Link do problema para dispositivos da Apple.
Problema
(A partir da 2ª série do E. M. – Nível de dificuldade: Médio)
Calcule S, sabendo-se que
S=218×207×196×(185)(185)+(1812)+(1912)+(208).
Observação: A notação (np) indica o número binomial de n por p.
Lembrete
Se n e p são números naturais tais que p⩽, definimos o número binomial \dbinom{n}{p} por:
\qquad \boxed{\dbinom{n}{p}=\dfrac{n!}{\left(n-p\right)!\, p!}}\, .
Solução 1
\textcolor{#800000}{(1)} Calculemos o numerador.
- \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*}
\textcolor{#800000}{(2)} Calculemos, agora, o denominador.
- Observe que:
\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*} - Observe também que:
\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*}
Portanto:
\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*}
\textcolor{#800000}{(3)} Por \textcolor{#800000}{(i)}\, e \textcolor{#800000}{(iv)}\, , segue que:
\qquad \qquad \fcolorbox{black}{#eee0e5}{$S=\dfrac{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}=\dfrac{\dfrac{21!}{13!\, 8!}}{\dfrac{21!}{13!\, 8!}}=1$} .
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: \dbinom{n}{p}=\dbinom{n}{n-p}.
P2: \dbinom{n}{p+1}+\dbinom{n}{p}=\dbinom{n+1}{p+1}.
Solução 2
- Numerador:
- \\
\, \\
\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*}
- Denominador:
\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*} - Por \textcolor{#800000}{(i)} e \textcolor{#800000}{(ii)}, temos que:
\qquad \qquad \fcolorbox{black}{#eee0e5}{$S=\dfrac{\dfrac{21}{8} \times \dfrac{20}{7} \times \dfrac{19}{6} \times \dbinom{18}{5}}{\dbinom{18}{5}+\dbinom{18}{12}+\dbinom{19}{12}+\dbinom{20}{8}}=\dfrac{\dfrac{21!}{13!\, 8!}}{\dfrac{21!}{13!\, 8!}}=1$} .
- Denominador:
Solução elaborada pelos Moderadores do Blog.
Nível B – Questão Difícil
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