Give a story proof that   $$\sum_{k=0}^n k{n\choose k}^2 = {n{2n-1\choose n-1}}$$  Consider choosing a committee of size n from two groups of size n each , where only one of the two groups has people eligible to become president.   What I think:   On the left hand side..  $$\sum_{k=0}^n k{n \choose k}^2 = \sum_{k=0}^n k{n \choose k}{n \choose k} =\sum_{k=0}^n { k }\underbrace{{n \choose k}{n \choose n - k}}_{\text{${n\choose k}$ is equal to ${n\choose n-k}$}}$$ This is an application of the "choosing the complement identity" ${n\choose k} = {n\choose n-k}$. $$\sum_{k=0}^n k{n \choose k}^2 = \sum_{k=0}^n \underbrace{ k \ \ \ {n \choose k}}_{\text{Pres eligible group}}{n \choose n - k}$$ ${k}$ represents the number of people selected from the "President-Eligible" group of n people. If there are ${k}$ people selected from this group, there must be ${n-k}$ people selected from the second group (since the number of people selected in the final group is n.) The left hand side sums up the cases for ${k}$.  The right hand side multiplies the possibilities for "president" with possibilities for "group members".  One point of confusion I had is that I thought it should be (${n{2n \choose n}}$). But then I related this to the idea that when you use the multiplication rule to find the number of ice cream possibilities, the cone choices (president choices) are separate from the flavor choices (group member choices)! So, the president should not be counted in the group member choices. (let me know if this is the correct way to think about it). But what I'm confused about is why we don't take that into account on the left hand side $k$ possibilities for president is multiplied by ${n \choose k}$ possibilities for members.. so the 'president' is double counted as a member..?
I have to give a combinatorial proof of  $$\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}.$$  I find it difficult to solve such problems. I'm not a brilliant person and never will be so I need to have algorithms to solve most problems. Usually I won't just "see" solutions unless I have a thorough enough understanding of a theory. This is also the case here. I know what the symbols mean, but I don't see what I should count and how to obtain the identity.  I think the right-hand side counts $n$-element subsets of a $(2n-1)$-element set with one element chosen to be "special". I think I should find a partition of the set of such subsets into $n$ parts so that the left-hand side is the sum of the cardinalities of the parts. Is there $n$ of anything on the right-hand side? Well, there are $n$ choices for the "special" elements, but only after I've chosen the $n$-element subset. Otherwise I can choose the "special element" in $2n-1$ ways.  This is a homework problem so I probably shouldn't ask for a full solution. But I would like to have some guidelines for approaching such problems, perhaps based on this example.
Resistors in series vs parallel Check if url has domain name and not an ip I have 3 accounts that I am deleting - however, several hours after the 24 hour (+ grace period - end about 4-5 hours ago), the countdown completed, but disappeared and the accounts are still there.   I would rather not go through another 24 hours, and would like to have these accounts deleted (and yes, I have thought about it and this decision is not up for discussion).  Currently, the accounts are still there.  Could this please be seen to? Why does the max aperture change when use Nikon 60mm AF-D Micro on D90 camera? I have found that my Nikon 60mm Micro AF-D changes aperture from 2.8 to 3.2 when used with a D90. I have tried both A and M mode. Is it by design? Why?  I intend to buy  and it will lock the maximum aperture at 2.0 (my guess) instead 2.8. So should I buy it or not, or would it be better to buy the newer 50mm 1.4 AF-S version? Undefined reference to static class member Help with null hypothesis As , the retag privilege serves a lot more confusion and a lot less of its benefit due to the suggested edit system. It must be done away with. It will be done away with. ♪  In light of this decided course of action, we're going to need a new snack for the road between "Create Tags" at 300 reputation and "Established User" at 1000 reputation. We'd like your help in thinking about some ideas on what would be a nice privilege to sit at this level. Since we're... a bit dry on our end.  The primary guideline in deciding what kind of privilege to put here is acknowledging its position in the overall hierarchy of privileges. 500 reputation is an early-intermediate user, one who has started to really get the hang of the site and its function. It can be something that provides convenience in how they currently can act on the site, or it could be some new functionality that fits as an early ability (especially considering there's no new abilities until 2000 reputation's "Edit Everywhat" ability).  I look forward to hearing what creative ideas y'all can provide. Difference between "did save" and "saved" What is the difference between the two following sentences?     1) Where did you save that file?   And      2) Where you saved that file ? Sam Weller in The Pickwick Papers: What accent is Dickens portraying? In Dickens' Pickwick Papers, there's a character "Sam Weller". Weller's dialogue is written somewhat phonetically, I presume, but I'm struggling to understand what accent Dickens is trying to portray. The main peculiarities of Weller's speech are using "v" where there should be a "w" and a "w" when there should be a "v". For example, he says "wery" instead of "very" and "avay" instead of "away". 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How do you perform a left outer join using linq extension methods Assuming I have a left outer join as such:  from f in Foo join b in Bar on f.Foo_Id equals b.Foo_Id into g from result in g.DefaultIfEmpty() select new { Foo = f, Bar = result }   How would I express the same task using extension methods? E.g.  Foo.GroupJoin(Bar, f =&gt; f.Foo_Id, b =&gt; b.Foo_Id, (f,b) =&gt; ???)     .Select(???) Show that    $$\frac{\sum_{i=0}^n\binom{k}{i}\binom{M-k}{n-i}\frac{k-i}{M-n}}{\binom{M}{n}}=\frac{k}{M},$$ where $M-k&gt;n$ and $k&gt;n$   From Vandermonde's identity, I get $\sum_{i=0}^n\binom{k}{i}\binom{M-k}{n-i}=\binom{M}{n}$.   But what about the sum of the product in the numerator? Any hint will be helpful. I made a hair particle system and set the Dupli Group to a group of 3 objects. 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