1
vote
2answers
61 views

Finding $n$ in $n=ReverseFactorial(x)$ where $x$ is known.

My software application receives a series of very large integers (hundreds of decimal digits). So far, I have been using string/textual representation of decimal digits for very simple manipulation ...
0
votes
1answer
44 views

help on manipulating this algebraic expression

So I have something like: $\frac {k!}{(k-3)!3!}$ I'm going to add $\frac 12k(k-1)$ to this, and I want to obtain $\frac {(k+1)!}{(k-2)!3!}$ as the result. I'm having trouble with this since I need ...
2
votes
1answer
154 views

How do you solve an inequality with the factorial of a variable?

How do you solve an inequality with the factorial of a variable? Example: Determine the interval of $n \in \Bbb N$ for which the following inequality holds: $$n! \leq 157788 \cdot 10^{10} $$ Can ...
1
vote
2answers
62 views

How $\alpha(\alpha+1)\ldots(\alpha+k-1)=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}$?

Probability function of Negative Binomial Distribution, $NB(\alpha,p)$, is $$P(X=k)=\binom{\alpha+k-1}{k}(1-p)^{\alpha}p^k,\quad \alpha>0$$ Probability generating function of Negative Binomial ...
0
votes
3answers
37 views

Simlifying [(k+1)! - 1] + (k+1)((k + 1)!)

I'm afraid I've gotten a bit rusty on Math since I was last in university. I was looking at a problem in my text book that simplified ...
1
vote
1answer
30 views

Simplify the following problem

How $$\frac{1}{k}\sum_{m=r}^{k-1}\frac{m!}{(m-r)!}=\frac{(k-1)(k-2)\ldots(k-r)}{r+1};\quad r=1,2,\ldots$$ I have thought in two ways: ...
0
votes
1answer
18 views

Simple Calculation

How $$\sum_{m=r}^{k-1}m(m-1)\ldots(m-r+1)=\sum_{m=r}^{k-1}\frac{m!}{(m-r)!};\quad r=1,2,\ldots$$ ? i have ...
1
vote
3answers
78 views

Show that $\frac{1}{r!}-\frac{1}{(r+1)!}\equiv\frac{r}{(r+1)!}$.

Show that $\frac{1}{r!}-\frac{1}{(r+1)!}\equiv\frac{r}{(r+1)!}$. I get $$\frac{1}{r!}-\frac{1}{(r+1)!}=\frac{(r+1)-r!}{r!(r+1)!}$$ and in the numerator since $$(r+1)!-r!=r$$ so ...
5
votes
3answers
233 views

Dividing factorials is always integer

Is there a simple way to show that $$n!\over r!(n-r)!$$ is always an integer?
4
votes
4answers
5k views

What are the rules for factorial manipulation?

I know that $$(k+1)! - 1 + (k+1)(k+1)! = (k+2)! - 1$$ thanks to wolframalpha, but I don't understand the steps for simplification, and I can't seem to find any rules about factorial manipulations ...
2
votes
1answer
82 views

Proving with factorials

Let x and y be the postive integers. Show that : $\displaystyle\frac{(x + y)!}{ (x + y)^{(x + y)}} < \frac{x! y!}{ (x^x + y^y)}$ Are there any identities we can use to easily prove this?
3
votes
1answer
90 views

Factorial Identity - True or False?

Let $x$ and $y$ be positive integers. Then, is \begin{align} \frac{x^{xy}}{(xy)!} = \sum_{k_1+...+k_x = xy} \frac{1}{(k_1)!...(k_x)!} \end{align} true, where $k_1$, ..., $k_x$ are all positive ...
5
votes
4answers
777 views

Simplify a factorial

I have the problem to evaluate the following: $$ (2n)!\over 2^n(n!) $$ Does this reduce to anything in particular? I stuck it into a computer and it's ...
13
votes
4answers
2k views

Highest power of a prime $p$ dividing $N!$

How does one find the highest power of a prime $p$ that divides $N!$ and other related products? Related question: How many zeros are there at the end of $N!$? This is being done to reduce ...
2
votes
1answer
66 views

$n^s=(n)_s+f(s)$, what is $f(s)$?

In the following equation, $$n^s=(n)_s+f(s)$$ What is general form for $f(s)$? Understand that, $$(n)_s=n(n-1)(n-2)\cdots(n-[s-1])=\text{ The Falling Factorial }$$ I have experimented with this ...
1
vote
3answers
168 views

What makes $0!$ equal to 1? [duplicate]

Possible Duplicate: Prove $0! = 1$ from first principles I don't understand how it's equal to 1. Also, I found that $(-x)!$ is equal to complex $\infty$. How is this so?
3
votes
2answers
176 views

Simple factorial

I'm re-learning factorials, and I encountered this exercice, but the solution had a diferent result than I got, and no matter how much I try to search, I can't find an explanation to the last step of ...
3
votes
2answers
219 views

Help with factorials, permutations, and combinations

I am an eighth grader in need of some help. I was assigned a school project on making a java application that computes the total permutations of to given numbers where nPr and later on nCr. I ...
-1
votes
1answer
170 views

How to answer to this algebra problem?

Thanks for your attention to this question, here is the problem: Compute the number of positive integers $x$ less than or equal to $1000$ that satisfy the following condition: $$x! \text{ is ...
0
votes
2answers
578 views

Simplifying this factorial expression

I know that $\frac{(m-1)!}{(m-n)!(n-1)!} + \frac{(m-1)!}{(m-n-1)!(n)!} = \frac{m!}{(n)!(m-n)!}$, but I am not sure on the intermediate steps. The only solution I am seeing involves finding a common ...
1
vote
1answer
105 views

Combination Problem with a Variable

I have the following problem: $_xC_6$ = $_xC_4$ I expand both sides to: $$\frac{x!}{[(x-6)!]6!} = \frac{x!}{[(x-4)]!4!}$$ Next I multiply both sides by the denominator of the right-hand ...
4
votes
3answers
84 views

Reducing $\prod \limits_{0 \le j \ne i \le n} \frac{n+1-j}{i-j}$ to $\frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)}$

How could we show that: $$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)} .$$ The module suggest we could reduce it by simply writing ...
11
votes
2answers
415 views

Given $n! = c$, how to find $n$?

I'm dealing with a time-complexity problem in which I know the running time of an algorithm: $$t = 1000 \mathrm{ms} .$$ I also know that the algorithm is upper bounded by $O(n!)$. I want to know ...
2
votes
1answer
340 views

Use of algebra and factorials for a question related to proof by induction

$$ \begin{align*} &= (n+1)! − 1 + ( (n+1) · (n+1)! )\\ &= (n+1)! (1+n+1) − 1\\ &= (n+1)! (n+2) − 1\\ &= (n+2)! − 1\\ \end{align*} $$ I'm confused at how the first ...
7
votes
13answers
2k views

Prove $0! = 1$ from first principles

How can I prove from first principles that $0!$ is equal to $1$?
9
votes
2answers
294 views

Number of zeros not possible in $n!$ [duplicate]

Possible Duplicate: How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? The number of zeros which are not possible at the end of the $n!$ is: ...