When n and k are positive integers.   Would this be a correct proof?  Description: Suppose how many ways can we choose k people from a group of n + 1.  LHS: The left-hand side gives us the number of ways to do this by definition, on how to select k from n+1.   RHS: The right-hand side gives us the number of ways to do this as well, but instead, it first counts the number of elements plus one (n+1) times the number of ways to choose an element a subset of k - 1 in n over the number of elements k.
Suppose we want to prove $$ k \binom{n}{k} = n \binom{n-1}{k-1}$$   In the LHS we are choosing a team of $k$ players from $n$ players. Then we are choosing a captain. In the RHS we are choosing a captain from the $n$ players. Then we are choosing the remaining $k-1$ players from the $n-1$ players.   Is this a correct interpretation?
I would like to clarify the differences between by value and by reference.  I drew a picture     So, for passing by value,  a copy of an identical object is created with a different reference, and the local variable is assigned the new reference, so to point to the new copy  How to understand the words: " If the function modifies that value, the modifications appear also within the scope of the calling function for both passing by value and by reference "  Thanks! I think my MacBook Pro battery was replaced by Apple's service team. How do I check that? I handed my MacBook Pro (15" Touch bar) to the service centre in order to have the keyboard replaced. Now, when I check the battery cycle count, it shows "1". Before it was at "82".   Also, I'm getting a battery life of only 4.5 - 5 hours now.   Is there a way to find if my battery was replaced or not? Can we insert or update FieldSets using apex code? Rotate (group) objects outwards on spherical particle system? Please anyone tell me when we have to use "has been, have been , had been" and something  started in past and still continuing is it possible to use has been ? The  EU book The Rise and Fall of Darth Vader reveals that Vader personally gave the stormtroopers the order to kill Owen and Beru Lars, and that he knew who he was condemning to death before he gave the order:  “Lord Vader,” said one of the sandtroopers, adjusting a control on his helmet so that only Vader could hear his voice.  “The Jawas sold a protocol droid and an astromech to these moisture farmers, but both droids are gone.” Moisture farmers? Intrigued, Vader examined the holograms of the kneeling couple and said, “The farmers’ names?” “Owen and Beru Lars, sir,” the sandtrooper responded. “They say they don’t know where the droids are, but it looks like a landspeeder is missing from their garage.” Owen and Beru, Vader recalled... “Tell Mr. and Mrs. Lars that they seem to have trouble keeping protocol droids on their property.”1 Not certain if he had heard correctly, the sandtrooper said, “Sir?” “Then you may extend to them every courtesy that you showed the Jawas before you continue your search. Establish checkpoints to detain any droids entering Mos Espa or Mos Eisley spaceports. And one more thing.” “Yes, Sir?” “Do not stop transmitting until I break the connection.” “Understood,” said the sandtrooper. Vader watched the sandtroopers carry out his orders on their helpless victims. He found the sight of rising flames — even holograms of flames burning millions of light years away — to be most satisfying.  I had always assumed that the stormtroopers killed the Larses on their own initiative, or in accordance with a previously given order.  The version of events in this book is interesting, but it raises the question of why it took Vader so long to figure out who Luke was. In canon, who gave the order to kill Owen and Beru Lars?  If Vader was involved, did he recognize the couple before he gave the order?  1 A reference to the fact that Anakin took C-3PO from the Lars homestead the last time he was on Tatooine, without informing the Larses that he would do so. Consider an irreducible but possibly periodic Markov chain on a finite state space with transition matrix $P$. We know there exists a unique stationary distribution $\pi$. If the Markov chain were aperiodic, we would have $P^n_{ij} \to \pi(j)$ as $n \to \infty$. This fails if the chain is periodic, but we have convergence of the Cesaro averages: $$\frac{1}{n}\sum_{k=1}^n P^k_{ij} \to \pi(j) \text{ as } n \to \infty.$$ Can anyone point me to a reference that states this fact? Every reference I've seen only considers convergence for aperiodic chains, or "fixes" periodicity by considering a lazy version of the chain. Alternatively, is there a simple way to obtain this result using the result for aperiodic chains? Why does adding a DOCTYPE affect my CSS? How does one account for the momentum of an absorbed photon? How to prove that $\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0$ & $\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0$? Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$? Unbiased estimator for geometric distribution parameter p Inequality involving the integral of vector valued function Suppose $f :[a,b]\to \mathbb{R}^n$ is given by $f(t)=\langle f_1(t), \ldots, f_n(t)\rangle$, where each of the $n$ component functions is integrable over $[a,b]$. I think the following inequality   $$\left|\int _a^b f(t) \, dt\right|\leq\int_a^b\left|f(t)\right| \, dt$$  still hold but I don't know how to prove it. Any idea?  Thanks Let $x &gt; 1$ and $a$, $b$ be positive integers. I know that $a$ divides $b$ implies that $x^a - 1$ divides $x^b - 1$. If $b = qa$, then  $$x^b - 1 = (x^a)^q - 1^q = (x^a - 1)((x^a)^{q-1} + \ldots + x^a + 1).$$  I'm interested in the converse of the statement. If $x^a - 1$ divides $x^b - 1$, does this imply that $a$ divides $b$? Choosing between meaningful and meaningless hostnames I have a batch file which is in a directory and must be run from there as well because it updates files within this directory. This works perfectly fine, except when the user runs the batch file as administrator (required on Vista). Then the starting directory is C:\Windows\System32. Is there any way to still be able to know from which directory the batch file was run? I dont want the user to enter the directory manually. I understand that division by zero isn't allowed, but we merely just multiplied $f(x) = 1$ by $\frac{x-1}{x-1}$ to get $f(x) = \frac{x-1}{x-1}$ and $a\cdot 1 = 1\cdot a = a$ so they're the same function but with different domain how is this possible?  Or in other words why don't we simplify $f(x) = \frac{x-1}{x-1}$ to $f(x) = 1$ before plotting the points. Is it just defined this way or is there a particular reason ?  Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of  $f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$. Find records from one table which don't exist in another I've got the following two tables (in MySQL): Phone_book +----+------+--------------+ | id | name | phone_number | +----+------+--------------+ | 1  | John | 111111111111 | +----+------+--------------+ | 2  | Jane | 222222222222 | +----+------+--------------+  Call +----+------+--------------+ | id | date | phone_number | +----+------+--------------+ | 1  | 0945 | 111111111111 | +----+------+--------------+ | 2  | 0950 | 222222222222 | +----+------+--------------+ | 3  | 1045 | 333333333333 | +----+------+--------------+  How do I find out which calls were made by people whose phone_number is not in the Phone_book? The desired output would be: Call +----+------+--------------+ | id | date | phone_number | +----+------+--------------+ | 3  | 1045 | 333333333333 | +----+------+--------------+ How do Zombies differentiate between a zombie and a human being? Zombies are always after human beings. How do Zombies differentiate between a zombie and a human being? Why don't zombies eat each other? Real valued analytic function defined on a connected set is constant Let $G$ be a connected set and $f : G \rightarrow \mathbb{C}$ a real valued analytic function. Prove that $f$ is constant.    My idea to prove the result is to prove a subset $A \neq \varnothing$ of the connected set $G$ is both open and closed. So $G=A$ Take $f(w) = a$ $$A = \{z\colon z \in G, f(z) = a\}$$ Now I want to show that $A$ is infinite. How to do it? After that it is easy to prove $A=G$. Unable to login to account in ubuntu 18.04 vmware workstation 15, after update I installed the latest updates from the software update center today (I update when ubuntu displays a notification), restarted the machine and now it's unable to boot properly. The best screen I can get is displayed below. If I resize the screen then it goes to black. If I restart the machine then it displays just purple.    Note that the screen is unresponsive, I can't click anything on it.  Other than SMBus Host Controller not enabled! I don't see any other warnings. Note that I don't know if this warning existed before the update.  I'm running ubuntu 18.04 in vwmare workstation 15 player (latest update). Any ideas on how to begin debugging this?    Update: Trying to boot Ubuntu, with Linux 4.15.0-50-generic instead of Ubuntu, with Linux 4.15.0-51-generic didn't help.  Update: Blacklisting i2c-piix4 according to  answer didn't remove the warning or fix the issue.  Update: Blacklisting i2c-piix4 according to  answer removed the warning but didn't fix the issue.  Update: The answer from Uzair helped me log into the account (run dpkg and update grub from recovery mode).  Answer:  answer helped me fully to resolve the issue. I don't understand the fix. If anyone does, put it as the answer and I'll mark it as such.  Regarding duplicity: The answer from another question did solve my issue but I don't think my question is a duplicate of that one. I began with a fresh install of 18.04 a few months ago, did several updates without an issue and only today experienced issues which are different than those described in the other question. I think we experienced different bugs with wayland and we both solved it by disabling it which is a destructive solution applicable to many questions.