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1) They are using 366 instead of 365 to take February 29th into account. 2) $\displaystyle\frac{366!}{(366-N)!}$ does correspond to what you have for the number of injective functions, since $\;\;\displaystyle\frac{366!}{(366-N)!}=\frac{366\cdot365\cdot364\cdots(366-N+1)(366-N)(366-N-1)\cdots3\cdot2\cdot1}{(366-N)(366-N-1)\cdots3\cdot2\cdot1}$ $\hspace{.8 ... 1 From wikipedia, I am picking out the relevant part that lends itself to proof by induction -$C_n$is the number of ways to correctly match$n$pairs of brackets (Dyck Language) in a string word consisting of these$n$pairs of brackets. This means that the number of opening brackets is never less than the number of closing brackets in any substring derived ... 3 See $$\frac{(2n)!}{n!(n+1)!} = \frac{(2n)!}{n!n!(n+1)} = {2n \choose n}\frac 1{n+1}$$ Then look at Wikipedia Catalan number. 3 The sum cannot converge, even for complex$a$. If an infinite series converges, the summands must tend to$0$. However, since each summand is the factorial of the previous summand, then the summands would also tend to$0!=1$, a contradiction. 1 To my knowledge, there is no way to express$(ab)!as something else in an exact manner. However for large numbers, Stirling's approximation of the factorial can be useful, if the sought result is numerical. The Stirling approximation reads $$n!\approx n^n\mathrm e^{-n} \sqrt{2\pi n}.$$ It is more handable in its logarithmic form $$\ln n!\approx n\ln ... -1 Here 34! contain 2^{32} and 5^7. So 2^{25} remaining and 2^{7}\cdot 5^7 = (10)^7 form 7 zeros at the end. So 34! = 295232799cd9604140809643ab \times 10^7 now after deleting 7 zero,s , So we get 34! = 295232799cd9604140809643ab Now we use divisibility test for last 7 digits using 2^7. Here (a,b)\in \{0,1,2,3,4,5,6,7,8,9\} So ... 4 You know that 34! is divisible by 9, so, because the sum of all other digits is 141, you know that c+d=3 or c+d=12. Now do the same for divisibility by 11 (remember the alternating sums criterion?). 14 HINT: The number is a multiple of 9 and a multiple of 11. 3 Obviously, n\ne0, and we can simplify:$$(n-1)!=1.$$By inspection, n-1=0 and n-1=1 are solutions. All other factorials have a factor that makes them exceed 1. 1 To prove that n>2\implies n!>n you can use induction. Base case: 3!=6>3 Induction step: if n!>n and n>2 then (n+1)!=n!(n+1)>n(n+1)>n+1. 3 If n!>n, then$$ (n+1)! = (n+1)n! > (n+1)n > n+1 $$as long as n>1. Now, 3!>3, and by induction we have proved that n!>n for all n \geq 3. Since 1!=1 and 2!=2, the assertion is proved. -1 This seems shorter: If n = 1 or n = 2, then n! / n = 1 and thus n! = n. If n > 2, then n! / n = (n-1)! \geq 2 and thus n! \neq n. Therefore, n \in \mathbb{N} \backslash \{ 0 \}, n! = n \Rightarrow n = 1 \vee 2. 2 n! = n(n-1)! So, start simplifying the first part of your equality. \frac{n!}{(n-1)!} = \frac{n(n-1)!}{(n-1)!} = n. Do you see how to continue? 1 Stirling's formula unfortunately has potentially big errors - the approximation does not mean that$$n! -\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n\to 0,$$only that$$\frac{n!}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}\to 1.$$So an approximation of n! is not of much use. Basically, the above formula is "trivial" one's you know Wilson's theorem: ... 1$$n!-(n-1)!=n(n-1)!-(n-1)!=(n-1)!\left( n-1 \right) $$1 n! is defined inductively by 0!=1 and n!=n\cdot(n-1)! for n\geqslant 1. Hence$$n! - (n-1)! = n\cdot(n-1)! - (n-1)! = (n-1)(n-1)! $$9 Hint: n! = n \cdot (n - 1)!. 0 There is no such integer. If you generalize to reals with Gamma function, x!\to\Gamma(x+1), then you have an ugly transcendental equation which you need to attack numerically, no way around it. Bisection is probably the safest bet. If it happens to be integer, I'd do a linear search... just increase x in x! until you get to the solution. If you jump ... -5 I was trying to get a simple proof and I came up with this one. Let us consider the function f:R->R, f(x)=(x^2) (^ means square; x^2=x*x). Now consider the convex region f(x)>(x^2) and lets call it something say, H. For n>=4, by using induction we can show that n!>(n^2). Therefore for all n belonging to N (N is the set of all natural numbers including 0), ... 1 Here, the factorials don't really matter because 2!=2, but in general, we do need to take factorials. Consider a simpler problem: How many anagrams does the word "AAAB" have? The answer is clearly 4, since once we know where the "B" is, we know the rest of the letters are "A". To get there, we have to consider that there are 4!=24 ways to arrange 4 ... 2 We are given that$$n!= \left\{\begin{align}1\quad \text{for}\quad n=0\\n\cdot(n-1)!\quad \text{for}\quad n>0\end{align}\right.$$and wish to show that$$n! = \prod_{i=1}^ni$$We start with n=1 \implies 1! = 1\cdot(0!) = 1 = \prod_{i=1}^1 i Now, let's assume (n-1)! = \prod_{i=1}^{n-1} i, then$$n! = n\cdot(n-1)! = n\prod_{i=1}^{n-1} i = ... 0 Induction step: If $$(n-1)!=(n-1).(n-2)\dots3.2.1$$ then by definition $$n!=n.(n-1)!=n.(n-1).(n-2)\dots3.2.1$$ 0 Hint: Let's assume you have 15 positions and you need to distribute the letter a5$times. Then you can achieve this in the following way: Distribute the letter "a" 5 times on 11 positions. Now set an empty space between every pair of consecutive letters. A specific example: The first distribution might yield -a-aaa--a-- Adding the spaces [here denoted by ... 4 A proof I found a while ago entirely relies on creative telescoping. Since$\frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)}$, $$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq ... 0 20!=1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20 20!=2^{1+2+1+3+1+2+1+4+1+2=18}*3^{1+1+2+1+1+2=8}*5^{1+1+1+1=4}*7^{1+1=2}*11^{1}*13^{1}*17^{1}*19^{1} 20! \gt 2^{18+8+4+2+1+1}*17^{1}*19^{1}\gt2^{34}*16*16=2^{34}*2^4*2^4>2^{42}>2^{40} Hence 20!\gt2^{40} I think this is very easy to think... 1 I like Knuth's notation for falling factorials better:$$ \alpha^{\underline{h}} = \alpha \cdot (\alpha - 1) \cdot \dotsm \cdot (\alpha - h + 1) $$First note that:$$ \Delta \alpha^{\underline{h}} = (\alpha + 1)^{\underline{h}} -\alpha^{\underline{h}} = (\alpha + 1) \cdot \alpha^{\underline{h - 1}} - \alpha^{\underline{h - 1}} \cdot (\alpha ... 7 Look closer at your graph! (I am going to use$k$and$n$here.) I get e.g. $$\frac{1}{100000}\left(\sum_{k=1}^{100000}\ln(k)\right)\approx \ln(100000)-0.999933$$ This suggests an asymptotic of$\ln(n)-1. Indeed: $$\begin{array}{ll} \displaystyle \frac{1}{n}\sum_{k=1}^n\ln(k) & \displaystyle =\ln(n)+\frac{1}{n}\sum_{k=1}^n \big(\ln(k)-\ln(n)\big) \\ ... 1 If I properly read the question, you are concerned by$$a_n=\frac{(n!)^{\frac{1}{n}}}{n}$$So$$\log(a_n)=\frac 1n \log(n!)-\log(n)$$As usual when working with factorials, Stirling approximation is useful. You will in particular find this very nice inequality$$\sqrt{2\pi}n^{n+\frac 12}e^{-n} \leq n! \leq en^{n+\frac 12}e^{-n}$$which would give you good ... 0 This is known as Brocard's problem. It has been shown there are only finitely many solutions given the ABC conjecture is true (this I believe was proved recently), and it is conjectured the 3 solutions you have given are the only solutions. Calculations up to n=10^9 show no other solutions. 0 I believe you mean permutations, rather than combinations. I am unaware of any single formula, per se. However, there is an algorithm, that you could use. It makes use of the factoradiac number system (also known as the factorial number system) and the Lehmer code. https://en.wikipedia.org/wiki/Factorial_number_system & ... 2 This provides another way. Let a_{n} > 0 for all n \geq 1. Then it can be shown that$$\liminf a_{n+1}/a_{n} \leq \liminf (a_{n})^{1/n} \leq \limsup (a_{n})^{1/n} \leq \limsup a_{n+1}/a_{n}.$$Thus if there is some l \in \mathbb{R} such that a_{n+1}/a_{n} \to l then (a_{n})^{1/n} \to l. Let a_{n} := n^{n}/n! for all n \geq 1. Then ... 2 Stirling's formula says the ratio of$$ \sqrt{2\pi x\,{}} \, \frac{x^x}{e^x} $$to x! approaches 1 as x\to\infty. You seem to have a crude approximation of Stirling's formula. (I think Stirling's contribution to this may have been the value of the constant factor. de Moivre earlier showed that the ratio of \sqrt{x}\ \dfrac{x^x}{e^x} to x! ... 5 Using the fact that n!\sim_{\infty }\sqrt{2\pi n}\left(\frac{n}{e}\right)^n,$$\lim_{n\to\infty }\frac{n!}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\sqrt{\frac{2\pi}{3}}\underbrace{\left(\frac{2}{e}\right)^n}_{\to 0}=0$$and thus ... 1 The correct constants are twice the value of \pi=3.1415\ldots and e=2.718281828459045\ldots, not 3 and 2. That's why it might seem close, that is,$$n!\sim n^n e^{-n} \sqrt{2\pi n}$$0 It doesn't look that close to me... but if you want to compare them, consider the asymptotic behavior of their logarithms. First,$$ \log x! = x \log x - x + O(\log x) $$by Stirling's approximation, and second,$$ \log\left(\sqrt{3x}\left(\frac{x}{2}\right)^x\right) = x \log x - x \log 2 + O(\log x). $$So the ratio between the two functions grows as ... 7 We will use the following facts: (i) The extension, to a larger domain, of a non-elementary function is also non-elementary; (ii) The derivative of an elementary function is also elementary; (iii) The product of finitely many elementary functions is also elementary; (iiii) The product of an elementary function times a non-elementary ... 0 If you take the log, it is:$$\frac{1}{n}\sum_{k=1}^n \log\left(\frac{k}{n}\right)$$Which is a Reimann sum for \int_{0}^1 \log x. The indefinite integral is F(x)=x\log x-x and \lim_{x\to 0} x\log x -x =0, and F(1)=-1. You have to deal with the fact that this integral is an improper integral, but it "just works." 1 They are not equal; {n \choose 2} = \frac{n(n-1)}{2}. But the difference between them is n/2, which is small enough that it can be rolled into the correction term in your link. 1 Since 2011 is prime, Wilson's Theorem applies. Modulo 2011 we have$$\eqalign{ (1\times3\times\cdots\times2009)^2 &=1\times3\times\cdots\times2009\times 1\times3\times\cdots\times2009\cr &\equiv1\times3\times\cdots\times2009 \times(-2010)\times(-2008)\times\cdots\times(-2)\cr &\equiv(-1)^{1005}(2010)!\cr &\equiv1\ ... 1 The factors that you are looking for are not divisible by 2 (because they are odd) and not divisible by 3 (because they are divisible by 3). We have 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9. If we remove the factors 2 and 3, then all that is left is 5 x 7, and the only factors that are odd and not divisible by 3 are 1, 5, 7 and 35. Of these, only 5 and 7 ... 39!$=$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$. Of all the odd factors, only$5$is of the form$3m + 2$. We can have$5$, and the only possible number to combine it is$7$to give$35$which is also of the form$3m + 2$. So these are the only 2 factors and their sum is$40$4 Notice, $$9!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$$$=362880=2^7\times 3^4\times5\times 7$$ total number of odd factors $$=1\times5\times 2\times 2=40$$ But, the odd factors of form$(3m+2)$are only two which can be found by putting$m=1$&$m=11$which are$3\times 1+2=5$&$3\times 11+2=35$Hence the sum of the odd ... 3 The possible numbers are$5,11,17,23,29,35$. So, we have at least$5+35=40$. From the given options, the answer is$(A)\$.