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2 votes
1 answer
34 views

Connection Between Derivations of Finite and Infinite Binomial Expansion

At first when learning the binomial expansion you learn it in the case of working as a shortcut to multiplying out brackets - anti-factorising if you will. In these cases what you are expanding takes ...
Ardavan Hamisi's user avatar
0 votes
1 answer
35 views

Binomial identity involving square binomial coefficient [closed]

I want to prove this identity, but I have no idea... Could someone please post a solution? Thank you. $$\sum_{k=0}^{n} \binom{-1/2}{n+k}\binom{n+k}{k}\binom{n}{k}= \binom{-1/2}{n}^2$$ (Maybe -1/2 can ...
anonymoususer's user avatar
3 votes
1 answer
45 views

Steps on solving part b of the bit-string exercise?

This is the exercise: How many bit strings of length $77$ are there such that a.) the bit string has at least forty-six $0$s and at least twenty-nine $1$s, and also the bit string corresponding to ...
zaxunobi's user avatar
  • 131
2 votes
2 answers
91 views

Compute the value of a double sum

I need some help computing a(n apparently nasty) double sum: $$f(l):=\sum_{j = \frac{l}{2}+1}^{l+1}\sum_{i = \frac{l}{2}+1}^{l+1} \binom{l+1}{j}\binom{l+1}{i} (j-i)^2$$ where $l$ is even. I'm not ...
Matt M's user avatar
  • 39
3 votes
3 answers
65 views

Prove that $\frac{(n + 1)!}{((n + 1) - r)!} = r \sum_{i=r - 1}^{n} \frac{i!}{(i - (r - 1))!}$

I Need Help proving That $$\frac{(n + 1)!}{(n - r + 1)!} = r \cdot \sum_{i=r - 1}^{n} \frac{i!}{(i - r + 1)!}$$ Or in terms of Combinatorics functions: $P_{r}^{n+1} = r \cdot \sum_{i = r-1}^{n} {P_{r-...
BGOPC's user avatar
  • 179
0 votes
1 answer
42 views

Binomial coefficient inequality ${n+x \choose x} > {{m+q-x} \choose q-x}$

Let $m,n,q$ be positive integers and $0\leq x\leq q$ where $x$ is an integer. When does the inequality $$ {n +x \choose x} > {{m+q-x} \choose q-x} $$ hold? Using the Hockey-Stick identity, $$ {n+x \...
Steve Cooper's user avatar
3 votes
0 answers
162 views

Closed form for $\sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$?

I've found this sum: $$S(n) = \sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$$ The inner sum is elliptical iirc, but perhaps the double sum has a nice expression. We can ...
hellofriends's user avatar
  • 1,940
0 votes
1 answer
33 views

How to upper bound $\sum_{m=2}^{d-1}\binom{n}{d+1+m} n^{-\alpha 2m} $ [closed]

As I'm saying in the title, I am looking for an upper bound (or an identity) for this: $$ \sum_{m=2}^{d-1}\binom{n}{d+1+m} n^{-\alpha 2m} $$ where $\alpha \in [0,\infty)$. Any ideas/suggestions? ...
mat95's user avatar
  • 341
2 votes
1 answer
72 views

Express Lucas numbers as a sum of binomial coefficients

So here's the question, prove that $$L_n = \sum_{k=0}^{n} \frac{n}{n-k} \binom{n-k}{k}$$ where $L_n$ is the $n$-th Lucas number. It really resembles the Fibonacci identity: $$F_n = \sum_{k=0}^{n} \...
Haimu Wang's user avatar
1 vote
0 answers
47 views

Find the number of lattice paths weakly under a slope $y = \mu x$

How many lattice paths are there from an arbitrary point $(a,b)$ to another point $(c,d)$ that stay weakly (i.e. it can touch the line) under a slope of the form $y = \mu x$, with $\mu \in \mathbb{N}$?...
alteredpulse's user avatar
0 votes
0 answers
34 views

Probability involving identical objects. I am not able to understand how the Ncr formula is being applied below for counting identical objects.

Question: A bag contains 5 identical red coins, 6 identical yellow coins and 8 identical blue coins. If 3 coins are picked up randomly from the bag, what is the probability that there is at least one ...
Vasu Gupta's user avatar
  • 1,050
2 votes
2 answers
64 views

binomial distribution but sometimes the last outcome doesn't matter

Here's the motivation for my question: I'm designing an RPG. To simplify as much as possible, lets say my enemy has $h = 4$ HP and I deal $a = 1$ damage with every attack. However, there's also a $p$ ...
waf9000's user avatar
  • 23
3 votes
3 answers
158 views

Iterated rascal triangle row sums

In this manuscript the authors propose the following conjecture (1) \begin{align*} \sum_{k=0}^{4i+3} \binom{4i+3}{k}_i &= 2^{4i+2} \end{align*} where $\binom{4i+3}{k}_i$ is iterated rascal ...
Petro Kolosov's user avatar
2 votes
2 answers
47 views

Given a set of integers, and the number of summations find the resulting frequencies

Given a set $X = \{x_1,x_2,...x_m\}\subset \mathbb{Z}$ and the number of allowed addends $N$. How can I find the frequency of every possible sum? Example: $X = \{-1, 2\}$ and $N=3$ then every ...
haifisch123's user avatar
1 vote
1 answer
38 views

Summation of n-simplex numbers

Gauss proved that every positive integer is a sum of at most three triangular(2-simplex) numbers. I was thinking of an extension related to n-simplex. Refer: https://upload.wikimedia.org/wikipedia/...
Shivang Gupta's user avatar
0 votes
0 answers
45 views

Closed form for nested sum involving ratios of binomial coefficients

I ran into the following nested sum of binomial coefficients in my research, but I couldn't find the closed form expression for it. I looked at various sources and still couldn't find the answer. So I ...
Weishun Zhong's user avatar
3 votes
1 answer
84 views

Combinatorial Proof of the $\sum_{k=1}^{n} k^2 \binom{n}{k} = n(n + 1) 2^{n - 2}$ [duplicate]

How to prove that $$\sum_{k = 1}^{n}k^2\binom{n}{k} = n(n + 1)2^{n -2}$$ in an combinatorial way? I have a algebraic proof method, but I don't know how to use a combinatorial proof method to do it.
crimsnow's user avatar
0 votes
0 answers
43 views

Evaluating Difference of Product of Binomial Coefficients

As part of my project I'm asked to evaluate the positivity of the following difference: $$\binom{l_1+t-1}{l_1}\binom{l_2+m+t-1}{l_2+m}\sum_{j=0}^{t}\binom{t-j+m}{m}\binom{j+l_1}{j}\binom{j+l_2}{j}-\...
Haimu Wang's user avatar
0 votes
1 answer
128 views

Spivak Exercise, Prove Vandermonde's Identity $\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$ [duplicate]

Prove that $$\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$$ Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$ Proof: Following Spivak's advice, we have $\sum_{k=0}^{n}\binom{n}{k}x^k=(...
Edward Falls's user avatar
0 votes
0 answers
44 views

Choosing an ordered triplet of non-negative integers $(m_1, m_2, m_3)$ such that $m_1 + m_2 + m_3 = n$.

Define $$A = \{ (m_1, m_2, m_3) : m_1 \geq 0, m_2 \geq 0, m_3 \geq 0, m_1 + m_2 + m_3 = n\}$$ Given that $n \geq 0$ and $n, m_1, m_2, m_3 \in \mathbb{Z}^+$, then why it is the case that $$\vert A \...
microhaus's user avatar
  • 934
0 votes
2 answers
45 views

Trying to prove equivalence of combinatorial formula and nested summations

I’m sorry if this is a dumb problem, but I’m trying to get into mathematics and prove this but I’m only in 9th grade and haven’t found any sources on this: Prove that $$ {n \choose r} = \overbrace{ \...
Joproblox Bardouha's user avatar
4 votes
3 answers
70 views

$(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$

Let $n$ and $a$ be natural numbers. How to prove the following for $x \in [0, 1)$? $$ (1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j $$...
ploosu2's user avatar
  • 9,748
4 votes
1 answer
141 views

What is the name of this combinatorial identity?

In the course of my physics research, I appear to have stumbled onto the following combinatorial identity: $${dn\choose m}=\sum_{\vec k} {n \choose \vec k}\,\prod_{j=0}^d {d \choose j}^{k_j},$$ where $...
David Raveh's user avatar
  • 1,835
0 votes
2 answers
73 views

Lower Bound on the ratio of binomial coefficients

Let $k,n,m$ be integers such that $k>n>m$. I am interested in providing a tight lowerbound on $$ A(k,n,m)=\frac{\binom{k-m}{n-m}}{\binom{k}{n}} $$ This term arises in a probability problem that ...
MMH's user avatar
  • 714
5 votes
2 answers
164 views

Estimate a sum of binomial coefficients

I should know this by the time, but: can someone tell me how to rigorously compute the leading order (including the constant) of the following sum: $$\sum_{ 1\leq k \leq n/3 } {2 k \choose k} {n-2k-1 ...
Olivier's user avatar
  • 1,363
4 votes
1 answer
78 views

Identity regarding the sum of products of binomial coefficients.

Consider the following toy problem Person A and Person B have $n$ and $n+1$ fair coins respectively. If they both flip all their coins at the same time, what is the probability person B has more ...
Demetri Pananos's user avatar
2 votes
1 answer
69 views

Closed form for a sum of binomial coefficients

Let $m,n,r\in\mathbb{N}\cup\{0\}.$ I am interested in finding a closed form for the sum $$\sum_{i=0}^m{{n+i}\choose{r+i}}.$$ Let $f(m,n,r)$ denote the above sum. We may make a few trivial observations....
aqualubix's user avatar
  • 2,145
2 votes
3 answers
84 views

What is the number of lattice paths of length 16 from the point (0,0) to (8,8) that go through (4,4) but don't go through (1,1), (2,2), (3,3)

what is the number of lattice paths of length 16 from $(0,0)$ to $(8,8)$ that go through $(4,4)$, don't go through $(1,1), (2,2), (3,3)$, and don't go over $y=x$? Here's what I tried: since we can't ...
user avatar
1 vote
2 answers
92 views

Alternating sum involving binomial coefficients

I want to prove that $$ \sum_{i=0}^{n}{n\choose i} \frac{\left(1 + \alpha i\right)^{n} \left(-1\right)^{n - i}}{n!} = \alpha^{n}. $$ This is a guess based on the computations for $n = 0,1,2,3$. Do you ...
stackQandA's user avatar
0 votes
1 answer
34 views

How to Derive the Binomial Coefficient Upper Bound and Final Inequality in "Scheduling Multithreaded Computations by Work Stealing"?

In the paper Scheduling Multithreaded Computations by Work Stealing under the section "Atomic accesses and the recycling game", it mentions the binomial coefficient approximation: $$ \binom{...
grzhan's user avatar
  • 3
1 vote
0 answers
61 views

Closed expression for $\sum _{m=1}^n{{n+m}\choose n} $ [duplicate]

I was given a problem in University: Let $n\in \mathbb N $. Find a closed expression for the following sum $\sum _{m=1}^n{n+m\choose n}$. I expressed binomials ${n+m}\choose {n}$$=\frac{(n+1)(n+2)......
Hrackadont's user avatar
1 vote
3 answers
146 views

Evaluate: $\sum_{i=1}^{\lfloor (n+1)/2\rfloor}i\binom{n-i+1}{i}$

Is there a closed form of the expression $$ \sum_{i = 1}^{\left\lfloor\left(n + 1\right)/2\right\rfloor} i\binom{n - i + 1}{i} $$ My Attempt: From what I observe it is a situation where there are $n/...
Maverick's user avatar
  • 9,599
4 votes
0 answers
128 views

Calculating $\sum\limits_{k=0}^n\binom{n}{k}/\left(2^k+2^{n-k}\right)$

I am trying to find a closed form for $\sum\limits_{k=0}^n\frac{\binom{n}{k}}{2^k+2^{n-k}}$. I saw on quora that integration can be used to rewrite portions of such equations, and so I attempted this. ...
plywood98's user avatar
2 votes
1 answer
62 views

How to apply Vandermonde's Identity regarding summation bounds ? $\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}=\binom{-k-j-2}{n-j-k}$

from the answer: Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ this happens from step 4 to 5: $$ \sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}=\...
Mr. Doge's user avatar
  • 123
2 votes
2 answers
85 views

Limit of $\sum_{j=0}^n (1-j){n\choose j} \frac{1}{(n-1)^j}\left( \frac{n-j}{n} \right)^k$ as $n\to \infty$ (and $k=n$)

Question Find $\lim_{n\to \infty} P(n,n)$ where $$ P(n,k) = \sum_{j=0}^n (1-j){n\choose j} \frac{1}{(n-1)^j}\left( \frac{n-j}{n} \right)^k. $$ The origin of this sum is from this question. The ...
ploosu2's user avatar
  • 9,748
2 votes
1 answer
75 views

Prove the following combinatorics equality [duplicate]

$$\mbox{Prove the equality:}\quad \binom{n}{k + m + 1} = \sum_{j = k + 1}^{n - m} \binom{j - 1}{k}\binom{n - j}{m} $$ for $m + k < n$. I tried to prove this using combinatorics and doing some ...
perenqi's user avatar
  • 169
4 votes
1 answer
47 views

Combinatorial proof that $\sum_{k=1}^{n} k {2n \choose n+k}=\frac{1}{2}n{2n \choose n}$ [duplicate]

I'd like to find a combinatorial/algebraic proof of the identity: $$\sum_{k=1}^{n}k{2n \choose n+k}=\frac{1}{2}n{2n \choose n}$$ The only proof of this that I've been able to find on the Internet, ...
N. S.'s user avatar
  • 81
2 votes
4 answers
155 views

A Combinatoric Identity Involving Binomial Coefficients: $\sum^{n}_{r=1} {n \choose r} (-1)^r \frac{2^r-1}{2r}= \sum^{n}_{r=1}\frac{(-1)^r}{2r}$

Recently I came across this combinatoric identity: $$ \begin{equation} \sum^{n}_{r=1} {n \choose r} (-1)^r \frac{2^r-1}{2r} \end{equation} = \sum^{n}_{r=1}\frac{(-1)^r}{2r} $$ I have verified that ...
Slice's user avatar
  • 23
1 vote
1 answer
54 views

Hint for a combinatorial proof of $n! \binom{n-1}{k-1} = \sum_{j=0}^n \overline{s}_{n,j} \tilde{S}_{j,k}$

To clarify notation, we have $\overline{s}_{n,j}$ the unsigned Stirling numbers of the first kind, meaning the number of permutations on an set with $n$ elements that has exactly $j$ cycles. $\tilde{S}...
Dominic Michaelis's user avatar
1 vote
2 answers
204 views

How to prove that for $a,b \in \mathbb{C}$, $\sum\limits_{k=0}^n \binom{a}{k}\binom {b}{n-k}= \binom{a+b}{n}$?

This exercise came from Complex Analysis by Freitag and Busam 1.2.11. $$\binom{a}{n}:= \prod_{j=1}^n\frac{a-j+1}{j} $$ Show: $\sum\limits_{\nu=0}^{\infty}\dbinom{\alpha}{\nu} z^\nu$ is absolutely ...
pie's user avatar
  • 6,620
0 votes
1 answer
78 views

Expressing a combination as sums of three terms or more, analogous to Pascal's identity

Are there any ways to express binomial coefficients as the sum of three or more terms, similar to how Pascal's Identity breaks a combination into the sum of two terms by relating two adjacent binomial ...
someone's user avatar
  • 11
6 votes
2 answers
267 views

Counting bit strings with given numbers of higher-order bit flips

Background information Bit flips Given a bit string, we say that bit flip happens when $0$ changes to $1$ or $1$ changes to $0$. To find bit flips, we can shift the string by $1$ and xor that new ...
Valeriy Savchenko's user avatar
1 vote
3 answers
79 views

Combinatorical identity $\sum_{k = d-i}^d (-1)^{k-d+i} \binom{k}{d-i} \cdot \binom{n}{d-k} = \binom{n-d+i-1}{i}$

How do we proove the following ugly identity of binomial coefficients? $$\sum_{k = d-i}^d (-1)^{k-d+i} \binom{k}{d-i} \cdot \binom{n}{d-k} = \binom{n-d+i-1}{i}$$ First I thought we could use ...
Lukas's user avatar
  • 141
0 votes
0 answers
38 views

Understanding the Derivation of a Formula Involving Binomial Coefficients and Factorials

I'm studying a formula that involves binomial coefficients and factorials, and I'm struggling to understand how it was derived. The image below is a screenshot from the paper. They are taking the ...
Dotman's user avatar
  • 326
3 votes
2 answers
113 views

An unusual binomial coefficient identity

I derived the following family of binomial coefficient identities rather indirectly for natural $t \ge 2$ (though it seems to hold more generally). I was hoping someone out there might know if this ...
Dean Rubine's user avatar
9 votes
3 answers
339 views

How can I simplify this combinatorics expression?

I got a expression that is related to the combinatorics, and it looks like this: $$\sum_{k=1}^n \frac{2^{2k-1}}{k}\binom{2n-2k}{n-k}\cdot \frac{1}{\binom{2n}{n}}$$ it is from a question I've been ...
madala's user avatar
  • 93
6 votes
2 answers
221 views

Is this identity I found playing around with generating function with coefficients $I(n) = \int_{0}^\pi \sin^n(x) dx$ useful and or reducible?

Let $I(n) = \int_{0}^\pi \sin^n(x) dx$ , using $\sin^2(x) = 1-\cos^2(x)$ and integrating by parts we get. $$ \begin{align} I(n) = \dfrac{n-1}{n} I(n-2) \end{align} $$ With $I(0) = \pi$ and $I(1) = 2$ ...
Sam's user avatar
  • 61
0 votes
0 answers
52 views

Sum of every $k$th binomial coefficients over upperindex

From the Hockey-Stick Identity, we have for $n\geq r \geq 0$: $$\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}.$$ Now, I am wondering whether an identity exists for any fixed ( $k > 0$ ): $$\sum_{i=...
Nebiyou Yismaw's user avatar
3 votes
5 answers
218 views

Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$

Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$ I tried splitting the terms inside the bracket into two parts $1+x+\dots+x^9$ and $x^{10}$, and then tried binomial theorem, but that ...
math_learner's user avatar
2 votes
2 answers
74 views

Correctness of Solution for Forming a Committee with More Democrats than Republicans

I recently encountered a problem and derived a solution, but I am uncertain about its correctness. Here's the problem: At a congressional hearing, there are 2n members present. Exactly n of them are ...
AzharKhan's user avatar

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