Tagged Questions
2
votes
0answers
33 views
Characterization of $\lambda,\mu\vdash n$ for which $\displaystyle{n\choose\mu}\mid{n\choose\lambda}$
In view of this question, I was wondering about general characterizations of $\mu,\lambda\vdash n$ for which
$${n\choose\mu}\,\left|\,{n\choose\lambda}\right.,$$
(see multinomial coefficient and ...
0
votes
2answers
84 views
How to show $\binom{2p}{p} \equiv 2\pmod p$?
how to prove $\forall p$ prime :
$\binom{2p}{p} \equiv 2 \pmod p$ we have:
$\binom{2p}{p} = \frac{2p (2p-1)(2p-3)...1}{p!p!}$ but how to continue?
6
votes
2answers
194 views
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 ...
0
votes
0answers
63 views
When inequality for binomial coefficients is true?
I've asked similar question here Inequality for binomial coefficients, but with slightly different assumptions. I am curious what happend if $m, k$ are fixed.
Let $m \leq n, n \leq N$ and $0\leq k ...
1
vote
1answer
115 views
Inequality for binomial coefficients
Let $m \leq n, n \leq N$ and $0\leq k \leq m$.
I am wondering what is the dependence of $n$ and $N$ that for all $m, k$
$$
\frac{{N-m \choose n-k}}{{N \choose n}}\leq 1.
$$
Thank you for your help.
3
votes
2answers
475 views
Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
So I'm trying to prove that for any natural number $1\leq k<p$, that $p$ prime divides:
$$\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$$
Writing these choice functions in ...
2
votes
1answer
56 views
Criterion for Wolstenholme Primes
Wolstenholme Theorem is a nice theorem that states that every prime $p >3$ satisfies:
$$\binom{2p}{p} \equiv 2 \pmod {p^3}$$
A Wolstenholme prime is a prime $p$ such that $\binom{2p}{p} \equiv 2 ...
5
votes
1answer
125 views
Primes Not Dividing $\binom{2n}{n}$
Let $n \geq 3$, show ${2n \choose n}$ is not divisible by $p$ for all primes $\frac{2n}{3} <p\leq n$
Note: This fact along with other facts about ${2n \choose n}$ are used in a proof of Bertrand's ...
3
votes
0answers
95 views
Prove: $\frac{(2px)!}{((px)!)^2}\equiv\frac{(2x)!}{((x)!)^2}\pmod{p^2}$
How can I prove the following, where $p$ is a prime and $x$ a positive integer?
$$\dfrac{(2px)!}{((px)!)^2}\equiv\dfrac{(2x)!}{((x)!)^2}\pmod{p^2}$$
I'm not sure if it is actually true, but I tested ...
6
votes
4answers
255 views
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 ...
1
vote
1answer
67 views
Does $k$-th power of $p$ divide ${}_n\!C_r$ if the previous divides $n$?
Does $p^k$ divide ${}_n\!C_r$ for all integer r if $p^k|n$ where $0\leq r \leq n$ and $p$ is prime?
3
votes
2answers
129 views
Proving ${p-1 \choose k}\equiv (-1)^{k}\pmod{p}: p \in \mathbb{P}$ [duplicate]
Possible Duplicate:
Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$
The question is as follows:
Let $p$ be prime. Show that ${p \choose k}\bmod{p}=0$, for $0 \lt k \lt p,\space ...
5
votes
2answers
235 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
0answers
157 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
vote
2answers
134 views
Hundreds’ place digit of $1993^3 – 913^3 – 1083^3$
How can I find the hundreds’ place digit of the following number:
$$1993^3 – 913^3 – 1083^3$$
I have not tried this one since I don't know how to begin.
I can tell the units digit of this but ...
-4
votes
2answers
236 views
Find the last digit of this series, for any value of $n$ and $m$,
Find the last digit of this number:
$$({}_{4n+1} C_0 )^{4m+1} + ({}_{4n+1} C_1 )^{4m+1} +({}_{4n+1} C_2 )^{4m+1} + \cdots + ({}_{4n+1} C_{4n+1} )^{4m+1}\;,$$
where $n$, $m$ belong to the holy set of ...
0
votes
0answers
41 views
Solving the equation $n(n-1)\cdot\cdot\cdot(n-k+2)(n-k+1) = a$ [duplicate]
Possible Duplicate:
How to reverse the $n$ choose $k$ formula?
I want to calculate reverse binomial coefficients. Given a number $m$, I want to compute all possibilites how $m$ could be ...
1
vote
2answers
221 views
Number of nonnegative integral solutions of $x_1 + x_2 + \cdots + x_k = n$
To find all solutions greater than or equal to $1$ of a linear equation in the form
$$x_1+x_2+x_3+\cdots+x_k=n ,$$
the number of them is $\binom{n-1}{k-1}$.
If I need all solutions to be greater or ...
15
votes
6answers
2k views
prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared.
My thought process: The numerator is the product of the first n even ...
0
votes
1answer
78 views
Divisibility question
Let $r$ be an integer greater than $2$.
Is there a simple way of showing that $2^r$ divides
$\left(\begin{array}{c}
{2}^{r-2} \\
k
\end{array}\right) 2^{2k}$
but it does not divide
...
1
vote
1answer
437 views
Proving that $n \choose k$ is an integer [duplicate]
Possible Duplicate:
Proof that a Combination is an integer
I can't think how to prove that ${n\choose k} \in\mathbb{Z}$.
I've played with it for a while, using the factorial definition for ...
9
votes
3answers
855 views
Elementary central binomial coefficient estimates
How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
Does anyone know any better elementary estimates?
6
votes
1answer
621 views
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 ...
13
votes
5answers
2k views
Prime dividing the binomial coefficients
It is quite easy to show that for every prime $p$ and $0<i<p$ we have that $p$ divides the binomial coefficient $\large p\choose i$; one simply notes that in $\large \frac{p!}{i!(p-i)!}$ the ...
