$\suṃ̣_{k=m}^{n}  (-1)^k {n \choose k} {k \choose m}$ for all positive integers such that n > m.   My attempt, since k=m. I get $\suṃ̣_{k=m}^{n}  (-1)^k {n \choose k} {m \choose m} = \suṃ̣_{k=m}^{n}  (-1)^k {n \choose k}$.  What do I do from this step?
I got quite some trouble trying to prove this.  $$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n&gt;m\geq 0$$  I tried using $$\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$$ and then changing the summation boundaries as $k=0, n=n-m$, but that took me nowhere... I would really appreciate help. Thank you.
Find count of all combination of numbers whose sum is x I understand that in a lot of big-lab fields it is common for the principal investigator to append their name to a paper even if they did not write the paper, design the experiment, or collect data since they spend energy securing funding, and managing the whole lab. What about for small labs?  What are the requirements for a supervisor to be included as an author on a paper, as opposed to just appearing in the acknowledgements? If you are working on your own projects independently of your supervisor, but using funding provided by your supervisor (how does this change when the funding provides resources versus just your salary), are you suppose to add them as authors or just acknowledge the source of funding? C# Settings.Default.Save() not saving? Prove that for any natural number $n>1$, we have $(2n)!<(n(n+1))^n$ Prove that for any natural number $n&gt;1$, we have $(2n)!&lt;(n(n+1))^n$    I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you. Down vote changed to up vote caused reputation meter to pass daily maximum Reflexive and relation Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes. How do I remove a property from a JavaScript object? Say I create an object as follows: let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot;,   &quot;regex&quot;: &quot;^http://.*&quot; };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot; }; How to split larger files into smaller parts? How to relate combinatorial proof to bipartite graphs? The question is - Let $k \geq 1$ be an integer and consider a sequence $n_1,n_2,\cdots,n_k$ of positive integers. Use a combinatorial proof to show that  $\binom{n_1}{2} + \binom{n_2}{2} +\cdots+ \binom{n_k}{2} \leq \binom{n_1+n_2+\cdots+n_k}{2}$  How can i prove this without using induction?  Additionally, For each $i$ with $1 \leq i \leq k$,consider the complete graph on $n_i$ vertices. How many edges does this graph have?  I thought of making graphs and figuring out a relation but still wasnt able to. Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating:  In analytic (Lagrangian) mechanics, the derivation of the Euler-Lagrange equations from the principle of least action assumes that the start and end coordinates at the initial and final times are known. As a consequence, any variation on the physical path must vanish at its boundaries. This conveniently cancels out the contributions of the boundary terms after integration by parts, and setting the requirement for minimal action, we obtain the E.L. equations.  This is all nice and dandy, but our intention is finding the location of a particle at a time in the future, which we do not know a priori; after we derive any equations of motion for a system, we solve them by applying initial values instead of boundary conditions.   How are these two approaches consistent? From my (honestly limited) knowledge of what the Heat Death of the Universe is, I understand that HDotU is the point when energy levels across the entire Universe reach equillibrium, thus energy doesn't flow anymore absolutely anywhere, and thus nothing will ever happen anymore.  If that's the case, would time still be? Even when there's not a single event happening inside it and nobody's there to observe the ticking, does time (or space, come to think of it) exist? Or would HDotU cause something that ends the entire spacetime? How do I find my follower if and when they leave me? primes whose difference is a multiple of $n$ Hello everyone this is my first question here so if there are any suggested edits  feel free to participate! Is it true that for every $n$ there two prime numbers $p$ and $q$ with $n\mid p-q$? I have verified this empirically but  am not sure if there is a proof for this Why is super.super.method(); not allowed in Java? I read  and thought that would easily be solved (not that it isn't solvable without) if one could write:   @Override public String toString() {     return super.super.toString(); }   I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.  What do you guys think?  EDIT: To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it. Deep in my families history (1600's or there abouts) our family split in two. About three or four generations later two distant cousins got married creating a split / rejoin.                G.G.G.Parents A        +---------+---------------------------+     G.G.Grand Parent                      G.G.Grand Parent        |                                     |     G.Grand Parent                        G.Grand Parent        |                                     |     Grand Parent                          Grand Parent        |                                     | +------+-------+                     +-------+-------+ Aunt       Parent A - - -+- - - Parent B         Uncle                          |                   +------+------+                Child A        Child B                  ...           ...   So in this small example, the family splits in two at G.G.Grandparents and then rejoins later when Parent A and Parent B marry.  About 8 years ago I tried to find a software package which could model this and failed. Has anything happened in the intervening time?  Sorry I'm just asking in SE without doing much research, but I spent a lot of effort last time without any success and I am sure there is someone out there who can point me to an excellent package!    @ColeValleyGirl: I'm looking for a piece of software that is able to print out such a tree/DAG (thanks @ACProctor) in high enough quality to display/frame. All software I tried before didn't seem to be able to display this as they seemed to support mainly variations of Ancestors / Descendants rather than an output of all the people that had been inputed.  @AdrianB38: Essentially, I want to see "everyone". The tree I have was put together by my grandfather, who probably had a lot more data/people, but I've only got the hand written tree which highlights this split. It's essentially a tree of my grandparents ancestors (including this split which is joined at a common ancestor) and descendants (me + my family). I'd like to reproduce this in the computer (there is about 400 years - hundreds of people).  @AdrianB38: Re your point about generations, yes, I have that issue. I think it's only one generation, might be two - i've not looked at the tree in a while. The split between families happened over about 150-200 years iirc.  I can write an SVG generator if I really need to (I'm a Software Engineer) albeit I'm aware such a "loop"/"dependency" is painful to program for, but I was hoping that there was something which could handle this. What is the idiom for "as anyone can see" or "clearly"? I keep wanting to say "the handwriting on the wall," but that's not correct. The place I'm trying to use it is, "Alice likes Bob, Carol likes Bob, Alice thinks Bob likes her, but Bob dates Carol, and so as anyone can see, Bob did not like Alice."  What is the appropriate idiom here? In the Android app, left sidebar, there is an editable list of "Your Sites". In the editing pane, the sites are re-orderable (although awkwardly), and the order one places the sites in there is preserved, even after quitting the app and returning to this pane. One expects that order to be reflected in the "Your Sites" portion of the left sidebar when the editing pane is closed, but it is not.  I'm not sure what is determining the order of the sites in the left sidebar, it doesn't seem to be by date they were "pinned" into the list one wants to show, either. Edited photos have reverted to originals on iphone Review counts in top-bar and /review don't match So far, I thought that the number shown in the orange box in the top-bar was the number of items available for review. However, when I acually visit /review, the numbers don't match:    Note that top-bar shows 15 items (?) but the queues only add up to 8.  Since I can't think of a reason for this being by design (which, as experience shows, is not a good indicator for "this is not an SE feature) I'm tentatively labelling this a (UI) bug. PHP Error - Uploading a file