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13

A combinatorial answer. The right-hand side is the number of permutations of $\{0, 1, \dots, k-1\}$. I'll let $k=5$ for concreteness. We count the permutations in another way. Take a $5$-tuple drawn from $\{0,1,\dots,4\}$ - that is, a member of $5^5$. This is probably not a permutations - for example, it might be $00000$. Remove anything which does not ...


12

Let $L$ be the collection of sequences indexed by $\mathbb{N}$ taking values in $\mathbb{R}$. $$L = \{ (x_0, x_1, x_2, \ldots ) : x_i \in \mathbb{R} \}$$ $L$ will be a vector space with respect to componentwise addition and scalar multiplication. Define a linear map $D$ on $L$ by $$L \ni x = (x_0, x_1, x_2, \ldots ) \quad\mapsto\quad Dx = (x_1, x_2, \ldots ) ...


11

The binomial theorem states that $$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$ Putting $x=1$ gives $$\sum_{k=0}^n \binom nk=2^n\\ \sum_{k=1}^n\binom nk=2^n-1$$


6

$34! = 295232799cd96041408476186096435ab000000$ $\left \lfloor \dfrac{34}{5} \right \rfloor = 6$ $\left \lfloor \dfrac{6}{5} \right \rfloor = 1$ So there are $6+1 = 7$ zeros at the end of $34!$. Hence $$\color{red}{b = 0}$$ THEOREM: Compute the following $N = 5q_1 + R_1$ $q_1 = 5q_2 + R_2$ $q_2 = 5q_3 + R_3$ ... $q_{n-1} = 5q_n + R_n$ where $0 \le ...


6

Let's consider the meaning of $\sum_{k=1}^n \left(\begin{array}{c} n \\ k \end{array} \right).$ You are interested in the number of ways to construct a non-empty subset from the original set of $n$ elements. Consider the set with $n$ elements, for each element, we have two options, to include it in one of the subset or not. If we allow our subset to be ...


6

Taking all the rounds together (including the $0^{th}$), you have formed all combinations with any of the five letters taken or not, which you can do in $2\cdot2\cdot2\cdot2\cdot2$ ways. (Equivalently, all five bits numbers from $00000$ to $11111$, which is exactly $2^5$.) For perfect rigor, one should show that there are no repetitions nor omissions.


6

As lulu said, this can be proved by inclusion-exclusion theorem. We have \begin{equation} |\cup_{i=1}^k A_i|=\sum_{j=1}^k(-1)^{j+1}(\sum_{1\leq i_1<...<i_j\leq k}|A_{i_1}\cap...\cap A_{i_j}|). \end{equation} Now let $A_i$ be the subset of function from $[k]$ to $[k]$ such that $i$ is not an image. Then lhs of the equation is equal to $k^k-k!$, which ...


5

In this answer, it is shown that for any $l$, $$ \begin{align} \sum_{r=0}^n\binom{n}{r}(-1)^r(l-r)^n &=\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}(r-l)^n\\ &=n! \end{align} $$ The given identity is this identity using $l=0$.


5

$(n+1)!=2\cdot 3\cdot \dots\cdot(n+1)$ here a product of $n$ numbers all are at least 2 so the result follows...


4

HINT: $$\frac{(n+1)!}{2^n}=\frac{2}{2}\frac{3}{2}\frac{4}{2}\cdots \frac{n-1}{2}\frac{n}{2}\frac{n+1}{2}$$


2

Hint: You can construct such sequence in the following consecutive steps: Step 1: construct a sequence of $3$ Heads and $4$ Tails; Step 2: Now put a block of extra $5$ Tails in the sequence.


2

It's not always an integer. Consider the case $N=p=2$. You obtain $3!/2!1!2!=3/2$.


2

Evidently computations show that there is a primorial between $n!$ and $(n+2)!$ Furthermore, those $n$ such that there is not a primorial between $n!$ and $(n+1)!$ are quite regularly spaced, either four or five apart. This suggests a relatively short proof should be available, based mostly on the idea that, when $n!$ and $p\#$ are of similar size, that $p ...


2

Here is a variant based upon the coefficient of operator $[z^k]$ used to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} [z^j](1+z)^k=\binom{k}{j}\qquad\qquad\text{or}\qquad\qquad k![z^k]e^{jz}=j^k \end{align*} We obtain \begin{align*} \sum_{j=1}^k(-1)^{k-j}j^k\binom{k}{j} ...


2

Use Stirling's approximation $$ n!\sim \sqrt{2\pi n} (n/e)^n $$ for large $n$ to write $$ \frac{(3n)!(1/27)^n}{(n!)^3}\sim\frac{\sqrt{2\pi 3 n} (3n/e)^{3n} (1/27)^n }{(\sqrt{2\pi n} (n/e)^n)^3 }=\frac{\sqrt{3}}{2n\pi}\to 0 $$ for large $n$.


2

This is a partial solution... Use all the divisibility rules. But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$. $34!$ is divisible by $9$, so: $$4+a+c+d=0\pmod 9.$$ It's divisible by $7$, so: $$000-000+5a0-643+609-618+847-140+604-cd9+799+327-952+2=0\pmod 7,$$ so ...


2

$34!=295232799039604140847618609643520000000$. Therefore $a=2, b=0, c=0, d=3$.


1

As WolframAlpha states: $$\frac{\Gamma(x+n)}{\Gamma(x)}=x(x+1)(x+2)(x+3)\dots(x+n-1)=\sum_{k=0}^n(-1)^{n-k}s(n,k)x^k$$ where $s(n,k)$ is a Stirling number of the first kind. Rewritable as $$x(x+1)(x+2)(x+3)\dots(x+n-1)=(x+n-1)(x+n-2)(x+n-3)\dots x=\frac{\Gamma(x+n)}{\Gamma(x+n-n)}$$ Use substitution $x+n=u$ $$=\frac{\Gamma(u)}{\Gamma(u-n)}$$


1

Check out tetration: https://en.wikipedia.org/wiki/Tetration#Iterated_powers_vs._iterated_exponentials The actual function you are describing is on that page as well, and is called a nested exponential. Tetration is a special case of a nested exponential. For the record, pentation is the next operation up (it's tetration nested like tetration is ...


1

Hint: \begin{align} \sum_{k=0}^{2n+2}\frac{(-1)^k}{k!(2n+2-k)!} =\frac{1}{(2n+2)!}\sum_{k=0}^{2n+2}(-1)^k{2n+2\choose k} =\frac{1}{(2n+2)!}(1-1)^{2n+2}=0. \end{align}


1

Both $a_n$ and $b_{n}$ are given by convolutions: $$\begin{array}{cclcl} a_n &=& \displaystyle\sum_{a+b=n}\frac{1}{(2a+1)!}\cdot\frac{1}{(2b+1)!} &=& \displaystyle[x^n]\left(\sum_{c\geq 0}\frac{x^c}{(2c+1)!}\right)^2 \\ b_n &=& \displaystyle\sum_{a+b=n}\frac{1}{(2a)!}\cdot\frac{1}{(2b)!} &=& ...


1

Perhaps consider this. Say you multiply $(n-1)!$ by $n$ to get $n!$. Then, if you continue multiplying by $n+1$ and $n+2$, etc, you won't multiply by another multiple of $n$ until you get to $2n$. $2n-n$ is n, so the $n^1$ factor will be show up $n$ times before being duplicated.


1

Hint $1:$ For the first two, observe that the product of $n$ consecutive numbers is divisible by $n!$ Hint $2:$ $\frac{(a \cdot b)!}{a! \cdot b!}$ is a multiple of $\binom{a+b}{a}=\binom{a+b}{b}$, both of which clearly are natural numbers, as $ab \geq {a+b}$, for all $a,b \geq 2$ Hint $3:$ $a=b=2$ acts as a simple contradiction as pointed in comments by ...


1

You've miscounted. I see that you are essentially taking the number of solutions to $x_1+x_2+\cdots+x_p=N$ and dividing by $p!$. But if $x_i=x_j$ then not all permutations give a different solution. The number of solutions to $x_1+x_2+\cdots +x_p=N$ with $x_1\leq x_2\leq \cdots \leq x_p$ is a much more complicated value. These are called "integer ...


1

The number of ways to distribute $N$ indistinguishable objects amongst $p$ distinguishable containers is the number of weak compositions of $N$ into $p$ parts, which is $$\binom{N+p-1}{p-1}=\frac{(N+p-1)!}{N!(p-1)!}\;.$$ However, you can’t simply divide by $p!$ to get the number of distributions into $p$ indistinguishable containers: the divisor for a ...


1

To promote my comment to an answer: You've gotten to the point of comparing two quantities with equal denominator, so it suffices to show that $$(n^2 + 3n + 2)n! = (n + 2)!$$ But this is equivalent to seeing that $$n^2 + 3n + 2 = (n + 2)(n + 1)$$ which is clearly true.


1

Write down the LHS as $${n \choose k} + {n \choose k+1} + {n \choose k+1} + {n \choose k+2}.$$ Now expand the first two terms to get ${n+1\choose k+1}$. The summation of the second two terms evaluates to ${n+1\choose k+2}$. The addition of the resulted two terms is the RHS.


1

If we line up the 3 heads, this creates 4 gaps into which we can put the tails. 1) There are 4 ways to select the gap which will contain at least 5 tails, so put 5 tails in this gap. 2) There are $\dbinom{7}{3}$ ways to distribute the remaining 4 tails, since there are 4 tails and 3 dividers (the heads). This gives $4\dbinom{7}{3}$ possibilities.


1

We define that $0! = 1$, and for $n > 0$, that $n! = n \cdot (n-1)!$. This means $1! = 1 \cdot 0! = 1$. For a combinatorial interpretation, $1!$ is the number of ways to create an ordered list containing a single object, so $1!=1$.


1

$$\lim_{n\to \infty} \frac{(3n)! (1/27)^n}{(n!)^3} = \lim_{n\to \infty} \frac{(3n)!}{(3^n n!)^3}$$ Let's solve it with a ram. It's obvious that $1 < (3n)! < 3^{3n} * n^n $ and $n^n < n! n!$ so we can use the Squeeze theorem: $$\lim_{n\to \infty} \frac{1}{(3^n n!)^3} = 0; \lim_{n\to \infty} \frac{3^{3n} * n^n}{(3^{n} n!)^3} = \lim_{n\to \infty} ...



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