How do you simplify $\frac{(n+2)!}{n!}?$

Painevg
2021-12-19
Answered

How do you simplify $\frac{(n+2)!}{n!}?$

You can still ask an expert for help

Juan Spiller

Answered 2021-12-20
Author has **38** answers

Explanation:

We can rewrite the numerator as:

$\frac{(n+2)\cdot (n+2-1)\cdot (n+2-2)!}{\left(n\right)!}$

$=\frac{(n+2)\cdot (n+1)\cdot \left(n\right)!}{\left(n\right)!}$

We can cancel$\left(n\right)!$ and $\left(n\right)!$ out:

$=\frac{(n+2)\cdot (n+1)\cdot 1}{1}$

$=(n+2)\cdot (n+1)$

$={n}^{2}+3n+2$

Thus, solved

We can rewrite the numerator as:

We can cancel

Thus, solved

esfloravaou

Answered 2021-12-21
Author has **43** answers

I wanted to write here with the same question, help me solve

RizerMix

Answered 2021-12-29
Author has **438** answers

Thanks for the detailed answer.

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I have been reading a paper: Wang X, Golbandi N, Bendersky M, Metzler D, Najork M. Position bias estimation for unbiased learning to rank in personal search.

I am wondering how they derive the M-step in an EM algorithm that has been applied to the problem in their paper.

This is the data log-likelihood function:

$\mathrm{log}P(\mathcal{L})=\sum _{(c,q,d,k)\in \mathcal{L}}c\mathrm{log}{\theta}_{k}{\gamma}_{q,d}+(1-c)\mathrm{log}(1-{\theta}_{k}{\gamma}_{q,d}).$

where observed data are (c:click, q:query, d:document, k:result position) rows, and ${\theta}_{k}$ is the probability a search result at rank k is examined by the user (assumed to be independent of query-document pair), and ${\gamma}_{q,d}$ is the probability a search result, i.e., a query-document pair is actually relevant (assumed to be independent of result position).

Here are some hidden variable probability expressions (E means examination and R means relevant, they assume that a click event implies a search result is both examined and relevant):

$\begin{array}{l}P(E=1,R=1\mid C=1,q,d,k)=1\\ P(E=1,R=0\mid C=0,q,d,k)=\frac{{\theta}_{k}^{(t)}(1-{\gamma}_{q,d}^{(t)})}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\\ P(E=0,R=1\mid C=0,q,d,k)=\frac{(1-{\theta}_{k}^{(t)}){\gamma}_{q,d}^{(t)}}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\\ P(E=0,R=0\mid C=0,q,d,k)=\frac{(1-{\theta}_{k}^{(t)})(1-{\gamma}_{q,d}^{(t)})}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\end{array}$

The paper has derived the M-step update formulas:

$\begin{array}{l}{\theta}_{k}^{(t+1)}=\frac{\sum _{c,q,d,{k}^{\mathrm{\prime}}}{\mathbb{I}}_{{k}^{\mathrm{\prime}}=k}\cdot (c+(1-c)P(E=1\mid c,q,d,k))}{\sum _{c,q,d,{k}^{\mathrm{\prime}}}{\mathbb{I}}_{{k}^{\mathrm{\prime}}=k}}\\ {\gamma}_{q,d}^{(t+1)}=\frac{\sum _{c,{q}^{\mathrm{\prime}},{d}^{\mathrm{\prime}},k}{\mathbb{I}}_{{q}^{\mathrm{\prime}}=q,{d}^{\mathrm{\prime}}=d}\cdot (c+(1-c)P(R=1\mid c,q,d,k))}{\sum _{c,{q}^{\mathrm{\prime}},{d}^{\mathrm{\prime}},k,}{\mathbb{I}}_{{q}^{\mathrm{\prime}}=q,{d}^{\mathrm{\prime}}=d}}\end{array}$

however, how to get these two formulas?

Note: M-step is usually derived from this formula:

$Q(\mathit{\theta}\mid {\mathit{\theta}}^{(t)})={\mathrm{E}}_{\mathbf{Z}\mid \mathbf{X},{\mathit{\theta}}^{(t)}}[\mathrm{log}L(\mathit{\theta};\mathbf{X},\mathbf{Z})]\phantom{\rule{thinmathspace}{0ex}}$

$\mathit{\theta}}^{(t+1)}=\underset{\mathit{\theta}}{\mathrm{a}\mathrm{r}\mathrm{g}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{a}\mathrm{x}}\text{}Q(\mathit{\theta}\mid {\mathit{\theta}}^{(t)})\phantom{\rule{thinmathspace}{0ex}$

where θ are parameters (i.e., ${\theta}_{k}$ and ${\gamma}_{q,d}$ in this case), X are data (c,q,d,k in this case), and Z are hidden variables (E and R?).

I am wondering how they derive the M-step in an EM algorithm that has been applied to the problem in their paper.

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where observed data are (c:click, q:query, d:document, k:result position) rows, and ${\theta}_{k}$ is the probability a search result at rank k is examined by the user (assumed to be independent of query-document pair), and ${\gamma}_{q,d}$ is the probability a search result, i.e., a query-document pair is actually relevant (assumed to be independent of result position).

Here are some hidden variable probability expressions (E means examination and R means relevant, they assume that a click event implies a search result is both examined and relevant):

$\begin{array}{l}P(E=1,R=1\mid C=1,q,d,k)=1\\ P(E=1,R=0\mid C=0,q,d,k)=\frac{{\theta}_{k}^{(t)}(1-{\gamma}_{q,d}^{(t)})}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\\ P(E=0,R=1\mid C=0,q,d,k)=\frac{(1-{\theta}_{k}^{(t)}){\gamma}_{q,d}^{(t)}}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\\ P(E=0,R=0\mid C=0,q,d,k)=\frac{(1-{\theta}_{k}^{(t)})(1-{\gamma}_{q,d}^{(t)})}{1-{\theta}_{k}^{(t)}{\gamma}_{q,d}^{(t)}}\end{array}$

The paper has derived the M-step update formulas:

$\begin{array}{l}{\theta}_{k}^{(t+1)}=\frac{\sum _{c,q,d,{k}^{\mathrm{\prime}}}{\mathbb{I}}_{{k}^{\mathrm{\prime}}=k}\cdot (c+(1-c)P(E=1\mid c,q,d,k))}{\sum _{c,q,d,{k}^{\mathrm{\prime}}}{\mathbb{I}}_{{k}^{\mathrm{\prime}}=k}}\\ {\gamma}_{q,d}^{(t+1)}=\frac{\sum _{c,{q}^{\mathrm{\prime}},{d}^{\mathrm{\prime}},k}{\mathbb{I}}_{{q}^{\mathrm{\prime}}=q,{d}^{\mathrm{\prime}}=d}\cdot (c+(1-c)P(R=1\mid c,q,d,k))}{\sum _{c,{q}^{\mathrm{\prime}},{d}^{\mathrm{\prime}},k,}{\mathbb{I}}_{{q}^{\mathrm{\prime}}=q,{d}^{\mathrm{\prime}}=d}}\end{array}$

however, how to get these two formulas?

Note: M-step is usually derived from this formula:

$Q(\mathit{\theta}\mid {\mathit{\theta}}^{(t)})={\mathrm{E}}_{\mathbf{Z}\mid \mathbf{X},{\mathit{\theta}}^{(t)}}[\mathrm{log}L(\mathit{\theta};\mathbf{X},\mathbf{Z})]\phantom{\rule{thinmathspace}{0ex}}$

$\mathit{\theta}}^{(t+1)}=\underset{\mathit{\theta}}{\mathrm{a}\mathrm{r}\mathrm{g}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{a}\mathrm{x}}\text{}Q(\mathit{\theta}\mid {\mathit{\theta}}^{(t)})\phantom{\rule{thinmathspace}{0ex}$

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Better understanding of the pre sheaf kernel of $\phi $

I'm currently reading chapter II of Hartshorne's algebraic geometry and am comfortable with the definition of a sheaf. I am new to the idea of the presheaf kernel of $\phi $ however.

Let $\phi :\mathcal{F}\to \mathcal{G}$ be a morphism of pre sheaves of abelian groups on a topological space X.

Am I correct in defining the kernel pre sheaf on φ to be the contra variant functor that consists of the data

For every open set $U\subset X$, we have an abelian group $\mathrm{\Gamma}(\mathrm{ker}\phi (U),\mathcal{F})$

For every inclusion $V\subset U$ open sets in X, we have a morphism ${\rho}_{UV}:\mathrm{\Gamma}(\mathrm{ker}\phi (U),\mathcal{F})\to \mathrm{\Gamma}(\mathrm{ker}\phi (V),\mathcal{F})$ that satisfies the usual three conditions (empty set, identity map and composition condition)?

Moreover,

Hartshorne remarks that while the pre sheaf kernel is a sheaf, in general, the pre sheaf cockernel and image are not sheaves. Is there any intuition for why this would be so?

I'm currently reading chapter II of Hartshorne's algebraic geometry and am comfortable with the definition of a sheaf. I am new to the idea of the presheaf kernel of $\phi $ however.

Let $\phi :\mathcal{F}\to \mathcal{G}$ be a morphism of pre sheaves of abelian groups on a topological space X.

Am I correct in defining the kernel pre sheaf on φ to be the contra variant functor that consists of the data

For every open set $U\subset X$, we have an abelian group $\mathrm{\Gamma}(\mathrm{ker}\phi (U),\mathcal{F})$

For every inclusion $V\subset U$ open sets in X, we have a morphism ${\rho}_{UV}:\mathrm{\Gamma}(\mathrm{ker}\phi (U),\mathcal{F})\to \mathrm{\Gamma}(\mathrm{ker}\phi (V),\mathcal{F})$ that satisfies the usual three conditions (empty set, identity map and composition condition)?

Moreover,

Hartshorne remarks that while the pre sheaf kernel is a sheaf, in general, the pre sheaf cockernel and image are not sheaves. Is there any intuition for why this would be so?