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2

You can show this directly by noting that $n!$ always divides $n!\binom{(a+1)n}{n} = (an+1)(an+2) \dots (an+n) = \frac{((a+1)n)!}{an!}$ for $a = 0, 1, 2, \ldots, (n-1)!$. This step relies upon knowing that for any $i,j\ge 0$ that $\binom{i+j}{i} = \frac{(i+j)!}{i!j!}$ is an integer. This is a standard result, and can be proven in several ways, ...


2

Hint: from $p$-adic analysis we know, that if $v$ is the largest number, such that $p^v$ divides $n!$ (where $p$ is prime), it is equal to $$v_p(n!) = \sum_{i=1}^\infty \left \lfloor \frac n{p^i}\right\rfloor.$$ Apply twice.


2

My favorite use of the factorial function for non-integer arguments is the formula for the volume of an $n$-dimensional ball of radius $r$: $$ \frac{\pi^{n/2}r^n}{(n/2)!}. $$


7

The factorial function is extended by the $\Gamma$ function. The relation is $$(n-1)! = \Gamma(n) = \int_0^\infty t^{n-1} e^{-t}\, dt$$ This can be analytically continued as a meromorphic function in the complex plane. Ref: John Conway's book on Complex Analysis.


1

One way is to compute it! In[2]:= Count[IntegerDigits[100!], 0] Out[2]= 30 This computation took 0.000033 seconds in Mathematica.


2

Consider the matrix $${\bf A_n} = \left(\begin{array}{cccc}1&0&\cdots&0\\0&2&\cdots&0\\\vdots&\vdots&\ddots&0\\0&0&0&n\end{array}\right)$$ It's determinant is $n!$. Furthermore $$\det({\bf A_m}+k{\bf I_m}) = \frac{(m+k)!}{k!}$$ Which illuminates some relations to combinatorics (permutations).


6

$2010!$ ends with $501$ zeroes since: $$\sum_{n=1}^{5}\left\lfloor\frac{2010}{5^n}\right\rfloor=501.\tag{1}$$ For the same reason, $$ \nu_2(2010!)=\sum_{n=1}^{10}\left\lfloor\frac{2010}{2^n}\right\rfloor=2002\tag{2}$$ so $\frac{2010!}{10^{501}}$ is an even number, and since: $$ 1\cdot 2\cdot 3\cdot 4\equiv -1\pmod{5} \tag{3}$$ it follows that: ...


6

Although not 100% about what you are asking (perhaps 87%), consider an $n$-dimensional cube with side $x$. The $n$-th derivative of the box's volume is $n!$. That is $$\frac{d^n}{dx^n} x^n=n!$$ Curiously enough, this is saying that the rate of the rate of the...of the rate of increase of the volume of an $n$-dimensional cube is precisely the volume of a ...


1

Finally, I solved it and wanted to share with you! $$LHS=\sum _{i=0}^m\binom{m}{i}p^i(1-p)^{m-i}\sum _{j=0}^n \binom{n}{j}\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}(j)$$ and let $$B=\sum _{j=0}^n \binom{n}{j}\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}(j)=\sum _{j=0}^n\frac{n(n-1)!}{(n-j)!(j-1)!}\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}$$ Now we can simplify B by using the ...


0

More generally, since $n! \sim \sqrt{2\pi n}(n/e)^n$, $\begin{array}\\ \binom{an}{bn} &=\frac{(an)!}{(bn)!((a-b)n)!}\\ &\sim\frac{\sqrt{2\pi an}(an/e)^{an}}{\sqrt{2\pi bn}(bn/e)^{bn}\sqrt{2\pi (a-b)n}((a-b)n/e)^{(a-b)n}}\\ &=\sqrt{\frac{a}{2\pi b(a-b)n}}\frac{(an)^{an}}{(bn)^{bn}((a-b)n)^{(a-b)n}}\\ &=\sqrt{\frac{a}{2\pi ...


0

A more straightforward approach using factorials and Sterling: ${{8n}\choose{4n}} = \frac{8n!}{4n!4n!}$ We know that $n!$ is approximately $\sqrt{2\pi n}(\frac{n}{e})^n$ So plugging this in for both the numerator and the denominator and you have your result.


1

From here, we have $$\dbinom{2m}m \sim \dfrac{4^m}{\sqrt{m\pi}}$$ Taking $m=4n$, we obtain $$\dbinom{8n}{4n} \sim \dfrac{4^{4n}}{\sqrt{4n\pi}} = \dfrac1{2\sqrt{\pi}} \cdot \dfrac{2^{8n}}{\sqrt{n}}$$


0

You can measure the error in your guess as follows: $$(2n+1)!^2=\frac{(4n+2)!}{\binom {4n+2}{2n+1}}$$ where $\binom {4n+2}{2n+1}$ is one of the central binomial coefficients. If you know anything about Pascal's Triangle, you should know that the central coefficients are the largest ones.


1

You can also "simplify" it this way \begin{equation} n!^2 = (2n)!\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\frac{\sqrt{\pi}}{2^{2n}}. \end{equation} Clearly this is not really a simplification, because you still have a product of factorials. The nice thing here is that the ratio of gamma functions is a relatively small number, so it gives you a precise ...


0

Well, I finally found the formula in S. Ramanujan, The Lost Notebook and other Unpublished Papers. S. Raghavan and S. S. Rangachari, editors. Narosa, New Delhi, 1987. page 339.


1

I will use $n$ instead of $x$. Note that if $n\ge 1$ then $$\frac{n^n}{(2n)!}=\frac{1}{n!}\left (\frac{n}{n+1}\cdot\frac{n}{n+2}\cdot\frac{n}{n+3}\cdots \frac{n}{2n} \right).$$ But it is clear that $$\frac{n}{n+1}\cdot\frac{n}{n+2}\cdot\frac{n}{n+3}\cdots \frac{n}{2n} \lt 1.$$ Thus if $n\ge 1$ then $$0\lt \frac{n^n}{(2n)!}\lt \frac{1}{n!}.$$ As ...


0

I have an idea(it might be wrong). According to Brocard's problem $$x^{2}-1=n!=5!*(5+1)(5+2)...(5+s)$$ here,$(5+1)(5+2)...(5+s)=\mathcal{O}(5^{r}),5!=k$. So, $$x^{2}-1=k *\mathcal{O}(5^{r})$$ Here, $\mathcal{O}$ is Big O notation. For every rational number $x$ , there is a rational $r$( since $k$ is a constant, if $x$ is increased, $r$ has to be ...


1

We know from the Hypergeometric Distribution and the Vandermonde Identity that $$\large\sum_{j=0}^{n}\frac{\binom nj\binom m{r-j}}{m+n\choose r}=1$$ which is independent of $r$. Summing the above from $r=0$ to $m+n$ gives $$\large\sum_{r=0}^{m+n}\sum_{j=0}^{n}\frac{\binom nj\binom m{r-j}}{m+n\choose r}=m+n+1$$ Put $i=r-j$ and note that values of $i>m$ ...


4

For integers $a\ge b\ge 0$, we have ${a\choose b}=\frac{a!}{b!(a-b)!}$ (see wiki). Take the special case $a=2b$ and you get ${2b\choose b}=\frac{(2b)!}{b!b!}$, then cross-multiply.


2

Use the sum of product of binomials formula $\sum _{j=0}^k \begin{pmatrix}k-j \\r \end{pmatrix}\begin{pmatrix}i+j \\s \end{pmatrix}=\begin{pmatrix}k+i+1 \\r+s+1 \end{pmatrix}$ Let $k=m+n-i$, $r=m-i$, and $s=i$ $\sum _{j=0}^{n} \begin{pmatrix}m+n-i-j \\m-i \end{pmatrix}\begin{pmatrix}i+j \\i \end{pmatrix}=\begin{pmatrix}m+n+1 \\m+1 \end{pmatrix}$ where ...


8

Ok, so I'm writing this as an answer because it doesn't fit in the comments: We can give a proof of $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ using a volume interpretation but it's not that illuminating and it is not interesting enough (maybe). Even worse, I will have to rewrite the equation as $$n(n-1)\cdots (n-k+1)=k\cdot (n-1)(n-2)\cdots ...


-1

Sorry if I am taking this is too literally, I hope you find my findings about this space an interesting as I found it to explore: To think about factorial as being about the volume of an oblong, as in your question, and further build to thinking about rules and relationships, I started by using thease shapes, with an area described with factorial, such as ...


1

It is possible to proof this by induction, however I don't consider it nice and as shown by others, there really are nice solution: Let's say the inequality holds for 1 to n: So it is $ (n-1)! ^{1 \over n-1} < n!^ {1 \over n} ; $ Now you want to proof the inequality for $ n+1; $ $ n! ^{1 \over n} < (n+1)!^ {1 \over (n+1)} ; $ Now we are pushing ...


2

No need to use induction. $$(n!)^{\frac{1}{n}}<((n+1)!)^{\frac{1}{n+1}}\impliedby n!<((n+1)!)^{\frac{n}{n+1}}\impliedby (n!)^{n+1}<((n+1)!)^n$$ The last inequality holds since for all $1\le i\le n$ and $i\in \mathbb{N}$ we have $i<(n+1)$.


3

HINT: Rearranging the inequality to get rid of the fractional exponents is the easiest approach, but you can also look at the ratio of consecutive terms: $$\left(\frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}\right)^{n+1}=\frac{(n+1)!}{n!\cdot n!^{1/n}}=\frac{n+1}{n!^{1/n}}>\frac{n+1}{(n^n)^{1/n}}>\ldots$$


1

This is not an answer but it seems to be too long for a comment. In the same spirit as user17762's answer, we can "see" that we have polynomials and, as a result, $$S(k,x)=\frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i)$$ $$S(k,x-1)=\frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i-1)$$ So, in other words, for given values of $k$ and $y$, you search for an integer $x$ such ...


1

If $x \in \mathbb{Z}$, we have $$\sum_{i=0}^x \dbinom{k+i}{k} = \dbinom{k+x+1}{k+1}$$ I trust this should help you. For smaller values of $k$, say for instance, $k=2$, we have $$S(2,x)=\dbinom{k+x+1}{k+1} = \dbinom{x+3}3 = \dfrac{(x+3)(x+2)(x+1)}6$$ a polynomial of degree $3$. In general, $S(k,x)$ is a polynomial in degree $k+1$.


2

You have a geometric sum. The $n$th term is given by $$\sum_{m=1}^n i^m =\frac{i-i^{n+1}}{1-i}=i\frac{1-i^n}{1-i}$$


0

Working modulo $4$ does not tell you anything about the quadratic character modulo $p$. For example, no square is $3$ modulo $4$, but $3$ is a square modulo $11$, $13$, $23$, $37$ ... Similarly, $7$ is a square modulo $3$, $19$, $29$, ... (see here) $\newcommand\leg[2]{\left(\frac{#1}{#2}\right)}$ In fact the Chinese Remainder Theorem tells us that, in some ...


0

You know that: $$60!\equiv(55)! *(-1)(-2)(-3)(-4)(-5)\mod 61$$ hence $2x\equiv -120x\equiv -1 \mod 61$ you need only to find $x$


0

We have $60! = 55! \times 56 \times 57 \times 58 \times 59 \times 60$. We have $60! \equiv -1 \pmod{61}$ from Wilson's theorem. Hence, $$55! \equiv - 56^{-1} \times 57^{-1} \times 58^{-1} \times 59^{-1} \times 60^{-1} \pmod{61}$$


8

Doing arithmetic modulo $\;61\;$ all along: $$55!=\frac{60!}{(-5)(-4)(-3)(-2)(-1)}=\frac{-1}{-120}=-\frac1{2}=-31=30$$


1

Lets use mathematical induction. For $n=2$, we see that $\frac{2!}{2^{2}}=\frac{1}{2}=\big(\frac{1}{2}\big)^{1}$. For $n=3$, we see that $\frac{3!}{3^{3}}=\frac{6}{27}=\frac{2}{9}\leq\frac{1}{2}=\big(\frac{1}{2}\big)^{1}$. Assume now that for $n=2k$, we have $\frac{(2k)!}{(2k)^{2k}}\leq\big(\frac{1}{2}\big)^{k}$ and for $n=2k+1$, we have ...


4

$\displaystyle\frac{n!}{n^n}=\frac{n(n-1)(n-2)\cdots2\cdot1}{n\cdot n\cdots n\cdot n}=\frac{n}{n}\frac{n-1}{n}\cdots\frac{k+1}{n}\frac{k}{n}\frac{k-1}{n}\cdots\frac{2}{n}\frac{1}{n}$ $\hspace{.4 in}\displaystyle\le \frac{k}{n}\frac{k-1}{n}\cdots\frac{2}{n}\frac{1}{n}\le\left(\frac{1}{2}\right)^k$ since $\displaystyle l\le k\implies l\le \frac{n}{2}\implies ...


0

Let's focus on the number of ways the girls can be selected. Let event $P$ be the event that Priya is selected to be one of the girls on the committee. Let event $R$ be the event that Roberta is selected to be one of the girls on the committee. The Inclusion-Exclusion Principle states that the number of committees that contain Priya or Roberta is $$|P ...


1

What you want to use is something called the Principle of Inclusion/Exclusion. Consider first the committees that Roberta is on. We include all such committees, giving $\binom{14}{3}\binom{16}{2}$. But we don't want all of these, since Priya could be on some. So let's exclude the committees in which both Roberta and Priya are on. There are ...


11

$$\frac{(n+1)!}{(n-2)!} = 990$$ $$ \frac{(n+1)(n)(n-1)(n-2)!}{(n-2)!} = 990$$ $$(n+1)(n)(n-1) = 990$$ $$(n+1)(n)(n-1) = 99 \times 10$$ $$(n+1)(n)(n-1) = 11 \times 10 \times 9$$ thus $n=10$


0

Hint: Cancellation by division of factorial common factors in numerator and denominator should leave you with three terms, giving a cubic polynomial.


1

The factorial function $x \mapsto x!$ is by convention written in "postfix" notation, like exponentiation $x \mapsto x^n$. The convention is that these postfix operators have high precedence (i.e., you evaluate them before other operators unless parentheses tell you otherwise): $3x^2$ means $3 (x^2)$ not $(3x)^2$ and $2n!$ means $2(n!)$ not $(2n)!$.


1

The standard notation for the factorial function is rather unusual. Usually we define a function $f$ from space $A$ to space $B$ by saying $f: A\rightarrow B$ and then whenever we write the function we (usually) require parentheses to describe precisely what the function operates on, so $2f(a)$ and $f(2a)$ mean clearly different things. In the order of ...


11

$$\frac{2017!+2014!}{2016!+2015!}=\frac{2014!(1+2017\times2016\times2015)}{2014!(2015+2016\times2015)}=\frac{1+2017\times2015\times2016}{2015+2016\times2015}=\frac{1+2017\times2015\times2016}{2015(1+2016)}=\frac{1+2017\times2015\times2016}{2015\times2017}\simeq\frac{2017\times2015\times2016}{2015\times2017}=2016$$


5

$$\Gamma(3/2) = \Gamma(1/2)/2 = \frac{1}{2}\int_0^{\infty} x^{-1/2} e^{-x} \, dx $$ Change variables to $y=x^{1/2}$, so $dy=dx/(2x^{1/2})$, and we have the integral $$ I = \int_0^{\infty} e^{-y^2} \, dy. $$ You may or may not have met this one before. Basically the easiest idea for evaluating this is to change coordinates to polars after squaring it. $$ I^2 ...


1

By definition of the Gamma function, we have $$\Gamma\left(\frac{1}{2}\right)=\intop_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}\ dt.$$ Now let $u=\sqrt{t}$, so that $du=\frac{1}{2\sqrt{t}}\ dt$. Then we have $$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}2e^{-u^{2}}du=\intop_{-\infty}^{+\infty}e^{-u^{2}}du.$$ This last integral is the Gaussian integral, ...


1

To some extent it seems like you'll need relationships between, for example, $x^2!$ and $(x!)^2$. That might be a tough nut to crack, or not. I tried a few and any pattern I could see broke down quickly: $$0^2! = (0!)^2 \\ 1^2! = (1!)^2 \\ 2^2! = (2!)^2 \cdot 3! \\ 3^2! = (3!)^2 \cdot \frac{8!}{4!} \\ 4^2! = (4!)^2 \cdot \frac{15!}{6!} \cdot 5 \cdot 4 \\ ...


1

I'm able to solve (2) and (3). (2) for its trivialness and (3) for its generalization properties. $$x!=e^x$$ Since, $$0!=1=e^0$$ $0$ is a solution to this equation. $$(2x!)^2+x!-1=0$$ Set $u=x!$ and then solve the resulting quadratic equation... $$4u^2+u-1=0$$ $$u=x!={{\pm \sqrt{17}-1} \over 8}$$ You mentioned that the ! could be interpreted as the gamma ...


2

$$P_o=\frac{1}{\sum_{k=0}^\infty\left(\frac{\alpha}{u}\right)^k.\frac{1}{k!}}$$ Substitute $\frac{\alpha}{u}=x$. The denominator is $$f(x)=1+x+\frac{x^2}{2!}+\dots$$ Notice that this is the Taylor expansion for $e^x$, and this yields the answer. $$f(x)=e^x$$ $$P_0= \frac{1}{f(x)} = e^{-\frac{\alpha}{u}}$$


0

I presume you came here from the relationship $(2n)!=2^nn!(2n-1)!!$ where the double factorial (double exclamation point) on the right says take every other term. $9!!=9\cdot 7 \cdot 5 \cdot 3 \cdot 1$ for example. You can prove this by looking at the terms that make up $(2n)!$. The $2^nn!$ takes care of the even ones and the $(2n-1)!!$ takes care of the ...


1

No, compare $$\frac{n!}{3^n}=\left(\frac{1\cdot2\cdot3}{3\cdot3\cdot3}\right)\left(\frac{4\cdot5\cdot6}{3\cdot3\cdot3}\right)\left(\frac{7\cdot8\cdot9}{3\cdot3\cdot3}\right)\cdots\left(\frac{(n-2)\cdot(n-1)\cdot n}{3\cdot3\cdot3}\right)$$ and $$\left(\left(\frac ...


1

The quantities are not equal; try $n=3$. As I pointed out in my comment above, you should get $1 = \frac{6}{27}$ which is clearly wrong. Factorial works nicely for whole, positive integers, but that's about the extent of it. In general it is true that $$\left(\frac{m}{n}\right)! \neq \frac{m!}{n!}$$ Factorial for fractions gets very tricky. Just read a ...


0

No, factorial function (or Gamma function) has no properties like this one which holds only for $n=0$ I guess.



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