§7不作解答，以下内容如有错误，欢迎在评论区或私信我指出。
\(Page.44\\1.\ 100!\approx \sqrt{200\pi}\,100^{100}e^{-100}=\sqrt{2\pi}\,10^{201}\,10^{\lg e^{-100}}=\sqrt{2\pi}\,10^{157}\approx 2.5*10^{157}\)?
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\(2.\ (1)\pi =\bigl(\begin{smallmatrix}1&2&3&4&5&6&7&8\\2&3&4&5&1&7&6&8\end{smallmatrix}\bigr)=(1\ 2\ 3\ 4\ 5)(6\ 7)(8)\quad LCM(5,2)=10\\\quad (2)\pi =\bigl(\begin{smallmatrix}1&2&3&4&5&6&7&8\\3&6&8&2&1&4&5&7\end{smallmatrix}\bigr)=(1\ 3\ 8\ 7\ 5)(2\ 6\ 4)\quad LCM(5,3)=15\)
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\(3.\ m‘=n-\sum_{k=1}^{m}l_{k}\Leftrightarrow n-m‘=\sum_{k=1}^{m}l_{k}\\\quad \varepsilon _{\pi}=(-1)^{\sum_{k=1}^{m}(l_{k}-1)}=(-1)^{\sum_{k=1}^{m}l_{k}-m}=(-1)^{n-m‘-m}=(-1)^{d(\pi)}\)
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\(4.\ \pi=\bigl(\begin{smallmatrix}1&2&3&···&n-1&n\\n&n-1&n-2&···&2&1\end{smallmatrix}\bigr)=\left\{\begin{array}{l}(1\ n)(2\ n-1)···(\frac{n-1}{2}\ \frac{n+3}{2})(\frac{n+1}{2})\quad n为奇数\\(1\ n)(2\ n-1)···(\frac{n}{2}\ \frac{n+2}{2})\quad n为偶数\end{array}\right.\\\quad \varepsilon _{\pi}=\left\{\begin{array}{l}(-1)^{\frac{n-1}{2}}\quad n为奇数\\(-1)^{\frac{n}{2}}\quad n为偶数\end{array}\right.\)
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\(5.\ 设k_{n}为\sigma (n)逆序的总数,则k=k_{1}+k_{2}+···+k_{n}\\\quad 即对于\sigma (i),存在k_{i}个\sigma (j),s.t.\ j<i且\sigma (j)>\sigma (i),记\varsigma _{i}=(\sigma (i) \ \sigma (j_{1}))(\sigma (i) \ \sigma (j_{2}))···(\sigma (i) \ \sigma (j_{k_{i}}))\\\quad 如此便可从n开始,依次构造\varsigma _{n},\varsigma _{n-1},···,\varsigma _{1},s.t.\ \varsigma _{1}\varsigma _{2}···\varsigma _{n}\sigma =e\\\quad 展开为k个对换\ \tau _{i},s.t.\ \tau _{k}\tau _{k-1}···\tau _{1}\sigma =e\ 成立\)
原文：https://www.cnblogs.com/cihua/p/Dsxyl_1_1_8.html