Tag Info

Hot answers tagged

22

What's the number of ways to split $1000$ people into $10$ teams of $100$ ?


18

Since you tagged this abstract algebra, I will give a hint based on it: the group $$\underbrace{S_{100}\times\cdots\times S_{100}}_{10\text{ copies}}$$ has an obvious embedding as a subgroup of $S_{1000}$.


11

$$ \dfrac{1}{2\pi} \int_0^{2\pi} e^{e^{i\theta}} e^{-in\theta}\; d\theta $$


11

Note \begin{align}\frac{(n!)^{1/n}}{n} &= \left[\left(1 - \frac{0}{n}\right)\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)\right]^{1/n}\\ &= \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1} \log\left(1 - \frac{k}{n}\right)\right\} \end{align} and the last expression converges to $$\exp\left\{\int_0^1\log(1 - ...


11

Let $a_n = \frac{4^n}{n+1}$ and $b_n=\frac{(2n)!}{n!^2}=\binom{2n}{n}$. Then $a_1=b_1$ and: $$ \frac{a_{n+1}}{a_n} = 4 \frac{n+1}{n+2},\qquad \frac{b_{n+1}}{b_n} = 2\,\frac{2n+1}{n+1}\tag{1} $$ hence we just need to check that: $$ \forall n\geq 1,\qquad \frac{2n+2}{n+2}< \frac{2n+1}{n+1} \tag{2} $$ holds to prove our claim by induction. Also notice ...


8

Here’s the abstract of Florian Luca, On factorials which are products of factorials, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 143, Issue 03, November 2007, pp 533-542: In this paper we look at the Diophantine equation $$n!=\prod_{i=1}^ta_i!\qquad n>a_1\ge a_2\ge\cdots\ge a_1\ge 2\;.$$ Under the $ABC$ ...


8

Following @RobertIsrael, we have $$\frac{1}{2\pi}\int_0^{2\pi}e^{e^{i\phi}}e^{-in\phi}d\phi=\oint_{|z|=1}e^{z}z^{-n}\frac{dz}{iz}\tag 1$$ We note that the integrand on the right-hand side of $(1)$ has a pole of order $n+1$ at $z=0$. The residue is given by $$\text{Res}\left(-i\frac{e^z}{z^{n+1}},z=0\right)=\frac{1}{n!}\lim_{z\to ...


7

Use Stolz Cezaro: $$\ln \lim _{ n\to \infty } \frac { ({ n!) }^{ 1/n } }{ n } =\lim _{ n\to \infty } \ln \left( \frac { ({ n!) } }{ n^n } \right)^\frac{1}{n} =\lim _{ n\to \infty } \frac{\ln(n!)- n \ln(n)}{n} $$ Now by SC we get $$\ln \lim _{ n\to \infty } \frac { ({ n!) }^{ 1/n } }{ n } =\lim _{ n\to \infty } \ln((n+1)!)- (n+1) ...


6

Consider the binomial coefficents ${200 \choose 100},{300 \choose 100},{400 \choose 100},{500 \choose 100},{600 \choose 100},{700 \choose 100},{800 \choose 100},{900 \choose 100},{1000 \choose 100}$ For example, $${700 \choose 100} = \frac{700!}{600!100!} = \frac{700\cdot 699\ldots 602\cdot 601}{100!}$$ and this is an integer...


6

You don't even have to use l'Hopital's rule; you can just plug in Sterling's formula and divide by $n$, then take limits. Another way would be to use arithmetic-geometric means: $${1 \over (n!)^{1 \over n}} = (\prod_{k=1}^n {1 \over k})^{1 \over n} \leq {1 \over n}\sum_{k = 1}^n {1 \over k}$$ Since $\sum_{k = 1}^n {1 \over k}$ grows as $\ln n$ the limit is ...


5

Use Stirling $n!\sim n^ne^{-n}\sqrt{2\pi n}$ to see that $$ \lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{1/n} =\lim_{n \to \infty} \left(\frac{n^n e^{-n}\sqrt{2\pi n}}{n^n}\right)^{1/n} =\lim_{n \to \infty} \frac1e \left({\sqrt{2\pi n}}\right)^{1/n}=\frac1e $$


4

Using the definition you provided:$$ !1 = 1! \sum_{i=0}^1\frac{(-1)^i}{i!} = 1 \left(\frac{1}{1} + \frac{-1}{1}\right) = 0 $$Then putting this in the other recursive equation you gave:$$ !1 = !(1-1)(1) - 1 \implies 0 = !0 - 1 \implies !0 = 1 $$ Considering the definition provided on Wikipedia: In combinatorial mathematics, a derangement is a permutation ...


4

The basic result is that the product of $k$ consecutive numbers is divisible by $k!$. This is the fact that ${{n}\choose{k}}$ is an integer when $n \ge k$. Now, $1000!$ can be decomposed into $10$ products of $100$ consecutive numbers.


4

Here’s a non-combinatorial, more number-theoretic proof. For any prime $p$ the number of factors of $p$ in $n!$ is $$\sum_{k\ge 0}\left\lfloor\frac{n!}{p^k}\right\rfloor\;,$$ so it suffices to show for an arbitrary prime $p$ that $$10\sum_{k\ge 0}\left\lfloor\frac{100}{p^k}\right\rfloor\le\sum_{k\ge 0}\left\lfloor\frac{1000}{p^k}\right\rfloor\;.\tag{1}$$ ...


4

If $x\ge 4$, $$\begin{align}(x+1)(x+2)&=\frac{(x+1)(x+2)(2x)!}{(x+2)!x!}\\&={2x\choose x}\\&>{2x\choose3}\\&=\frac{2x(2x-1)(2x-2)}{6}\\&\ge\frac x3(x+3)(x+2)\end{align}$$ quickly leads to a contradiction. Manually check all cases with $x\le 3$.


4

If $1\le n\le 5$, then $(n,m)=(2,1),(3,1),(5,2)$. Assume $n\ge 6$. $$(n-1)!+1=n^m$$ $$\iff (n-2)!=\frac{n^m-1}{n-1}=1+n+n^2+\cdots+n^{m-2}+n^{m-1}$$ $$\iff (n-2)!-m=(1-1)+(n-1)+(n^2-1)+\cdots+(n^{m-1}-1)$$ $2,\frac{n-1}{2}$ are different and $\le n-2$, so $n-1\mid (n-2)!$. Then $n-1\mid m$, so $m\ge n-1$ and $(n-1)!=n^m-1\ge n^{n-1}-1$, impossible ...


4

ok the reasoning goes as follows, whatever $S^3$ is, we know that $$ S^3\ge n*n*n*(n-3)*(n-3)*(n-3)\dots $$ but we also know that $n>(n-1),n>(n-2)$ and $(n-3)>(n-4),(n-3)>(n-5)$ and therefore we just plug in and get the following inequality $$ S^3\ge n*n*n*(n-3)*(n-3)*(n-3)\dots \ge n*(n-1)(n-2)*(n-3)(n-4)(n-5)\dots=n! $$ and thats it. bests


4

A simple proof is based on the observation that $\dfrac{(2n)!}{(n!)^2}$ is the central binomial coefficient $\displaystyle{ {2n} \choose n}$. Look at row $2n$ in the Pascal triangle. The sum of all $n+1$ terms is $2^{2n}= 4^n$. Now, the central binomial coefficient is the largest number in that row and so $4^n \le (n+1){{2n} \choose n}$. [I've used $a_1 ...


4

Note $a_n = \frac{x^n}{2^n \cdot n!}$. So we have: $a_{n+1} = \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!}$ Thus we see: $\frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!} \cdot \frac{2^n \cdot n!}{x^n} $. Can you simplify further? EDIT: Note that $n! = 1 \cdot 2 \cdot 3 \: \cdot \: ... \: \cdot \: (n-1) \cdot n$. So that means that $(n+1)! = 1 \cdot ...


3

By rearranging terms, we can see that $$(n!)^2=[1\cdot n][2\cdot (n-1)][3\cdot (n-2)] \cdots [(n-1)\cdot 2][n\cdot 1].$$ Each of the $n$ products $k\cdot (n-k)$ is $\ge n$. Thus $$(n!)^2 \ge n^n \quad\text{and therefore}\quad (n!)^{1/n}\ge \sqrt{n}.$$


3

First note that $$ 0<\zeta(n)<\zeta(2) $$ for $n>2$. Thus $$ 0\leq\lim_{n\to\infty}\left|\frac{\zeta(n)}{n!}\right|<\lim_{n\to\infty}\frac{\zeta(2)}{n!}. $$ Can you finish from here?


3

It is trivially zero, because $\zeta$ is a positive decreasing function over the integers $n\geq 2$.


3

Take Cauchy's differentiation formula: $$ f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz $$ and plug a holomorphic $f$ such that $f^{(n)}(a)=1$. For example, $f(z)=\exp(z)$, $a=0$, and $\gamma$ the unit circle: $$ \frac{1}{n!} = \frac{1}{2\pi i} \oint_\gamma \frac{e^z}{z^{n+1}}\, dz $$ Does that count?


3

You can write the inverse Laplace transform of $1/s^{n+1}$, evaluated at $t=1$, as $1/n!$. The integral is $$ \int_{c-i\infty}^{c+i\infty}\frac{1}{2\pi is^{n+1}}e^s\,ds=\frac{1}{n!}, $$ for suitable real $c$.


3

This is not a complete answer, but hopefully it explains why your observation is unsurprising. Note that for $n \geq 4$, $c(n)$ is a multiple of every $m \leq n$. So, if $c(n) \pm k$ is prime, every prime factor of $k$ must be greater than $n$. This means that it is very likely that the smallest $k>1$ such that $c(n) \pm k$ is prime will itself be ...


2

I have a stronger statement: $$\binom{1000}{100}=\frac{1000!}{100!^{10}\binom{900}{100}\binom{800}{100}\binom{700}{100}\binom{600}{100}\binom{500}{100}\binom{400}{100}\binom{300}{100}\binom{200}{100}}$$ is an integer. More generally: $$\binom{ab}{b}=\frac{(ab)!}{b!^{a}\binom{(a-1)b}{b}\binom{(a-2)b}{b}\cdots\binom{2b}{b}}$$ is an integer ...


2

It is a plot of $\Gamma(x+1)$, known to equal $x!$ at integers.


2

Rewriting the factorial as the Gamma function and Stirling's approximation we get what I think is the closest possible approximation that you could do by hand: $$n! \approx \sqrt{2 \pi n} \cdot \left( \frac{n}{e} \right)^n$$ Where $e = 2.71828\dots$. Unfortunately, this might not be quicker than multiplying all the numbers together by hand, but it's ...


2

As long as you haven't made an algebra mistake, stirlings approximation should work. Separate $n!$ into two parts. Assign $1$ to everything below $n/2$. Assign $n/2$ to everything above $n/2$ So you get $(n/2)^{n/2}<n!$ Apply the root and you get: $(n/2)^{1/2}<n!^{1/n}$ $(n/2)^{1/2}$ clearly aproaches infinity so it will make the limit zero.


1

Since $\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=m$, we have: $$ n!=\prod_{m=2}^{n}m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}$$ so: $$ n!\geq \frac{n^n}{e^{n-1}} $$ and the claim easily follows.



Only top voted, non community-wiki answers of a minimum length are eligible