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0

Both of the previous answers are logically flawed, though both can be repaired. It is not enough to show that the target equality implies an identity: if it were, we could argue that $-1=1$ implies that $(-1)^2=1^2$, which is true, so $-1=1$. It’s crucial that the steps be reversible, and this point should be made explicitly. To avoid this problem I prefer ...


0

I think you can write (n+1)! as (n+1)n! so this will result in (n+1)n! − 1 + (n+1)(n+1)n! = (n+2)(n+1)n! − 1 -1 on left side will be annihilated by -1 on right side resulting in (n+1)n! + (n+1)(n+1)n! = (n+2)(n+1)n! dividing all by (n+1)n! will result in 1 + (n+1) = (n+2) taking off parenthesis 1 + n + 1 = n + 2 so the final result will be n+2 ...


0

Hint: Your equation simplifies to this: $$(n+1)!(1+(n+1))-1=(n+2)!-1\Rightarrow (n+1)!(n+2)-1=(n+2)!-1$$ $$\Rightarrow (n+2)!-1=(n+2)!-1$$ See if you can figure out what I did in the last step.


3

We just have to prove that $\binom{m+n}{m}$ divides $\binom{2m}{m}\cdot\binom{2n}{n}$. So, we may imagine to have a parliament, with $2m$ members in the right wing and $2n$ members in the left wing. We may choose a committee with $n$ people from the left wing and $m$ people from the right wing in $\binom{2n}{n}\cdot\binom{2m}{m}$ ways, then we may choose ...


0

Yes, use Bertrand's postulate. This is a well known way to prove your question. The postulate states that for n>3, there exists a prime number such that n>p>(n/2). Let's assume for now that there exists one value of p such that p^2 divides n. Now, if p^2 divides n, then there must be some n-1>K>p such that p divides K. So, K>(or)=2p>n (from Bertrand's ...


1

In answer to your specific question there are not only 12 zeros at the end of the number but an additional 7 zeros occurring amongst the other non-zero digits


1

number of zeros in $n!$ equal to number of $5$ in $n!$ =$$\sum_{k=1}^{\infty}\left \lfloor \frac{n}{5^k} \right \rfloor\\=\left \lfloor \frac{50}{5} \right \rfloor+\left \lfloor \frac{50}{5^2} \right \rfloor+\left \lfloor \frac{50}{5^3} \right \rfloor+...=10+2+0=12$$it means 12 zero in front of 50!


1

The number of 0's is equal to the powers of 5 in the expansion of 50!. This is because the prime decomposition of 50! will have more factors of 2 than factors of 5, and whenever we have a factor of 2 and 5 we can combine them and tack on a 0 at the end of the number. The number of powers of 5 is $\lfloor{\frac{50}{5}}\rfloor + \lfloor\frac{50}{25}\rfloor = ...


0

Much more than $5$, surely. Since $50!$ is divisible by $2\cdot 5 \cdot 10\cdot 20\cdot 30\cdot 40\cdot 50$, and this number already has $6$ zeroes, you can be sure that $50!$ has at least $6$ zeroes. In fact, you need to count up how many twos and how many fives appears in the factorization of $50!$. Then, the number of zeroes is the smaller of these two ...


1

Let $\displaystyle P(x)=\sum_{r=0}^n (-1)^r\binom{n}{r}(x-r)^n$; we want to show that $P(x)=n!$. Since $\displaystyle P(x)=\sum_{r=0}^n(-1)^r\binom{n}{r}\bigg[\sum_{j=0}^n\binom{n}{j}(-r)^jx^{n-j}\bigg]=\sum_{j=0}^n (-1)^j\binom{n}{j}\bigg[\sum_{r=0}^n(-1)^r\binom{n}{r}r^j\bigg]x^{n-j}$, it suffices to show that ...


1

using the nested sums the general case can be reduced to the $k=0$ case with an interesting modification ... $$ \begin{align} &\sum_{r=0}^{n} { \binom{n}{r} (-1)^r(k-r)^n } \\ \\=&\sum_{r=0}^{n} { \binom{n}{r} (-1)^r \sum_{i=0}^{n} { \binom{n}{i} k^i(-r)^{n-i} }}\\ \\=&\sum_{i=0}^{n} {\binom{n}{i} k^i \sum_{r=0}^{n} { \binom{n}{r} (-1)^r ...


1

Your last step is incorrect. In fact, $$ \frac{(n+1)^{n+1}}{n^n(n+1)} = \frac{(n+1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n = \left(1+\frac1n\right)^n $$ which you might recognize. In your last step, on the bottom you replaced $n+1$ with $n$, which is okay here because their ratio is $\frac{n+1}{n} = 1+\frac1n \to 1$; but on the top you replaced ...


1

$$\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\frac{(n+1)(n+1)^n}{(n+1)n!}\frac{n!}{n^n}=\frac{(n+1)^n}{n^n}=(1+\frac{1}{n})^n\rightarrow e>1$$


2

Hint: Show by induction or inspection that $n^n \geq n!$. Then your terms do not converge to zero so the series necessarily diverges.


3

Stirlings approximation of the factorial: $$ \sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n $$ 5 and 3 are constant factors so put them before the limit. Now we have to concern ourselves with: $$ \frac{5}{3}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n+3^n} = ...


2

Use equivalents: First $3n!+3^n=3n!+o(n!)$ so: $$ 3n!+3^n\sim_\infty3n!\sim_\infty 3\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n$$ and $$\frac{5n^n}{3n!+3^n}\sim_\infty\frac{\mathrm e^n}{3\sqrt{2\pi n}}\xrightarrow[n\to\infty]{}\infty.$$


2

I think the limit diverges as consider the inverse of limit i.e $$\lim_{n\to\infty} \frac{3n! + 3^n}{5n^n}$$ $$\lim_{n\to\infty} \frac{3}{5}*\frac{3^n}{n^n}$$ tends to zero as 3/n tends to zero as n tends to infinity. Also $$\lim_{n\to\infty} \frac{3}{5}*\frac{n!}{n^n}$$ also tends to zero as $$ e\left(\frac{n}{e}\right)^n \leqslant n! \leqslant ...


2

Use Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} .n^n e^{-n}$$ $$ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $$ $$ =\lim_{n\to\infty} \frac{2 \pi n .n^{2n} e^{-2n}}{\sqrt{2 \pi 2n} .(2n)^{2n} e^{-2n}} $$ $$ =\lim_{n\to\infty} \frac{\sqrt{\pi n}}{4^n}=0$$


6

As all terms are positive, we have $$0 \leq \frac{(n!)^2}{(2n)!} = \frac{n!}{2n \cdot \dots \cdot (n+1)} = \prod_{k=1}^n \frac{k}{k+n} \leq \prod_{k=1}^n \frac{1}{2} = \left(\frac{1}{2}\right)^n$$ So then as $$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n = 0$$ It follows that $$\lim\limits_{n\rightarrow\infty} \frac{(n!)^2}{(2n)!} = 0$$ ...


8

Hint: for combinatorial reasons (in how many ways can you choose $n$ objects from $n$ pairs of objects?), $$\binom{2n}n\ge2^n\ ,$$ so $$\frac{(n!)^2}{(2n)!}=\frac1{\binom{2n}n}\le\frac1{2^n}\ .$$


2

Since each term is bounded above by $2^{-n}$, I guess the answer must be zero.


4

Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n} $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0. $$


1

As @Albert mentions the total number of passwords the user can generate is $3^n$. However, the total number of combinations an attacker must consider is wider: $x!\cdot3^n$ for an n length password and a $x$ element special character set.


1

Monk pretty much solved it. Each letter in the word can be represented by the lower caps, the capitalized letter or the symbol we chosed for it. $$\#\{passwords\ for\ an\ n-length\ word\}=3^n$$


2

Another way to do: Let $a_n=\frac{10^n}{n!}$. Clearly $$a_{n+1}=\frac{10}{n+1}a_n$$ and hence $\{a_n\}$ is positive and decreasing when $n\ge 10$. By the bounded Monotone Principle, $\lim_{n\to\infty}a_n=L$ exists. Thus $$ L=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{10}{n+1}a_n=0\cdot L=0.$$


2

If $n>20$, then $$ 0<a_n=\frac{10^n}{n!}<\frac{10^{20}}{20!} \left(\frac{10}{20}\right)^{n-20}=\frac{20^{20}}{20!}\cdot\frac{1}{2^{n}}\to 0, $$ as $w^n\to$, whenever $\lvert w\rvert<1$.


7

$$\frac{10^n}{n!} = \frac{10\cdot 10 \cdot 10 \cdot 10 \cdots 10}{1 \cdot 2 \cdot3 \cdots n} =\frac{10^{11}}{11!}\frac{10}{12}\frac{10}{13}\cdots\frac{10}n < \frac{10^{11}}{11!}\left(\frac{5}{6}\right)^{n-11} \to 0$$ as $n \to \infty.$


0

@kmitov gives a very good prove by using ratio test. () Also, you could always remember that factorial grows faster then exponential function. So the limit goes to 0. For a prove: Set $a_n=10^n/n!$ and take arbitrary $\epsilon>0$, you can always find a $N$ large enough such that for all $n>N$ such that $$ a_n<\epsilon $$ Together with the fact ...


4

The series $\sum_{n=0}^\infty \frac{10^n}{n!}$ is convergent and the sum is $e^{10}$. The convergence can be checked by ratio test $$\lim_{n \to \infty} \frac{10^{n+1}}{(n+1)!}\frac{n!}{10^n}=\lim_{n \to \infty}\frac{10}{n+1}=0<1$$ By the necerssery condition for the convergence of an infinite series the common term must go to 0.


1

The value of $(-2)!!!=1$ according to $$n!!!=n(n-3)!!!$$ $$1!!!=1(1-3)!!!$$ $$1=(-2)!!!$$


3

The "triple factorial" is defined for non-negative integers $n$ as $$ n!!! = n (n-3)(n-6) \cdots (n \bmod 3) \, , $$ compare "Multifactorials" in Wikipedia. For $n \ge 0$ and $j > 0$ $$ n!!! = \frac{(n+3j)!!!}{(n+3)(n+6)\cdots (n+3j)} \tag{1} $$ holds. This relation can be used to define the function for negative integers as well, as long as they are not ...


4

Hint: for any $\;x\in\Bbb R\;$ : $$\sum_{n=0}^\infty\frac{x^n}{n!}=e^x$$


-1

You must have made a typo with your second entrance into WA. To confirm your first answer: \begin{align}\require{cancel} \binom{n+1}{3}\cdot\frac{(n-1)!+(n-2)!}{(n+1)!}&= \frac{\cancel{(n+1)!}}{3!(n-2)!}\cdot\frac{(n-1)!+(n-2)!}{\cancel{(n+1)!}}\\[1em] &= \frac{(n-1)!+(n-2)!}{3!(n-2)!}\\[1em] &= ...


-1

$$n/6$$ as ${n+1\choose 3}$ can be reduced to $(n+1)(n)(n-1)/3!$, now $(n+1)(n)(n-1)$ is canceled from the denominator and the $(n-2)!$ is canceled too so we are left with $(n-1) + 1$ numerator and $3!$ denominator. So the answer is $n/6 $.


1

The inclusion-exclusion is around the number of pairs of adjacent people. How many ways are there to pick $n$ pairs of adjacent people? Either $2k$ and $1$ are a pair, and then pick $n-1$ pairs from a row of $2k-2$ people, or they aren't, and you want $n$ pairs from a row of $2k$ people. I did this to deal with the circular symmetry of the problem. Using ...


3

Oh, lol! $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n-1)! + (n-2)!}{3!(n-2)!} = \frac{(n-2)!((n-1) + 1)}{3!(n-2)!}=\frac{n}{3!} = \frac{n}{6}$ Correct? :) How come Wolfram Alpha gives two different results?


7

Well, for starters you can cancel the two $(n+1)!$s from the top and the bottom of the fraction. Also note that $(n-1)! = (n-1)((n-2)!)$ and then you can cancel an $(n-2)!$ from the top and the bottom.


2

This might help give you an intuitive idea of why those quantities are related: Say you have a set of $N$ things, and you're going to pick $2$ of them, where order doesn't matter. Give each thing a number from $1$ to $N$, and put them in order from thing $N$ down to thing $1$, like so: $$ N,N-1,N-2,...,2,1 $$ Now, we can pick the first thing that will be ...


8

Note that $$n!\,e=\sum_{k=0}^\infty\frac{n!}{k!}=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac{n!}{k!}$$ The first sum on the RHS is always an integer since $n\geq k$. The second sum satisfies $$\begin{align} \sum_{k=n+1}^\infty\frac{n!}{k!} &=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots\\ ...


4

Let's rewrite this as $\;\displaystyle S=\sum _{j=0}^{n} \frac{1}{j!}-2\;$ then prove that $$\sum _{j=0}^{n} \frac{1}{j!}=\frac{\lfloor e\; n!\rfloor}{n!}$$


5

Now we see that $(3!, 11)=(6,11)=1$. Hence by Fermat's theorem we have $(3!)^{10}\equiv 1[11]$ and hence $(3!)^{10m}\equiv 1[11]$. Moreover $10$ divides $5!$ so that $(((3!)^{5!})^{7!})^{...}\equiv 1[11]$. Required reminder is 1.


4

$$(3!,11)=1$$ By Fermat's Little Theorem, $$3^{11-1}\equiv1\pmod{11}$$ and $10|5!$


4

They are using the trapezoid rule over $n-1$ intervals: $(1,2), (2,3) \dots (n-1,n)$ The trapezoid rule for the first interval is $\int_1^2f(x)\;dx \approx \frac 12(f(1)+f(2))$ When you add these up over all the intervals, each internal point gets a coefficient $1$, with $\frac 12$ coming from the interval below and $\frac 12$ from the interval above but ...


3

This looks very similar to Pochhammer symbol used to represent the falling factorial $$(x)_n=x(x-1)(x-2)\cdots(x-n+1)$$ and when $x$ is a non-negative integer, $(x)_n$ gives the number of $n$ permutations of an $x$ element set, or equivalently the number of injective functions from a set of size $n$ to a set of size $x$ (this is a quote from the ...


0

The multifactorial notation I just found being used online is $$n!^{(b)}=\prod_{k=0}^{\lfloor{\frac{n}{b}\rfloor}} (n-bk)$$ Can someone confirm that this is standard?


2

I much prefer the combinatorial argument, but it’s useful to be able to manipulate summations and falling factorials, so here for the record is the induction step of the proof by induction on $k$. $$\begin{align*} \sum_{i=0}^{k+1}\binom{k+1}in^{\underline{k+1-i}}m^{\underline ...


3

How many ways can you pick $k$ balls from a set of $n$ different red balls and $m$ green different balls? Answer $$(n+m)^{\underline k}$$ But you can count them another way. First suppose that the $k$ balls are red, then $k-1$ are red and $1$ is green, etc. This gives $$\sum_{j=0}^k\binom kj n^{\underline j} m^{\underline {k-j}}$$


1

This is because, using factorial notation, $$(2n-1)\times(2n-3)\times\cdots\times1=\frac{(2n)!}{n!2^n}$$ and thus $$2^n(2n-1)\times(2n-3)\times\cdots\times1=\frac{(2n)!}{n!}=2n\times(2n-1)\times\cdots\times(n+1)$$


1

You can do this using generating functions. This is analogous to the classic coin change problem. In your case, you want to look at the term for $x^{133}$ in the expansion of $$ f(x) = \frac{1}{(1-x^1)(1-x^2)(1-x^3)}.$$ We may rewrite $f$ as $$ f(x) = \frac{1/6}{(1-x)^3} + \frac{1/4}{(1-x)^2} + \frac{17/72}{1-x} + \frac{1/8}{1+x} + \frac{-j/9}{x-j} + ...


2

That would be $$(1\cdot 2\cdot 3\cdot\ldots\cdot n)-1$$ If $n=0$ or $n=1$ this is zero.



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