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+%%=====================================================================================
+%%
+%%       Filename:  cours.tex
+%%
+%%    Description:  
+%%
+%%        Version:  1.0
+%%        Created:  03/06/2024
+%%       Revision:  none
+%%
+%%         Author:  YOUR NAME (), 
+%%   Organization:  
+%%      Copyright:  Copyright (c) 2024, YOUR NAME
+%%
+%%          Notes:  
+%%
+%%=====================================================================================
+\documentclass[a4paper, titlepage]{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{textcomp}
+\usepackage[french]{babel}
+\usepackage{amsmath, amssymb}
+\usepackage{amsthm}
+\usepackage[svgnames]{xcolor}
+\usepackage{thmtools}
+\usepackage{lipsum}
+\usepackage{framed}
+\usepackage{parskip}
+\usepackage{titlesec}
+\usepackage{newtxtext}
+
+% \renewcommand{\familydefault}{\sfdefault}
+
+% figure support
+\usepackage{import}
+\usepackage{xifthen}
+\pdfminorversion=7
+\usepackage{pdfpages}
+\usepackage{transparent}
+\newcommand{\incfig}[1]{%
+	\def\svgwidth{\columnwidth}
+	\import{./figures/}{#1.pdf_tex}
+}
+
+\pdfsuppresswarningpagegroup=1
+
+\colorlet{defn-color}{DarkBlue}
+\colorlet{props-color}{Blue}
+\colorlet{warn-color}{Red}
+\colorlet{exemple-color}{Green}
+\colorlet{corol-color}{Orange}
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+  \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
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+  \def\FrameCommand{{\color{corol-color}\vrule width 3pt} \hspace{10pt}}%
+  \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+
+\def \freespace {1em}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
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+ headpunct=,%
+% bodyfont=\sffamily,%
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+]{better-defn}
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+]{better-exemple}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
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+ preheadhook=\vspace{\freespace}\begin{corol-leftbar},%
+ postfoothook=\end{corol-leftbar},%
+]{better-corol}
+
+\declaretheorem[style=better-defn]{defn}
+\declaretheorem[style=better-warn]{warn}
+\declaretheorem[style=better-exemple]{exemple}
+\declaretheorem[style=better-corol]{corol}
+\declaretheorem[style=better-props, numberwithin=defn]{props}
+\declaretheorem[style=better-thm, sibling=props]{thm}
+\newtheorem*{lemme}{Lemme}%[subsection]
+%\newtheorem{props}{Propriétés}[defn]
+
+\newenvironment{system}%
+{\left\lbrace\begin{align}}%
+{\end{align}\right.}
+
+\newenvironment{AQT}{{\fontfamily{qbk}\selectfont AQT}}
+
+\usepackage{LobsterTwo}
+\titleformat{\section}{\newpage\LobsterTwo \huge\bfseries}{\thesection.}{1em}{}
+\titleformat{\subsection}{\vspace{2em}\LobsterTwo \Large\bfseries}{\thesubsection.}{1em}{}
+\titleformat{\subsubsection}{\vspace{1em}\LobsterTwo \large\bfseries}{\thesubsubsection.}{1em}{}
+
+\newenvironment{lititle}%
+{\vspace{7mm}\LobsterTwo \large}%
+{\\}
+
+\renewenvironment{proof}{$\square$ \footnotesize\textit{Démonstration.}}{\begin{flushright}$\blacksquare$\end{flushright}}
+
+\title{TD du 20 mars}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+	\maketitle
+	\subsection*{Exercice 1}
+	On n'oublira pas la proposition suivante~:
+	$$ |\mathcal{P}(E)| = 2^n $$
+	où $n$ est le cardinal de $E$ et $\mathcal{P}(E)$ désigne l'ensemble des parties de $E$.
+
+	\begin{enumerate}
+		\item C'est un quadruplet. Il y a donc $10^4$ arrangements possibles. Si on évite les répétitions, il y en a $\frac{10!}{6!}$
+		\item Il s'agit aussi d'un arrangement car l'ordre compte. Il y a donc $\frac{10!}{7!}$ possibilités.
+		\item Le nombre de chemin possible est $\binom{p+q}{p}$.
+		\item $12!$ manières de les aligner. Si $1$ et $2$ se suivent, alors il suffit de déterminer la place de $1$ pour déterminer $2$. On n'a donc que $11$ tomes à placer donnant ainsi $11!$ possibilités.
+		\item $\displaystyle\binom{p}{1}\times 2^{n-p}$ car on se retrouve à choisir une partie de $E\backslash A$ (on a donc $2^{n-p}$ choix)
+		\item $2^8 = 256$ car ordre avec répétition
+		\item $10^{14}$ car ordre avec répétition 
+		\item $\binom{11}{5}$. Si deux amis ne peuvent venir qu'ensemble, on a un ami en moins à choisir si un des deux amis arrivent. Si deux amis ne peuvent pas se voir, on a un ami de possible en moins.
+
+			Dans le premier cas, on a $\binom 93+\binom 95$ (car on somme les «~ou bien~»~!).
+
+			Dans le deuxième cas, on a $2\binom 94 + \binom 95$ (idem).
+		\item 
+	\end{enumerate}
+	\subsection*{Exercice 2}
+	Je sais faire, donc flemme (par contre j'ai calculé ce que ça valait en python)
+	\subsection*{Exercice 5}
+	\begin{enumerate}
+		\item $\displaystyle f(x) = \displaystyle\sum_{k=0}^{n-1}\binom nk\frac{2^{n-k}}{x^{n-k-1}} = 2\binom n1 + o_{x\to +\infty}(1) \xrightarrow[x \to \infty]{} 2n$
+		\item $\displaystyle\sum_{i=0}^{n} \binom ni x^i = (x+1)^n$
+
+			$\displaystyle\sum_{i=0}^{n} \binom ni = (1+1)^n = 2^n$
+
+			$\displaystyle\sum_{i=0}^{n} \binom ni(-1)^i = (1-1)^n = 0$
+	\end{enumerate}
+\end{document}