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\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}