# Questions tagged [factorial]

Questions on the factorial function, $n!=n\cdot(n-1)\cdot...\cdot1$. Consider using the tag (gamma-function) if dealing with noninteger arguments.

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### Factors of ${{x+n} \choose {n}}$

Let $v_p(n)$ be the highest power of $p$ that divides $n$. It seems to me that for any prime $p$, $v_p\left({{x+n} \choose {n}}\right) \le \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$ Am I ...
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### Finding $\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$

I'm looking for a way to find this limit: $\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$ I think I have found that it diverges, by plugging numbers into the formula and "sandwich" the result. However I ...
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### An inequality with exponents, factorials and nth roots!

Problem: Prove for natural numbers $n > 2$, $$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \frac{n!}{(n+1)^n}.$$ I am unable to do this one. ...
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### Showing $(n+1)^n<e^nn!$ by induction

Show $(n+1)^n<e^nn!$ I know why that would be the case using general knowledge and a bit of substitution but am clueless on how to prove it.
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### Inductive Proof that $k!<k^k$, for $k\geq 2$.

Call $P(k): k!<k^k$, for $k\geq 2$ Test it out with 2, and it's true ($2<4$). Assume that $P(k)$ is true for some $k\geq2$. Then show that $P(k+1)$ is true. $P(k+1): (k+1)!<(k+1)^{k+1}$ ...
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### How to prove the sum of combination is equal to $2^n - 1$

One of the algorithm I learnt involve these steps: $1$. define a set $S$ of $n$ elements $2$. form a subset $S'$ of $k$ choice from $n$ elements of the set $S$ ($k$ starts with $1$), which is ...
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### Dividing factorials is always integer

Is there a simple way to show that $$n!\over r!(n-r)!$$ is always an integer?
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### $n$ choose $k$ where $n$ is less than $k$

I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, ...
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### Factorial sum estimate $\sum_{n=m+1}^\infty \frac{1}{n!} \le \frac{1}{m\cdot m!}$

Prove that: $$\displaystyle \sum_{n=m+1}^\infty \dfrac{1}{n!} \le \dfrac{1}{m\cdot m!}$$ I have tried induction on $m$ but it does not work very well. Any suggestion?
### Prove $1(1!)+\dots+n(n!) = (n+1)!-1$ using induction
So I'm trying to prove this statement (through induction): $$1(1!)+2(2!)+\dots +n(n!)=(n+1)!-1$$ But I'm confused with the inductive step here: $$(n+1)!-1+[(n+1)(n+1)!] = (n+2)!-1$$ What do I do ...