I know the formula for finding the number of permutations of the $3$x$3$ cube, which is $$\frac{8! \times 3^8 \times 12! \times 2^{12}}{12}$$  But why must the division by $12$ be a part of the formula? From looking it up I know it has to do with some of the positions being impossible without disassembling the cube, but why specifically division by $12$ and not another number?
Could anyone explain why the number of legal or reachable combinations of a $3\times 3\times 3$ Rubik's Cube is $1/12\mbox{th}$ of the total. I understood the logic behind the total number of combinations: $8! \cdot 2^{12} + 3^{12} \cdot 12!$. What I am not able to understand is why do we divide this quantity by $2 \cdot 3\cdot 2$. I don't have any idea about parity or group/set theory. Is there another possible explanation for it?
What's the difference between the work function and ionisation energy? Why is sudoers NOPASSWD option not working? Is it possible to remove a terminal command? For example, if there was a command call lookup (not an alias), how would I go about removing it? Is there a term for grammatical mistakes as a result of trying too hard? How to remove edit from edit queue? I asked a question on SO:  and got an answer that, while good, I thought could be improved with a minor edit:   After a little while, a comment on the answer showed me that I was wrong to have edited the answer.  How do I undo it? In a non-inertial reference frame, the centripetal force is balanced by a centrifugal force. Therefore, shouldn't the net acceleration be zero? Yet, you still feel an acceleration outwards? Why is that? I'd like to do something like this to tick a checkbox using jQuery:  $(".myCheckBox").checked(true);   or  $(".myCheckBox").selected(true);   Does such a thing exist? Let $a,b$ be positive integers.  When $$k = \frac{a^2 + b^2}{ab+1}$$ is an integer, it is a square.    Proof 1: (Ngô Bảo Châu): Rearrange to get $a^2-akb+b^2-k=0$, as a quadratic in $a$ this has two values: $a$ and $kb - a = (b^2-k)/a$. (The second root is determined in two different ways from the expansion $(x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$.)  Now suppose we have $a,b$ such that $k$ is an integer but not a square, by the investigation about roots we have that the second root is a nonzero integer since $k,b,a$ are integers and $k \not = b^2$, futhermore it is positive which is easily seen from its defining equation.  WLOG assume $a \ge b$ so that the second root is strictly smaller than $a$. This leads to a decent, replacing $a$ with the second root.    Proof 2 (Don Zagier): Apply reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + kxy + y^2$, which is the unique reduced quadratic form in its equivalence class.    Note that Proof 2 is pretty much the same as Proof 1 when written out in explicit detail, but I could not read Zagier's book because I cannot read German.  I would like to know more approaches to this and other alternative proofs of this result if possible! Thanks in advance.  I would also be interested in related problems (especially easier ones of a similar nature) and texts which cover the reduction theory in English. Finding places to play Magic the Gathering games A friend of mine recently got MTG for the XBox and is talking us into getting real cards.  I'm all up for starting but I dread the idea of playing the same two people over and over with no change epically since one of them is over competitive and exploits every loophole he can to win.  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The comment adding interface is simply removed on closed questions and their answers &mdash; no Reply button in the toolbar, and no Add Comment button under the question.        Occurs on Android app v1.0.50 and v1.0.51, on a HTC One running Android 4.4.3. How do you run as a different user from the Start Menu in Windows 10? Running an application as a different user (e.g. domain admin account) from the start menu (by shift + right-clicking the application) used to be an option in Windows 7 &amp; XP.  However, I can't find that option in Windows 10. The workaround seems to be either 1) to find the application in Windows Explorer (shift + right-click) or 2) use runas.exe from the command line.  However, in order to use those workarounds, I have look up the executable name first. It's a bit difficult because I don't have the name of every RSAT tool or executable name memorized.  (e.g. "Active Directory Users and Computers" is dsa.msc, "Routing and Remote Acces" is rrasmgmt.msc)  Is there a simpler way to do this? Can you negate the Frightened condition by not looking at the source of your fear? As the title says, if you cause a creature to be Frightened of you, as in the Frightened condition, but they can somehow manage to not see you, do they stop rolling with disadvantage to attacks/ability checks?  Below are scenarios to illustrate my question.  Scenarios:   If a creature is frightened of you, and they turn their back against you to hit someone else, would they no longer have disadvantage to attack rolls/ability checks? If a creature is frightened of you and they have Blindsight, can they close their eyes to stop seeing you, and hence stop rolling with disadvantage? If a creature has 60 ft of movement and is frightened of you, and there is a corner they can reach in 15 ft that will block you from sight, can they run to that corner to stop being frightened, then run back to attack without disadvantage? Assume they would not provoke OA to do this, although the OA is immaterial to the question. Texture Channel animated, doesn't appear in viewport