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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 ...

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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}$$THEOREM: Compute the following N = 5q_1 + R_1 q_1 = 5q_2 + R_2 q_2 = 5q_3 + R_3 ... q_{n-1} = 5q_n + R_n where 0 \le ... 6 Let's consider the meaning of \sum_{k=1}^n \left(\begin{array}{c} n \\ k \end{array} \right). You are interested in the number of ways to construct a non-empty subset from the original set of n elements. Consider the set with n elements, for each element, we have two options, to include it in one of the subset or not. If we allow our subset to be ... 6 Taking all the rounds together (including the 0^{th}), you have formed all combinations with any of the five letters taken or not, which you can do in 2\cdot2\cdot2\cdot2\cdot2 ways. (Equivalently, all five bits numbers from 00000 to 11111, which is exactly 2^5.) For perfect rigor, one should show that there are no repetitions nor omissions. 6 As lulu said, this can be proved by inclusion-exclusion theorem. We have $$|\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}|).$$ Now let A_i be the subset of function from [k] to [k] such that i is not an image. Then lhs of the equation is equal to k^k-k!, which ... 5 In this answer, it is shown that for any l,$$ \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)!} &=& ...

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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.

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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 ...

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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 ...

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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 ...

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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.

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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.

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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.

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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 $1!=1$.

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$$\lim_{n\to \infty} \frac{(3n)! (1/27)^n}{(n!)^3} = \lim_{n\to \infty} \frac{(3n)!}{(3^n n!)^3}$$ 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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