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Problema
(A partir da 2ª série do E. M. – Nível de dificuldade: Médio)
Calcule [tex]S[/tex], sabendo-se que
[tex]\qquad \qquad \quad \quad S=\dfrac{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}.[/tex]
Observação: A notação [tex]\dbinom{n}{p}[/tex] indica o número binomial de [tex]n[/tex] por [tex]p[/tex].
Lembrete
Se [tex]n[/tex] e [tex]p[/tex] são números naturais tais que [tex]p \leqslant n\, [/tex], definimos o número binomial [tex]\dbinom{n}{p}[/tex] por:
[tex]\qquad \boxed{\dbinom{n}{p}=\dfrac{n!}{\left(n-p\right)!\, p!}}\, .[/tex]
Solução 1
[tex]\textcolor{#800000}{(1)}[/tex] Calculemos o numerador.
- [tex]\begin{align*}\boxed{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}&=\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times\dfrac{18!}{(18-5)!\, 5!}\\
&=\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times\dfrac{18!}{13!\, 5!}\\
&=\dfrac{21\times 20\times 19\times 18!}{13! \times 8\times 7\times 6\times 5!}\\
&=\boxed{\dfrac{21!}{13!\, 8!}} \qquad \textcolor{#800000}{(i)}
\end{align*}[/tex]
[tex]\textcolor{#800000}{(2)}[/tex] Calculemos, agora, o denominador.
- Observe que:
[tex]\begin{align*}\dbinom{18}{5}+\dbinom{18}{12}&=\dfrac{18!}{\left(18-5\right)!\, 5!}+\dfrac{18!}{\left(18-12\right)!\, 12!}\\
&=\dfrac{18!}{13!\, 5!}+\dfrac{18!}{6!\, 12!}=\dfrac{18!\times 6+18!\times 13}{13!\, 6!}\\
&=\dfrac{18!\times (6+13)}{13!\, 6!}=\dfrac{18!\times 19}{13!\, 6!}\\
&=\dfrac{19!}{13!\, 6!}\qquad \textcolor{#800000}{(ii)}
\end{align*}[/tex] - Observe também que:
[tex]\begin{align*}\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}&=\left[\dbinom{18}{5}+\dbinom{18}{12}\right]+\dbinom{19}{12}\\
&\stackrel{\textcolor{#800000}{(ii)}}{=}\dfrac{19!}{13!\, 6!}+\dbinom{19}{12}=\dfrac{19!}{13!\, 6!}+\dfrac{19!}{(19-12)!\, 12!}\\
&=\dfrac{19!}{13!\, 6!}+\dfrac{19!}{7!\, 12!}=\dfrac{19!\times 7+19!\times 13}{13!\, 7!}\\
&=\dfrac{19!\times (7+13)}{13!\, 7!}=\dfrac{19!\times 20}{13!\, 7!}\\
&=\dfrac{20!}{13!\, 7!}\qquad \textcolor{#800000}{(iii)}
\end{align*}[/tex]
Portanto:
[tex]\begin{align*}\boxed{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}&=\left[\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}\right]+\dbinom{20}{8}\\
&\stackrel{\textcolor{#800000}{(iii)}}{=}\dfrac{20!}{13!\, 7!}+\dbinom{20}{8}=\dfrac{20!}{13!\, 7!}+\dfrac{20!}{(20-8)!v8!}\\
&=\dfrac{20!}{13!\, 7!}+\dfrac{20!}{12!\, 8!}=\dfrac{20!\times 8+20!\times 13}{13!\, 8!}\\
&=\dfrac{20!\times (8+13)}{13!\, 8!}=\dfrac{20!\times 21}{13!\, 8!}\\
&=\boxed{\dfrac{21!}{13!\, 8!}}\qquad \textcolor{#800000}{(iv)}
\end{align*}[/tex]
[tex]\textcolor{#800000}{(3)}[/tex] Por [tex]\textcolor{#800000}{(i)}\, [/tex] e [tex]\textcolor{#800000}{(iv)}\, [/tex], segue que:
[tex]\qquad \qquad \fcolorbox{black}{#eee0e5}{$S=\dfrac{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}=\dfrac{\dfrac{21!}{13!\, 8!}}{\dfrac{21!}{13!\, 8!}}=1$}[/tex] .
Solução elaborada pelos Moderadores do Blog.
Ferramentas que podem ajudar na segunda solução
Duas propriedades dos números binomiais podem encurtar a solução deste problema:
P1: [tex]\dbinom{n}{p}=\dbinom{n}{n-p}.[/tex]
P2: [tex]\dbinom{n}{p+1}+\dbinom{n}{p}=\dbinom{n+1}{p+1}.[/tex]
Solução 2
- Numerador:
- [tex]\\
\, \\
\begin{align*}\boxed{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}&=\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times\dfrac{18!}{(18-5)!\, 5!}\\
&=\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times\dfrac{18!}{13!\, 5!}\\
&=\dfrac{21\times 20\times 19\times 18!}{13! \times 8\times 7\times 6\times 5!}\\
&=\boxed{\dfrac{21!}{13!\, 8!}} \qquad \textcolor{#800000}{(i)}
\end{align*}[/tex] - Denominador:
[tex]\begin{align*}\boxed{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}&\stackrel{\textcolor{#800000}{(P1)}}{=}\dbinom{18}{13}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{12}\\
&=\left[\dbinom{18}{13}+\dbinom{18}{12}\right] +\dbinom{19}{12}+\dbinom{20}{12}\\
&\stackrel{\textcolor{#800000}{(P2)}}{=}\dbinom{19}{13}+\dbinom{19}{12}+\dbinom{20}{12}\\
&=\left[\dbinom{19}{13}+\dbinom{19}{12}\right]+\dbinom{20}{12}\\
&\stackrel{\textcolor{#800000}{(P2)}}{=}\dbinom{20}{13}+\dbinom{20}{12}\\
&\stackrel{\textcolor{#800000}{(P2)}}{=}\boxed{\dbinom{21}{13}} \qquad \textcolor{#800000}{(ii)}
\end{align*}[/tex] - Por [tex]\textcolor{#800000}{(i)}[/tex] e [tex]\textcolor{#800000}{(ii)}[/tex], temos que:
[tex]\qquad \qquad \fcolorbox{black}{#eee0e5}{$S=\dfrac{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}=\dfrac{\dfrac{21!}{13!\, 8!}}{\dfrac{21!}{13!\, 8!}}=1$}[/tex] .
Solução elaborada pelos Moderadores do Blog.
Nível B – Questão Difícil
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