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

learn more… | top users | synonyms (1)

3
votes
2answers
148 views

Combinations and Gaussian function

I notice that the function $\binom{C}{x}$, where $C$ is some constant, resembles a Gaussian function; for example, here is the plot for $\binom{20}{x}$: This corresponds to the Gaussian function $a ...
0
votes
3answers
161 views

Binomial expansion and factorials

I have come across this question, the answer is simply stated as 36, and while I can see how 36 is gotten, I don't understand why? How is it meant to be known that b = 36, just looking at the ...
18
votes
3answers
13k views

Combinatorial proof of summation of $\sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$

Can you guys help me prove this? There is a way of proving this logically but I was hoping to find a more "mathematical" proof, if possible. $$\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n ...
5
votes
2answers
731 views

Odd Binomial Coefficients?

By Newton's Formula: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k $$ Proof that every $\dbinom{n}{k}$ is odd if and only if $n=2^r-1$. I have already shown that if $n$ is of the form $2^r-1$, ...
3
votes
2answers
369 views

Lower Bound of Central Binomial Coefficients

I would like to prove by induction the following inequality: $\frac{4^n}{n+1} < \binom{2n}{n}$, for all natural numbers n > 1. Any hints?
3
votes
1answer
497 views

A sum involving permutation

Does there exist a nice closed form formula for the sum $$\sum_{k=0}^m P(m,k)x^k$$ where $P(m,k)=C(m,k)*k!$, $C(m,k)$ being the "m choose k" number. Formula given by Maple 11 is complicated. I ...
0
votes
1answer
69 views

Get p from cumulative binomial function

I would like to isolate p in the following. I am not sure if it is even possible. a = B(1; 10, p) B(x;n,p) is the cumulative binomial function.
3
votes
1answer
403 views

Surprising approximation of weighted sum of binomial coefficients

The following sum appeared in connection to the problem addition of angular momentum in physics: $$ \frac{1}{2^{n+3}}\sum_{k=0}^n \left(\frac{n-2k-1}{\sqrt{k+1}}+\frac{n-2k+1}{\sqrt{n-k+1}}\right)^2 ...
4
votes
2answers
200 views

Why does $\binom{10}{7} = \frac{10!}{(10-7)!7!}$

We just learned that: $\dbinom{10}{7}= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$, so that: If you throw a dice 10 ...
16
votes
1answer
408 views

What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations: $$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = ...
2
votes
5answers
5k views

Sum from 0 to n of $ n \choose i $? [duplicate]

Possible Duplicate: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$ Evaluation $\sum\limits_{k=0}^n \binom{n}{k}$ Is there a simple proof for this equality: $$\sum_0^n {n ...
10
votes
3answers
531 views

proof that $1 = \sum\limits_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}$

I'm looking for a proof of this identity: $$ 1 = \sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k} $$ I'll take anything, but a combinatorial proof would be nice - all of the terms in the sum ...
3
votes
3answers
395 views

Proof of a Binomial Identity using a combinatorial argument

Question Prove that if $k$ and $l$ are two positive integer with $k ≥ l$, then $\binom{2k}{2} =\binom{k−l}{2}+ \binom{k+l}{2}+ k^2 − l^2$ using a combinatorial argument. I tried using Vandermonde's ...
3
votes
1answer
442 views

Sum of every $k$th binomial coefficient.

It is widely known that $$\sum_{m=0}^{n} {n\choose m} = 2^n$$ and that $$\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose 2m} = 2^{n-1}$$ Both results can be proven by exploting the nature of the roots ...
1
vote
1answer
147 views

Binomial probability

Suppose that the probability of a company supplying a defective product is $a$ and the probability that the supplied product is not defective is $b$. Before each product supplied is released for ...
6
votes
3answers
1k views

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 ...
11
votes
1answer
708 views

Factorial canceling on expansion of binomial coefficients on Concrete Mathematics

On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as: \[ \frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z} \] where \[ ...
2
votes
1answer
328 views

Evaluating 'combinatorial' sum

Help me please to calculate the following sum. I have seen such kind of formulas in the papers related to combinatorics, specifically 'trees'. I am curious how to calculate or approximate this sum: ...
4
votes
1answer
156 views

Calculate $\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n\choose i}$

It is known that $\sum_{i=1}^n {n \choose i}=2^n$. I am wondering what would be the sum if we change the upper limit to $\sqrt n/2$, i. e. How to calculate$$\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n \choose ...
5
votes
2answers
365 views

Combinatorial argument for $\sum_{k=0}^n k^2 \binom{n}{k} = n(n+1)2^{n-2}$

Can you please give a combinatorial argument for the argument below? $$\sum_{k=0}^n k^2 \binom{n}{k} = n(n+1)2^{n-2}$$ From RHS, I drew the following argument: There are $n+1$ people. In how ...
2
votes
1answer
397 views

Evaluation $\sum\limits_{k=0}^n \binom{n}{k}$

Please help me to evaluate combinatorially the following sum: $$\sum_{k=0}^n \binom{n}{k}$$ Thank you.
0
votes
1answer
109 views

Simple question about an asymptotic equality

Could someone please explain the second equality in Conjecture 1.1: http://arxiv.org/pdf/math/0501313v2.pdf ? (reproduced below) $(1+o(1))n^22^{1-n}=\left(\frac{1}{2}+o(1)\right)^n$ Initially, I ...
7
votes
1answer
2k views

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 ...
6
votes
2answers
438 views

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 ...
3
votes
2answers
162 views

Why is the binomial coefficient related to the binomial theorem?

The binomial coefficient basically provides the number of ways to choose a set of $k$ from $n$ sets. To me, it can be considered the number of unique ways to pick $k$ amount of "cards" from a deck of ...
3
votes
2answers
544 views

Sum of cubes of binomial coefficients

I reduced a homework problem in combinatorics to giving an asymptotic estimate for $\sum_{k=0}^n{n \choose k}^3$. I assume Stirling's approximation can help, but I'm not experienced with making ...
1
vote
3answers
164 views

Combinatorial inequality $\binom{n}{j}\leqslant 2^n$

I was trying to prove (or to find a counterexample) of the following inequality: $$\binom{n}{j}\leqslant 2^n$$ As I coudn't find a proof/counterexample, I tested some numbers and could see it ...
0
votes
1answer
136 views

Identity of binomial coefficients with a series

I never really used any series/infinite sums and now I should proove the following identity: $$\sum\limits_{k=0}^{\infty}\binom{m}{k}\binom{n}{l-k}=\binom{m+n}{l}$$ Can you please explain me, how to ...
2
votes
1answer
549 views

Evaluate the Binomial Coefficient

Please help me evaluate the following Binomial coefficient using any known properties of Pascal triangle: $$ \binom{52}{4} - \binom{47}{4} + \binom{47}{5} $$
2
votes
1answer
134 views

Comparing two binomial coefficient sums

Let $j \in N, n\in N, n>1, q\geq 2$. I would like to show that $$ \sum_{j=\frac{n}{\ln n}}^{\sqrt n/2}(2j-n)^q{n \choose j}<\sum_{j=\sqrt n/2}^{{\frac n2}}(2j-n)^q{n \choose j} $$ Any help would ...
2
votes
2answers
572 views

A sum involving powers of binomial coefficients.

Find the formula for the following sum of binomial coefficients: $$ \sum_{m\ge 0} (-1)^m {\binom{n}{m} }^3 .$$ Could you find the formula for $\sum\limits_{m\ge 0} (-1)^m{\binom{n}{m}}^4$?
2
votes
1answer
384 views

A Curious Binomial Coefficient Sum

Let $k, l \leq n$ be non-negative integers. Does the following identity simplify? \begin{align} \sum_{j = 0}^{k} \binom{k}{j} \binom{j + n -l + 1}{n} = \binom{n - l + 1}{n} ...
1
vote
0answers
293 views

Calculation of a 'double' sum

Let $n \in N$ and $q\geq 2$. I am trying to calculate the following sum: $$ \sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!} $$ Any help will be ...
4
votes
1answer
214 views

Evaluating a limit involving binomial coefficients.

If $N_c=\lfloor \frac{1}{2}n\log n+cn\rfloor$ for some integer $n$ and real constant $c$, then how would one go about showing the following identity where $k$ is a fixed integer: $$\lim_{n\rightarrow ...
5
votes
3answers
330 views

proof of formula and calculation sum

Show that following formula is true: $$ \sum_{i=0}^{[n/2]}(-1)^i (n-2i)^n{n \choose i}=2^{n-1}n! $$ Using formula calculate $$ \sum_{i=0}^n(2i-n)^p{p \choose i} $$
-1
votes
2answers
361 views

Prove a binomial inequality

I would like to prove the following inequality: $$ {m+n \choose m} \ge \frac{(n+1)^m}{m!} $$ Any hints?
3
votes
0answers
95 views

max number of times integer $>1$ appears in pascals triangle

$120, 210 ,3003$ appear $6$ times in Pascal's triangle. $120={10\choose3}={16\choose2}={120\choose1}\\$ $210={10\choose4}={21\choose2}={210\choose1}\\$ ...
1
vote
0answers
52 views

Inexpressibility of certain coefficients and discrete versions of Hölder's theorem

In an answer to a recent question, I noted that there were probably no explicit formulas for Stirling numbers (of the first kind, specifically) and speculated that this might be coupled to a sort of ...
3
votes
1answer
249 views

Summation: $ \sum \limits_{r=0}^n \frac{ \binom n r}{x+r} $

How to evaluate $$ \sum \limits_{r=0}^n \large \frac{\binom n r} {x+r} $$ I got this problem from a friend according to him, $ \binom n r$ is the coefficient of $(1+x)^n$. I am not sure how to ...
4
votes
2answers
382 views

computation of the sum

I am having trouble to compute the following sum: $$ \sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}} $$ Here $p\geq 2$. To simplify the question, we can even assume that ...
12
votes
2answers
266 views

Proving a certain binomial identity with three parameters

I would like to prove the following identity: $$\sum_{m\geq 0} (-1)^{i-m}{m+k \choose m} {i-1 \choose m-1}{m+k+1 \choose j} = \sum_{m\geq 0} {m+k \choose k}{k+1 \choose i-m}{k+1 \choose j-m}$$ for ...
0
votes
2answers
299 views

Prove that $n! = \sum_{i=0}^{n-1}{n \choose i}(n-i)^n (-1)^i$ [duplicate]

Possible Duplicate: Proof of a combination identity:$\sum \limits_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$ Prove that product of $n(n-1)(n-2)\dots(2)(1)$ ...
6
votes
1answer
197 views

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 $d \geq 1$, ...
3
votes
2answers
376 views

lacunary sum of binomial coefficients

I am sure there must be a known estimate for sums of the form: $$ S_d(n)=\sum_{j=0}^n \binom{dn}{dj} $$ where $d\ge 1$. For $d=1$, the answer is obvious. For $d=2$, the answer is here: Sum with ...
1
vote
1answer
87 views

Probability question involving binomial theorom

I'm having difficulty solving this question : In a multiple choice test there are 5 questions each with three possible answers. For each question a student chooses an answer at random. Find the ...
4
votes
3answers
637 views

Binomial Coefficients in the Binomial Theorem - Why Does It Work Question

to keep it simple: Given $(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$ Could you please give me an intuitive combinatoric reason to why the binomial coefficients are ...
7
votes
4answers
366 views

Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$.

I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble ...
2
votes
2answers
263 views

Combinatorial Proof for a $ p\mid\binom{p}{k} \ \ \ \ \ 0<k<p$ .

I'm looking for a combinatorial proof to the following statement: $$ p\mid\binom{p}{k} \ \ \ , \ \ 0<k<p \ \ \ \ \ \ \text{and} \ \ p \ \text{is prime}.$$ Thank you.
0
votes
2answers
432 views

Simple Question About Binomial Theorem

On one of my calculus lectures I've seen the lecturer write: $$(1+p)^n=1+np+\frac{n(n-1)}{2}p^2+\cdots+p^n$$ Could you please explain to me how did he get this equation? Thank you very much.
2
votes
3answers
1k views

Sum with binomial coefficients: $\sum_{k=0}^{n}{2n\choose 2k}$

I'm repeating material for test and I came across the example that I can not do. How to calculate this sum: $\displaystyle\sum_{k=0}^{n}{2n\choose 2k}$?