As it stands,  – a site for hobbyists and researchers, whose audience is not very likely to purchase Teams – is much, much harder to use, for no benefit that I can see. Can you roll back this change, please? Recently I had to change the folder permissions of /usr folder. Now the whole sudo is not working. If I try to update my system using this command:  sudo apt-get update   I get the following message:  sudo: /usr/lib/sudo/sudoers.so must be only be writable by owner sudo: fatal error, unable to load plugins   So I tried to change the folder permissions by opening Nautilus as root by using this command  gksudo nautilus   But nothing happened. Nautilus didn't open. This is an emergency as the whole of sudo is not working. when I tried searching the solution from Google many suggested re-installation. I don't prefer re-installation. so is there any other solution? Uniform convergence of derivatives, Tao 14.2.7. Converting vector layer to raster using QGIS? In qgis in the raster menu there is an item "convert vector to raster".  This behaves strangely - I do not understand how it is supposed to work.  I can select a vector file, an attribute and an output raster. But, for some reason the output file is only accepted if it exists and even if I select an existing raster nothing seems to happen.  Also I do not understand how the program knows what kind of raster to create, what cell-size, projection etc.  How can I convert vector to raster? Transforming geometry PostGIS column with unknown geometry type? I run PostgreSQL 9.1 and PostGIS 2.0. I have a table that contains one geometry column.   But that column contains point, polygons lines, etc.   So I set it like so,   ALTER TABLE jake ADD COLUMN jake_geom geometry;  and then I setted some constraints to make it geometry enabled, but not limited on a single geometry type  ALTER TABLE jake   ADD CONSTRAINT enforce_dims_geom CHECK (st_ndims(jake_geom) = 2);  ALTER TABLE jake   ADD CONSTRAINT enforce_srid_geom CHECK (st_srid(jake_geom) = 900913);   Now, I want to reproject the geometries in that column to 4326.   According to , for the transform to work, I need to set a geometry type.  (the TYPE geometry(MultiPolygon,2163) part)  but my column does not have certain geometry type. How do I proceed?  Or I can simply do  ALTER COLUMN geom USING ST_Transform(geom,2163);  avoiding to set a type? Is it possible to recover original content from an edited image file? OK possibly a random question here, but one that is relevant to the site and regarding a hot topic nowadays: privacy.  If I load an image file (in my case a PNG photocopy of a document which contains some personal details) into Gimp, and I 'fill' with black (i.e. as if to redact) the areas of the image that contain sensitive information, and I then overwrite the original file, is there any way that the original untouched image can be recovered or gleamed? Such as low level bit analysis or...?  If so what tools and techniques are behind that, and how can I mitigate this? Is it a simple case of not overwriting the original but simply creating a new file with the changes? "To do this" or "For doing this" to start explaining something? Many questions have been asked and answered about "to + bare infinitive" vs "for + gerund" in the middle of a sentence.  However, when one wants to start explaining something, what is better? For instance:  "Now, let us focus on baking the bread.   To do this, start by turning the oven on.", or For doing this, start by turning the oven on."?   Are there even better expressions? I have not found anything concerning this situation, and I feel that the answers about the same problem but in the middle of a sentence do not apply here. What is the different between an Normal user account and the Admin account in Ubuntu?  You must always enter your password if you want do something. Can you give a Normal account  "limitations" to do not install or remove anything or to not open any program? Can you also "hide/deactive" the terminal and console for a normal user? How can I install software or packages without Internet (offline)? I have a friend who has got a computer that is not connected to the Internet. Is there any way to install software offline easily? Can QGIS export Geospatial PDF for Avenza PDF map app? Since QGIS uses GDAL, does any map exported as a PDF become a Geospatial PDF? I am hoping to import maps from QGIS into the Avenza PDF Map app on my phone. Previous posts mentioned exporting as Geospatial PDF's and GeoTiff's, but none had an exact answer or process.