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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Approximating the logarithm of the binomial coefficient

We know that by using Stirling approximation: $\log n! \approx n \log n$ So how to approximate $\log {m \choose n}$?
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Permutations containing a given subsequence

Let $f(n)$ denote the number of $4n$-long strings formed from $2n$ a's and $2n$ b's, such that the string contains, as a (possibly non-consecutive) subsequence, a pattern containing $n$ a's and $n$ ...
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Proving $\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}$ algebraically

I am working through the exercises and have spent half a day on one problem so I decided to get some help because I can't figure it out. Show that if $n$ is a positive integer at most equal to $m$, ...
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1answer
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$\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{2}{n} \right) \cdot … \cdot \left( 1 - \frac{k-1}{n} \right) = \frac{n!}{n^k r! (n-k-r)!}$

I'm trying to understand a proof in "Interpolation and Approximation by Polynomials" by Phillips. Let me quote (page 253): "For $k\geq 1$ we begin with $$B_{n+k}^{(k)}(f;x)=\frac{(n+k)!}{n!} ...
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Mathematical Induction problem involving binomial coefficients [on hold]

I cannot solve it by using mathematical induction. Please help me with it. $$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{n}=2^n$$ ...
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No closed form for the partial sum of ${n\choose k}$ for $k \le K$?

In Concrete Mathematics, the authors state that there is no closed form for $$\sum_{k\le K}{n\choose k}.$$ This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the ...
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Lower bound for the chromatic number of $\mathbb{R}^n$

I'm going through a proof that of the following lower bound for the chromatic number of $\mathbb{R}^n$: $$\chi(\mathbb{R}^n) \geq (1.2 + o(1))^n$$ At some point in the proof we get that ...
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Is the given binomial sum almost everywhere negative as $K\to\infty$?

The binomial sum is as follows: $$\mathcal {L}^K(\theta)= \sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\theta^i\left((1-\theta)^{K-i}-\frac{1}{2}(1-\theta)^{-K}(1-2\theta)^{K-i}\right)$$ which can also ...
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Stuck in a problem in permutation and combination.

I am solving problems in permutation & combination and stuck in this problem. Two players $P_1$ and $P_2$ play a series of $2n$ games. Each game can result in either a win or a loss for $P_1$. ...
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Summation of $\binom{N}{K}$

I was working on a math problem that required me to figure out the general summation of $\binom{N}{0} + \binom{N}{1} + \ldots + \binom{N}{K}$. I know that if $k = N$, the answer is $2^N$. But is ...
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How to simplify $\sum_{r=1}^{y} \binom{x-1}{r}\binom{y-1}{r}$? [on hold]

To find sum of the product of two combination terms $$\sum_{r=1}^{y-1} \binom{x-1}{r}\binom{y-1}{r}$$
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1answer
151 views

How can I prove this combinatorial identity?

Let $n,m$ be non-negative integers. How can one prove the following identity? $$\sum_{j=0}^n j\binom{2n}{n+j}\binom{m+j-1}{2m-1}=m\cdot4^{n-m}\cdot\binom{n}{m}$$
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An identity involving Stirling numbers of the second kind and binomial coefficients

Need to prove: $$\sum\limits_{k=0}^{n} \binom nk k^r x^k = \sum\limits_{j=0}^{r} \binom nj j! (1+x)^{n-j} x^j S(r,j)$$ where $S (n, k)$ denotes a Stirling number of second kind, the number of ...
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1answer
25 views

Show that if $x\equiv 1 \pmod {m^k}, $then $x^m \equiv 1\pmod{m^{k+1}}$.

Let $k\ge 1, m\ge 1.$ Show that if $x\equiv 1 \pmod {m^k}, $then $x^m \equiv 1\pmod{m^{k+1}}$. First I noticed that the assumption would imply $x^m \equiv 1 \pmod{m^k}$, but that doesn't seem to ...
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1answer
19 views

What's the solution to this binomial?

what's the coefficient of $x^6$ in the expansion of $(1+X^2+X)^{-3}$? I have factorized the term to $\left(\frac{1-x^{-3}}{1-x}\right)^3$ after this I'm having problem solving it
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Combinatorial identity binomial coefficients [duplicate]

How to prove that $$ \binom{m}{p} = \sum_{j=0}^q \binom{q}{j}\binom{m-q}{p-j}\;?$$
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Formula for coefficient of Mahonian numbers

I recently came out with this article . It tells about triangle of mahonian number.The T(n,k) is coefficients in expansion of Product_{i=0..n-1} (1 + x + ... + x^i), where k ranges from 0 to n*(n + ...
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2answers
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Why General Leibniz rule and Newton's Binomial are so similar?

The binomial expansion: $$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$ The General Leibniz rule (used as a generalization of the product rule for derivatives): $$(fg)^{(n)} = ...
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1answer
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finding the value of a node in Pascal’s (a.k.a Yanghui's) triangle [closed]

Image the Pascal Triangle is on an x-y cartesian plane. so that the values of the nodes, by location are ...
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1answer
22 views

Unimodality of sequence

I have to show the following: a) was pretty easy to show, however, I am not able to get something useful out of the recursive definition in b) and I have no idea how to approach c). What bijection ...
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3answers
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Inequality of factorial - Binomial coefficient

my name is Rafał and I decided to create this thread because of my inability to find a solution. I have been fighting with this inequality for 1.5 week and I have a hope that you will give me any hint ...
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Curious Binomial Coefficient Identity

Consider the following set of identities: ${m+1\choose 1}={m\choose 1}+1$, ${m+1\choose 2}=2\binom m 2 - {m-1\choose 2}+1$, ${m+1\choose 3}=3\binom m3-3{m-1\choose 3}+{m-2\choose 3}+1$, ... This set ...
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1answer
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Complex Analysis proof of multinomial expression

I've recently come across the following identity $$ \displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n} $$ A nice complex analysis proof (by Felix Marin, here) follows as: ...
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1answer
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Proving $\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$

Prove the identity: $\displaystyle\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$ It looks a bit similar to the "no gets their own hat back" problem or inclusion exclusion ...
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coefficients of polynomial and binomial expressions

Let us say we are given a polynomial p(x)=$\sum_k a_k x^k$. In order to find $\sum_k a_k$ we simply need to evaluate p(1), and similarly there are many other tricks. Is there any trick to evaluate ...
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2answers
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Find the coefficient of $x^4$ in the expansion of $(1 + 3x + 2x^3)^{12}$?

I have not learnt the multinomial theorem yet, and was trying to approach this using the binomial theorem. I divided the terms as $a$ being $(1+3x)$ and $b$ being $2x^3$. I then used $${12\choose ...
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When is $n\choose k$ a multiple of $n$

While working through a question, the solution states that in the finite field $\mathbb{F}_p$ for $p$ prime, we have $(u+v)^p=u^p + v^p$ and since $(u+v)^p={p\choose 0}u^pv^0+{p\choose ...
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Final step of a random walk proof

I am working through the last bit of a random walk proof to show that a 3-d random walk is transient. The result I am looking for states that: $\frac {1}{2}^{2s} {{2s}\choose{s}} \sum_{j+k\leq{n}} ...
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Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$

I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$ Thanks in advance for any help.
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How to sum $\sum_{k=1}^n (k+1)(k)(k-1)$

Is there an intelligent way to do this sum without using sums of cubes and sums of squares? $$\sum_{k=1}^n (k+1)(k)(k-1)$$
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Combinatorial Interpretation of these two identities

Currently, I am trying to prove the following two identities, which arose as a result of my other question in the Math StackExchange recently: \begin{equation} ...
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Interpretation of a combinatorial identity

I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math ...
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1answer
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Tree addtion has to do with Pascal's Triangle, why?

Let me define tree addition of a list of numbers as follows: 4 3 2 1 7 5 3 12 8 20 I conjecture that it is true that the tree addition of n numbers ...
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1answer
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Where is my formula false??

I wrote a formula that returned how many numbers in a given row of pascals triangle are divisible by a given prime. This formula was created to answer https://projecteuler.net/problem=148. I was ...
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1answer
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Relationship between Factorial and Binomial coefficients

Over at this link, there is a claim that $(2n)! = n!n! {{2n} \choose {n}}$ - see Tom Boardman's answer, the second one down. I'm wondering why this is the case and if anyone can provide a proof. Is ...
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1answer
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Why can I not include unused cards into a second binomial coefficient?

In trying to count the number of 13-card hands where there is at least one ace and no J, Q, K, we can see one way is $$ \sum_{k=1}^4 \binom{4}{k}\binom{36}{13-k} = 9722433280. $$ However, I cannot ...
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What will be the sum of the series of binomial co-efficients?

What will be the sum of the following binomial co-efficent series $$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$ Thank you
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Finding the formula for summation of the series

I was just solving a competitive programming question, wherein I found out that a formula can be used for solving it efficiently. Problem statement: http://www.spoj.com/problems/TOHU/ I tried a lot to ...
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1answer
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Formula for combinatorial series sum [duplicate]

As a part of one computer algorithm, I want to find sum for $$n+ \frac{n(n+1)}{2!} + \frac{n(n+1)(n+2)}{3!}+....+ \frac{n(n+1)(n+2)...(n+r-1)}{r!} $$. I looked at $$\frac1{(1-x)^n}$$. But it is ...
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How to calculate sum of combinations with different n and k

Input: $[X,Y]$ and $L$ Output : no of increasing sequence of length L and all elements should be $X\le i \le Y$ e.g: for $[6,7]$ and $2$ sequences are $6,66,67,7,77.$ For the above question my ...
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Solving inequality equation involving sum of binomial coefficients

I have a function $f(k,\,i)$ involving binomial coefficients: $$f(k,\,i)\,=\left(\begin{matrix}k+i \\ k\end{matrix}\right)=\frac{(k+i)!}{k!\,i!}$$ And the following sum over this function (expansion ...
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Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$

While doing a combinatorial problem, with $n$ being even, I came up with the expression $$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$ for which I used wolfram to get a closed form expression of ...
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A reference for a combinatorial identity

I have come across this identity from study of species. I am not posting my method but I am interested in knowing whether it arises in some other contexts as well. The identity is: $$\sum ...
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1answer
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Where does this equation come from: $ (1+mx)^n = 1 + \sum_{n=1}^{\infty} {\binom{2n}{n} \over 4^n } x^n $

I have found the following problem here: https://brilliant.org/problems/intriguing-sum/?group=Km7yEIDGtHDa&ref_id=709399 In the solution a solver directly started with the equation given in the ...
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Does this sum of products of binomial coefficients have a simple closed form?

Let $c,m,k$ be positive integers. Is there a simple closed form for the following sum? $$ \sum_{i=1}^{c-1} (-1)^i {c \choose i} {im \choose k} $$ Mathematica finds nothing, and Maxima's implementation ...
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Find the limit of $4^n\cdot\binom{2n}{n}/\binom{4n}{2n}$

I am trying to prove that $$f(n)=4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}$$ converges as $n\rightarrow\infty$. I have already tried to use the fact that, if $n, k \in\mathbb{N}, n\geq k\geq1,$ then ...
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1answer
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Growth of modified binomial recurrence

The binomial coefficients $\binom{n}{r}$ satisfies $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$. This means it is a solution of the equation $f(n,r)=f(n-1,r)+f(n-1,r-1)$, with initial conditions ...
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1answer
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Growth of binomial recurrence with different initial conditions

The binomial coefficients $\binom{n}{r}$ satisfies $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$. This means it is a solution of the equation $f(n,r)=f(n-1,r)+f(n-1,r-1)$, with initial conditions ...
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2answers
378 views

Show $\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$ [duplicate]

My question is: show $$\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$$ $$n\geq m\geq 1$$ I tried to do this via induction and failed. there has to be another way of doing this. We could either ...