我们如何才能正确地在分数内进行选择格式化？

Question 1

就我个人而言，我将使用枚举来构造itemize。然后使用align*环境。我将使用\times而不是*。我会在问题和其解决方案之间留出一点空间。

\documentclass{article} 
\usepackage{amsmath}
\usepackage{showframe}
\begin{document}

A class has $2n$ students who must be split up into pairs. We consider two sets of pairs $S$ and $T$ different if at least one pair in the set $S$ isn't a pair in the set $T$. A pair is unordered, so we consider the pair $(1, 2)$ and $(2, 1)$ to be the same pair. 

How many different sets of pairs can the class be split up into, in terms of~$n$? 

\begin{itemize}
\item For example, with $n = 2$, the answer is $3$. The sets are $\{(1, 2), (3, 4)\}$, $\{(1, 3), (2, 4)\}$ and $\{(1, 4), (2, 3)\}$.

\item For the first pair choose $\dbinom{2n}{2}$.

\item For the second pair choose $\dbinom{2n-2}{2}$.

\item For the third pair choose $\dbinom{2n-4}{2}$.

\item The total number  of ways to choose is \[\dbinom{2n}{2}\times \dbinom{2n-2}{2}\times \dbinom{2n-4}{2}\times \dots \times \dbinom{2}{2}.\]
\end{itemize}

Therefore, the total number of different sets of pairs which the class may be split up into, in terms of $n$, is as follows:
\begin{align*}
\frac{ \dbinom{2n}{2} \times \dbinom{2n-2}{2} \times  \dots \times\dbinom{2}{2} }{n!} &= \frac{2n!}{(2n-2)!2!} \times \frac{(2n-2)!}{(2n-4)!2!} \times\dots\times \frac{2!}{0!2!}\times \frac{1}{n!}\\
&= \frac{2n!}{2!^nn!}.
\end{align*}
\end{document}

Answer

就我个人而言，我将使用枚举来构造itemize。然后使用align*环境。我将使用\times而不是*。我会在问题和其解决方案之间留出一点空间。

\documentclass{article} 
\usepackage{amsmath}
\usepackage{showframe}
\begin{document}

A class has $2n$ students who must be split up into pairs. We consider two sets of pairs $S$ and $T$ different if at least one pair in the set $S$ isn't a pair in the set $T$. A pair is unordered, so we consider the pair $(1, 2)$ and $(2, 1)$ to be the same pair. 

How many different sets of pairs can the class be split up into, in terms of~$n$? 

\begin{itemize}
\item For example, with $n = 2$, the answer is $3$. The sets are $\{(1, 2), (3, 4)\}$, $\{(1, 3), (2, 4)\}$ and $\{(1, 4), (2, 3)\}$.

\item For the first pair choose $\dbinom{2n}{2}$.

\item For the second pair choose $\dbinom{2n-2}{2}$.

\item For the third pair choose $\dbinom{2n-4}{2}$.

\item The total number  of ways to choose is \[\dbinom{2n}{2}\times \dbinom{2n-2}{2}\times \dbinom{2n-4}{2}\times \dots \times \dbinom{2}{2}.\]
\end{itemize}

Therefore, the total number of different sets of pairs which the class may be split up into, in terms of $n$, is as follows:
\begin{align*}
\frac{ \dbinom{2n}{2} \times \dbinom{2n-2}{2} \times  \dots \times\dbinom{2}{2} }{n!} &= \frac{2n!}{(2n-2)!2!} \times \frac{(2n-2)!}{(2n-4)!2!} \times\dots\times \frac{2!}{0!2!}\times \frac{1}{n!}\\
&= \frac{2n!}{2!^nn!}.
\end{align*}
\end{document}

Question 2

您的输入中存在一些不一致之处。例如，表示数学变量或常量的符号应该总是排版为数学。

因此使用A class has $2n$ students; 类似地，

the pair $(1, 2)$ and $(2, 1)$ to be the same pair

即使结果与使用获得的结果非常相似$，但省略它可能会导致问题，最糟糕的情况是(2, 1)可以跨行拆分，而 $(2, 1)$ 不能。

在算法的演示中使用它\dbinom只是浪费空间；仅在特殊情况下使用它，例如显示大分数。

别忘了关系：你不希望in the set S在之前拆分S，所以最好输入

in the set~$S$

（带有$，因为它是一个数学变量）。

在正式数学中，星号从不用于表示乘法。对于你的情况，没有哪个符号是最好的。

如果阶乘前面有一个乘法因子，则最好用一个细空格将其与阶乘分开\,。

这是我的建议。

\documentclass{article}
\usepackage{amsmath}

\begin{document}

A class has $2n$ students who must be split up into pairs. We consider 
two sets of pairs $S$ and $T$ different if at least one pair in the set~$S$ 
isn't a pair in the set~$T$. A pair is unordered, so we consider the pair
$(1, 2)$ and $(2, 1)$ to be the same pair. How many different sets of pairs 
can the class be split up into, in terms of~$n$? (For example, with $n = 2$, 
the answer is~$3$. Sets are $\{(1, 2), (3, 4)\}$, $\{(1, 3), (2, 4)\}$ and 
$\{(1, 4), (2, 3)\}$.)
\begin{itemize}
\item For the first pair choose $\binom{2n}{2}$,

\item for the second pair choose $\binom{2n-2}{2}$,

\item for the third pair choose $\binom{2n-4}{2}$,

\item and so on.
\end{itemize}

The total number  of ways to choose is
\[
\dbinom{2n}{2}\dbinom{2n-2}{2}\dbinom{2n-4}{2}\dotsm\dbinom{2}{2}.
\]
Therefore, the total number of different sets of pairs which the class may 
be split up into, in terms of~$n$, is as follows:
\begin{align*}
\frac{\dbinom{2n}{2}\dbinom{2n-2}{2}\dbinom{2n-4}{2}\dotsm\dbinom{2}{2}}{n!}
&=
\frac{(2n)!}{(2n-2)!\,2!}\frac{(2n-2)!}{(2n-4)!\,2!}\dotsm\frac{2!}{0!\,2!}\frac{1}{n!}
\\
&=\frac{(2n)!}{(2!)^n\,n!}
\end{align*}

\end{document}

通过一些其他修复：例如2n!与不一样，(2n)!而您想要后者。

Answer