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## Hot answers tagged factorial

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 $n=10$

11

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

9

hint: $\dfrac{n!}{n^n} < \dfrac{1}{n}$

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 $$n(n-1)\cdots (n-k+1)=k\cdot (n-1)(n-2)\cdots ... 8 Doing arithmetic modulo \;61\; all along:$$55!=\frac{60!}{(-5)(-4)(-3)(-2)(-1)}=\frac{-1}{-120}=-\frac1{2}=-31=30$$7 The factorial function is extended by the \Gamma function. The relation is$$(n-1)! = \Gamma(n) = \int_0^\infty t^{n-1} e^{-t}\, dt$$This can be analytically continued as a meromorphic function in the complex plane. Ref: John Conway's book on Complex Analysis. 6 A nice definition of n! is the number of bijections from \{1,2,\dots,n\} to itself. Thus 0! is the number of bijections from the empty set to itself. There is one such map, which can be thought of as an empty product of sorts. 6 Although not 100% about what you are asking (perhaps 87%), consider an n-dimensional cube with side x. The n-th derivative of the box's volume is n!. That is$$\frac{d^n}{dx^n} x^n=n!$$Curiously enough, this is saying that the rate of the rate of the...of the rate of increase of the volume of an n-dimensional cube is precisely the volume of a ... 6 2010! ends with 501 zeroes since:$$\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}$$it follows that: ... 5$$\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 ...

5

Hint: For $(1)$, we have $(2k+2)!=(2k+2)(2k+1)(2k)!$, which implies that $$\frac{(2k)!}{(2k+2)!}=\frac{(2k)!}{(2k+2)(2k+1)(2k)!}=\frac{1}{(2k+2)(2k+1)}.$$

4

$$\frac{1}{n^k}\binom{n}{k} = \frac{1}{k!}\frac{n!}{n^k(n-k)!} = \frac{1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$$ and every term of the last product tends to one as $n\to +\infty$. Moreover, $$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}.$$

4

You are on the right track, but you did not consider all pairs $(x,y)$ where $x!=y!=1$. Also, you can be more concise, as follows. Assume without loss of generality that $x\le y$. Then $z! = x!+y! \le 2y!$. Then $z!$ exceeds $y!$ by at most a factor of $2$. Since $x!$ is positive, $z!>y!$, and thus $z!$ must exceed $y!$ by more than a factor of $1$. Then ...

4

false: take: $n=6$ $\frac{n}{3}! = 2$ and $\frac{n}{3}! = 80/81$ In fact the diagram below shows that equality rarely happens

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 $\displaystyle l\le k\implies l\le \frac{n}{2}\implies ... 4 For integers$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. 3 André Nicolas gives the clue, "It counts in two different ways the number of ways to divide 2n people into n teams of 2 people." The left-hand side Consider the following procedure: Select the youngest person who has not yet been put in a two-person team. Match them with another so far unselected person. This produces ... 3 As suggested by Workaholic, $$I=\left(-\frac{1}{2}\right)!=\Gamma\left(\frac{1}{2}\right) = \int_{0}^{+\infty}x^{-1/2}e^{-x}\,dx = 2\int_{0}^{+\infty}e^{-y^2}\,dy = \int_{\mathbb{R}}e^{-y^2}\,dy$$ and by Fubini's theorem: $$I^2 = \int_{\mathbb{R}^2}e^{-(y^2+z^2)}\,dy\,dz = \int_{0}^{2\pi}\int_{0}^{+\infty}\rho\, e^{-\rho^2}\,d\rho\,d\theta = \pi ... 3 Consider the function$$ (1-x)^n = \sum_{k=0}^{\infty} \binom{n}{k} (-1)^k x^k. $$Then we can write$$ \sum_{k=0}^{n} \binom{n}{k} (-1)^k k x^k = x\frac{d}{dx}\sum_{k=0}^{n}\binom{n}{k} (-1)^k x^k = x\frac{d}{dx} (1-x)^n. $$Setting x=1 gives the result$$ \sum_{k=0}^{n} \binom{n}{k} (-1)^k k = 0, \quad (n>1) $$since$$ x\frac{d}{dx} (1-x)^n = ... 3 Note: I sympathize with your question because this used to bother me as well. It seemed absurd that some authors of textbooks could get by with letting$0!=1$because it was "convenient to define it to be so" (can't remember exactly which book that came out of). First note that the title of your question does not make a great deal of sense in the context of ... 3 As to why there is one way to arrange zero objects, I once read a very nice explanation: To count the number of ways to arrange something, you take a picture of every possible layout of the items, and then you count the number of distinct pictures. For zero items, there is going to be no items in the arrangements, but you will still have one single picture ... 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: $$\left(\frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}\right)^{n+1}=\frac{(n+1)!}{n!\cdot n!^{1/n}}=\frac{n+1}{n!^{1/n}}>\frac{n+1}{(n^n)^{1/n}}>\ldots$$ 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$i<(n+1)$. 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. $$f(x)=e^x$$ $$P_0= \frac{1}{f(x)} = e^{-\frac{\alpha}{u}}$$ 2 The equation$n! = n \cdot (n-1)!$only holds for$n \geq 1$, so you're not allowed to plug in$n=0$like you did in the question. Try plugging in$n=1$instead. We get: $$1! = 1 \cdot 0!$$ Hence $$1! = 0!$$ So if you agree that$1!$equals$1$, then: $$1 = 0!$$ 2 $$0! = \prod_{j \in \varnothing} j = e^{\sum_{j \in \varnothing}\log j} = e^0 = 1$$ 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$r$: $$\frac{\pi^{n/2}r^n}{(n/2)!}.$$ 2 You have a geometric sum. The$n$th term is given by $$\sum_{m=1}^n i^m =\frac{i-i^{n+1}}{1-i}=i\frac{1-i^n}{1-i}$$ 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 ... 2 Here is the simplest proof I know. It is a very straightforward application of the calculus of finite differences. Let$f(x)$be a polynomial. Consider its backward finite difference $$(\Delta f)(x) = f(x) - f(x - 1).$$ Key lemma: If$f$has leading term$a_n x^n$, then$\Delta f$has leading term$n a_n x^{n-1}\$. In particular, its degree is one less ...

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