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\[\begin{aligned}
1! &= 1\;;&
\\
2 ! &= 2\times 1 = 2\;;&
\\
3 ! &= 3\times 2 \times 1 = 6\;;&
\\
4 ! &= 4\times 3 \times 2 \times 1 = 24\;;&
\\
5 ! &= 5 \times 4 \times 3 \times 2 \times 1 = 120.&
\end{aligned}\]
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$\dfrac{6!}{3!} = \dfrac{6\times 5 \times 4 \times \cancel 3 \times \cancel 2 \times \cancel 1}
{\cancel 3 \times \cancel 2 \times \cancel 1}
=6\times 5 \times 4 = 120$.
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$\dfrac{9!}{11!} = \dfrac{\cancel{9!}}{11\times 10 \times \cancel{9!}} = \dfrac 1{11\times 10} = \dfrac 1{110}$.
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$\dfrac{20!}{3!\times 5! \times 2!} = \dfrac{20!}{3\times 2 \times 5 \times 4 \times \underbrace{3\times 2}_6 \times 2}$
$=\dfrac{20!}{6 \times 5 \times 4\times 3\times 2 \times 2}$
$=\dfrac{20!}{2\times 6!}$.
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Si $n \neq 0$, nos factorielles expriment des produits:
\[\begin{aligned}
&\frac 1{n!}- \frac 1{(n+1)!}
\\
&= \frac1{n\times \cdots \times 1} - \frac 1 {(n+1)\times n \times \cdots \times 1}&
\\
&=\frac{\underline{(n+1)} \times 1}{\underline{(n+1)} \times n \times \cdots \times 1} - \frac 1 {(n+1)\times n \times \cdots \times 1}&
\\
&=\frac{n+1 - 1}{(n+1)!}&
\\
&=\frac n {(n+1)!}.&
\end{aligned}\]
Si $n = 0$ alors, sachant que $0! = 1$:
\[\frac 1 {n!}- \frac 1{(n+1)!} = \frac 1 1 - \frac 1 1 = 0.\]
Or dans ce cas
\[\frac{n}{(n+1)!} = \frac 0 1 = 0.\]
Donc notre simplification reste vraie dans le cas particulier où $n = 0$.
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