Coefficients involved in the Binomial Theorem. $\binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

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Combinatorial Interpretation of Fractional Binomial Coefficients

My question is a bit imprecise - but I hope you like it. I even strongly think it has a proper answer. The binomial coefficient $\binom{\frac{1}{2}}{n}$ is strongly related to Catalan numbers - the ...
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237 views

Vandermonde identity in a ring

Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and ...
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Asymptotics for sum of binomial coefficients from Concrete Mathematics

Concrete Mathematics EXERCISE 9.25: Supposing \[ S_n = \sum_{k=0}^n \binom{3n}k \] Prove that \[ S_n = \binom{3n}{n}\left(2-\frac4n+O\left(\frac1{n^2}\right)\right) \] This sequence also ...
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Solution of $\large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ with $n\geq 3$

I found this question in an old problem set. There's no hint or solution mentioned. For $n \geq 3$, prove or disprove the existence of $(x,y,z) \in \mathbb N^3, ...
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Find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$?

I just want to ask you if my solution is correct. Here's the problem, Using the Binomial Theorem, find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$. Solution: The binomial theorem is, ...
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Simplify $\sum_{i=0}^n (i+1)\binom ni$

Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$ $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
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Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$

Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ ...
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What is the coefficient of $x^{25}$ in $(x^3 + x + 1)^{10}$?

Working on some contest problems and came across this question. Here's what I have so far on the off chance that my thinking is correct... So using Vieta's the coefficient of the $x^{25}$ should be ...
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251 views

Identity for $\sum\limits_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j}$?

I have run across the following multinomial series: $$ \sum_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j} $$ Here, $d>1$. This seems like a formula which has either a well-known identity, ...
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The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is

The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is? I tried this: ...
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Short and intuitive proof that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$

The simple inequality that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ has a number of different proofs. But is there a particularly intuitive, short and elegant proof that uses the natural ...
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Is there a binomial identity for this expression $\frac{\binom{r}{k}}{\binom{n}{k}}$?

What I'm trying to prove is this summation: $$\sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i = \dfrac{r}{n} \cdot k$$ I used induction on $k$ as follows: ...
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Bounds for $\binom{n}{cn}$ with $0 < c < 1$.

Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq ...
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Number of acyclic digraphs on $[n]$ with $k$ edges and each indegree, outdegree $\leq 1$

How many (labelled) acyclic digraphs are there on the vertex set $[n]$ with exactly $k$ edges and each indegree, outdegree $\leq 1$? The answer is $${n \choose k} {n-1 \choose k} k!.$$ Is there a ...
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How to understand this combinatorially: $\sum^{2k}_{i=0} \binom{4k}{2i} (-1)^{i}=2^{2k}(-1)^{k}$

The TAs in my department are stuck in assisting an undergraduate with the following problem: $$\sum^{2k}_{i=0} C^{4k}_{2i}(-1)^{i}=2^{2k}(-1)^{k}.$$ We tried to solve this via induction (obviously ...
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Inductive proof for the Binomial Theorem for rising factorials

I want to proove the following equality containing rising factorials $$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$ For $n=1$ this equality is ...
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How to simplify or calculate a formula with very big factorials

I'm facing a practical problem where I've calculated a formula that, with the help of some programming, can bring me to my final answer. However, the numbers involved are so big that it takes ages to ...
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What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$?

What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$ in terms of $d$? Does the limit exist? Is there a simple upper bound interms of $d$?
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Inequality involving factorial

I am trying to prove following inequality: $$\binom{n}{k}<(en/k)^k$$ I tried Stirling approximation but I could not get anything further. Then I get $$\binom{n}{k}\approx \frac{\sqrt{2\pi ...
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Truncated alternating binomial sum

It is easily checked that $ \sum_{i = 0}^n (-1)^i \binom{n}{i} = 0$, for example by appealing to the binomial theorem. I'm trying to figure out what happens with the truncated sum $\sum_{i=0}^{D} ...
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Help evaluating a limit

I have the following limit: $$\lim_{n\rightarrow\infty}e^{-\alpha\sqrt{n}}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}\sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}$$ where $\alpha>0$. ...
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Yet another sum involving binomial coefficients

Let $k,p$ be positive integers. Is there a closed form for the sums $$\sum_{i=0}^{p} \binom{k}{i} \binom{k+p-i}{p-i}\text{, or}$$ $$\sum_{i=0}^{p} \binom{k-1}{i} \binom{k+p-i}{p-i}\text{?}$$ ...
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How to solve 0.5 choose 4?

I was solving this problem for homework. It says, in the problem, that if n is positive you use the generalized definition of binomial coefficients. In my case, n is positive so I just plugged n= 0.5 ...
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Binomial Theorem past exam question, what do I do?

I have trouble understanding what I'm supposed to do in some of these math questions. Here's an exam question from an old exam: Let $A$ be a set with $n$ elements. The number of subsets of $A$ with ...
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Choice Problem: choose 5 days in a month, consecutive days are forbidden

I'm "walking" through the book "A walk through combinatorics" and stumbled on an example I don't understand. Example 3.19. A medical student has to work in a hospital for five days in January. ...
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Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$

Using the Taylor expansion for $${(1+x)}^{-1/2}$$ we have $${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$$ for $|x|<1$. But if $|a| <1$, how can we use the above fact to find ...
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Formula for $\sum_{k=0}^n k^d {n \choose 2k}$

If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$ We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = ...
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a problem about binomial coefficients sum

I want to know how to caculate $\binom{2n}{0}^3-\binom{2n}{1}^3+\cdots+(-1)^k\binom{2n}{k}^3 \cdots+\binom{2n}{2n}^3$? The sum equals $ (-1)^{n}\binom{3n}{2n}\binom{2n}{n} $, but I donot know how to ...
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Help with a Binomial Identity: $\sum_{k=0}^{\infty} \binom{n+k}{k}2^{-k} = 2^{n+1}$

The following is a problem from the 5th edition of Niven's An Introduction to the Theory of Numbers: Problem 23 of Section 1.4 asks us to prove that $$\sum_{k=0}^{\infty} \binom{n+k}{k}2^{-k} = ...
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Wanted: Insight into formula involving binomial coefficients

There is the formula $$\sum_i (-1)^i\binom{a}{k+i}\binom{l+i}{b} = (-1)^{a+k} \binom{l-k}{b-a}.$$ Only finitely many summands are non-zero (those for $i\in\{b-l,\ldots,a-k\}$), so the sum is finite. ...
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Combinatorial proof: $p^{r-n}$ divides $\binom{p^{r-2}}{n}$

Let $p$ be an odd prime. Then if $1<n<r$, $$p^{r-n}\,\left|\,\binom{p^{r-2}}{n}\right.$$ Does anyone have a clever combinatorial proof of this fact? There's an easy argument just by counting ...
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Fractional Binomial Coefficients

I recently examined the binomial coefficient $\binom{\frac{1}{2}}{k}$ and found that the denominator was always a power of two. The same is true of $\binom{\frac{1}{3}}{k}$, where the denominator is ...
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How does Combination formula relates in getting the coefficients of a Binomial Expansion?

Sorry for the very basic question. I'm a programmer, not a mathematician. The title says it all but I'll just include a little background on why I asked the (stupid?) question. I'd just finished ...
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Prove the identity $\sum\limits_{s=0}^{\infty} {p-s \choose m}{q+s \choose q} = {p+q+1 \choose p-m} $

I need to prove $$\sum_{s=0}^{\infty} {p-s \choose m}{q+s \choose q} = {p+q+1 \choose p-m} $$ using: $$(1-x)^{-m-1} (1-x)^{-q-1} = (1-x)^{-m-q-2} .$$ ok, generating function :$\frac1{(1-x)^{m+1}} = ...
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Binomial probability with summation

Show that $$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$$ Attempt: It becomes: $$\sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}}$$ Telescoping, pairing, binomial theorem don't ...
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How many solutions are possible to this equation?

Given $$A+2B+3C=N $$ where $N$ is a given positive integer. $A ,B,C\in\mathbb{N}$ vary from $0$ to $\infty$. How many solutions will be there to this equation?
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Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$

I am trying to evaluate the finite sum \begin{equation} f(u)=\sum_{k=0}^n{n\choose k}^2u^k,\quad 0<u\le1 \end{equation} In an first attempt, I think of the generating function \begin{equation} ...
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Show that $\sum_{k=2012}^{n} 2^k\binom{n}{k} = \Theta(3^n)$

In this question we are asked to show that $\sum_{k=2012}^{n} 2^k\binom{n}{k} = \Theta(3^n)$ What I did: $\sum_{k=2012}^{n} 2^k\binom{n}{k} = \sum_{k=2012}^{n} 2^k*1^{n-k}\binom{n}{k} \leq ...
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find the sum $\sum_{k=0}^{n} k\binom{n}{k}^2p^k$

I need to find the sum $$\sum_{k=0}^{n} \binom{n}{k}^2kp^k,$$ for an integer $n$ and $0<p<1$. Mathematica would only return $$\sum_{k=0}^{n} k p^k \binom{n}{k}^2 = n^2 p \ _2F_1(1-n, 1-n, 2, ...
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An enlightening proof of a specific combinatorial identity

Concerns about the arithmetic genus of projective hypersurfaces led me to make the following combinatorial conjecture: $${d-1\choose n+1} =\sum_{i=0}^{n+1} (-1)^{n+i+1} {d\choose i}$$ for $d \geq 1$, ...
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Partial sum of rows of Pascal's triangle

I'm interested in finding $$\sum_{k=0}^m \binom{n}{k}, \quad m<n$$ which form rows of Pascal's triangle. Surely $\sum\limits_{k=0}^n \binom{k}{m}$ using addition formula, but the one above ...
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Are these two binomial sums known? Proven generalization to the Hockey Stick patterns in Pascal's Triangle

English translation. You can see the original - deprecated - in Portuguese here Hi, I arrived at a generalization for the Hockey Stick Patterns, from our beloved Pascal's Triangle. This ...
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Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$ I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively ...
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Intuition behind negative combinations

Take $\binom{n}{r}$. It denotes how in how many different ways you can choose $r$ elements from a set of $k$ elements. For case $\binom{4}{3}$ which evaluates to $\frac{4!}{3!(4-3)!}=4$, it perfectly ...
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Elementary bound of binomial coefficient

I'm working my way through an Erdős paper from the sixties and some of the elementary bounds he claims seem to be just beyond my reach. The expression looks horrendous but maybe there is a clever ...
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190 views

A resemblance between 2 binomial identities - why?

Let $F$ be any field (or even ring). The following formal power series identity (i.e., equality in $F[[x]]$) holds for any $j \ge 0$: $$(1-x)^{-j} = \sum_{i \ge 0} \binom{i +j -1}{i} x^i $$ The ...
6
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Prove $(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j} $

I stumbled upon the identity $$(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}. $$ The right-hand side is a polynomial. ...
6
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1answer
157 views

How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$

How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$ This equation How prove it? Thank you I want take this ...
6
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1answer
375 views

Vandermonde's-like identity

The Vandermonde's identity gives $$\sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.$$ Here is an example of Vandermonde's-like identity: For all $0 \le m \le n$, $$\sum_{k=0}^{2m} ...
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Relation between different ways of accessing bernoulli numbers with matrices

First Variant: Bernoulli numbers can easily be expressed by linear algebra equations. For example just using the recursion formula $$\sum_{k=0}^{n-1}{n\choose k}B_k=0$$ which is equation (34) from ...