In a step in a proof that the probability to return to origin in a symmetric random walk is $1$ the following combinatorics result seems important:  $$\displaystyle \sum_{n=0}^\infty{2n\choose n}\,x^n = \frac{1}{\sqrt{1-4x}}$$  I understand that the proof involves the generalized binomial theorem, but I don't know how to get about solving the interplay between the $x^n$ in the power series and the binomial coefficients.
37474
How would you prove that the generating function of $\binom{2n}{n}$ is $\frac{1}{\sqrt{1-4x}}$?  More precisely, prove that( for $|x|&lt;\frac{1}{4}$ ):  $$\sum^{\infty}_{n=0}x^n\binom{2n}{n}=\frac{1}{\sqrt{1-4x}}$$  Background: I was trying to solve $$S=\sum^{\infty}_{n=0}\frac{(2n+1)!}{8^n(n!)^2}=\sum^{\infty}_{n=0}\frac{(2n+1)}{8^n}\binom{2n}{n}$$ Which if we let $f(x)$ be the generating function in question, would be simply $$f(x)+2xf'(x)$$ With $x=\frac{1}{8}$. Is there a simple proof of the first identity?  states it without a proper reference ( states it without proof). Is there an easier way of calculating $S$? (which is $\sqrt{8}$, by the way)
58852
has the identity $$S(n+1, k+1) = \sum_i {n \choose i} S(i, k)\;.$$ I can't seem to find a proof for this and was wondering what it entails. Is it a purely combinatoric argument, or an algebraic manipulation of the explicit formula $S(n, k) = \frac{1}{k!} \sum_{i=0}^k (-1)^i {k \choose i}(k-i)^n $ with the known recurrence formula $S(n+1, k+1) = S(n, k) + (k+1)S(n, k+1)? $
33366
I flip a coin until I get $m$ more heads than tails, or $n$ more tails than heads. Let the expected number of flips of the coin before stopping be $f(m,n)$.  I obtained $f(m,n)=mn$ from the recursion $f(m,n)=1+\frac{f(m-1,n+1)+f(m+1,n-1)}2$ with $f(k,0)=f(0,k)=0$ for all $k$.   Other than going through this recursion (and either solving by inspection or by writing as linear recurrence in single variable and solving brute force), is there an intuitive reason you should expect this process to take $mn$ flips? I was thinking about the more general problem with probability $p$ of getting heads and was struck by how simple the formula became when handling, what turned out to be a special case (general formula broke down) of $p=\frac12$.
119758
Is it true that $\sum_kk\binom{n}{k}^2=n\binom{2n-1}{n-1}$? (I proved it using generating functions). Could you help me to prove it combinatorially? please
109007
This is a homework problem. In the first part of the problem, I managed to use a combinatorial problem to prove the following identity:  $\sum_{k=0}^{n}(-1)^k {2n-k \choose k} 2^{2n-2k} = 2n+1$  But I actually have problem with the second part of the problem which asked me to prove this identity "directly", probably using some form of generating functions/algebra method.  There was a hint given but I got stuck: The hint says to calculate   $\Sigma_{n=0}^\infty\Sigma_{k=0}^{n}(-1)^k {2n-k \choose k} 2^{2n-2k} x^{2n}$  and that it would be useful to consider the fact that $ \Sigma(2n+1)x^{2n}= \frac{d}{dx}\frac{x}{1-x^2} $ and $(1-x)^{-a-1}=\Sigma_{j=0}^\infty {a+j \choose j}x^j$.  I am able to prove both these "possibly useful facts" in the hint, but I don't see how to calculate the suggested double sum or prove the identity. [I would have thought that the combinatorial proof is harder to find!] [You may assume that I am familiar with the formal treatment of power series.]  Any help would be greatly appreciated.
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In the first course of probability by sheldon ross on page 155 at the bottom. It is the proof for the expectation of Binomial distribution. It uses the identity:  $$i{n \choose i } =n {n-1 \choose i - 1}$$  Since learning combinatorics, I have been trying to think of this intuitively. So is there an intuitive proof for this identity, rather than just using brute force?
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Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
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In a proof I've come across the following identity:     $$ \sum_{l=0}^{n-j} \binom{M-1+l}{l} \binom{n-M-l}{n-j-l} = \binom{n}{j} $$   I see that it's right, when plugging in numbers, but I don't see the algebraic or combinatorial proof behind it. Can anyone help me with that?
110410
Prove that for any $0\lt r\lt n$ we have $$\binom nr=\binom{n-1}{r-1}+\binom{n-1}r.$$  How do prove this and what step do i take in order for it to be true?
122214
I believe the following is an identity (I've tested with a few random $m$ and $n$ values, could be wrong though):  $$\sum_{k= 0}^{\infty}{m \choose k}{n \choose k}k=n\binom{m+n-1}{m-1}$$ but I'm not sure how to prove it.   Initially I thought of considering the right side combinatorially: pick $m-1$ things from a bucket of $m+n-1$ things and repeat this $n$ times. But how do I think of the left side combinatorially, since there is an infinite sum? Or maybe this is the incorrect approach?
87923
Prove that $$\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$$  The exercise provides the following hint: $\,\,\displaystyle{n \choose k}={n\choose n-k}$.  Any help?
18019
I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.  I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.  I want to see an actual proof behind this equation.  Does anyone have any ideas?
23566
I read upon  Gambler’s Ruin problem and encountered this interesting question. Consider a simple random walk on $\{0, 1, \ldots, N\}$ with absorbing barriers at $0$ and $N$. What is the probability $u_k$ that the walk is absorbed at $N$ if it begins at a point $k$, $0 \leq k \leq N$?  If the question doesn't indicate anything, does it mean this is symmetric? If so, would it be $u_k=\left(\frac{1}{2}\right)^{(N-k)}$?
109886
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.
90426
I need to show that $$\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$$  I know that $\displaystyle \binom{n}{k}\binom{k}{i}$ is counting the number of ways to pick $k$ elements from $n$ and then $i$ elements from those $k$, so $\displaystyle \sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i}$ is counting the number of ways to do that $\forall k$.  I know that $\displaystyle \binom{n}{i} = \binom{n}{n-i}$ and that $2^{n-i}$ is the number of elements in the power set for $[n]\setminus[i]$.  How do I show that both sides are equivalent? I'm stumped on this one. Could anyone give me a pointer in the right direction?
41278
A summary of the given question:   A biased coin has a probability ($p$) of getting heads. Whats the probability of getting the 1st heads on an even numbered flip. $H_1$ is head of first flip. $E$ is first head on even numbered flip. $X^c$ means event $X$ didn't occur. For example: $H_1^c$ means tails on first flip      I don't understand the part  $$P(E | H_1^c) = P(E^c)$$  I seem to think in fact, $P(E | H_1^c) = P(E)$ since the only way $P(E)$ can occur is when $H_1^c$ has occurred.   Moving forward, how do I get   $$P(E) = (1-p)(1-P(E)) = \frac{1-p}{2-p}$$
17910
Show that $\displaystyle{n!=1+\left(1-\frac1{1!}\right)n+\left(1-\frac1{1!}+\frac1{2!}\right)n(n-1)+\cdots}$. I can't figure out how this can be solved. I tried to use the binomial theorem but I couldn't prove it. Any help will be greatly appreciated.