Google
Câu hỏi: Nếu $C_{n}^{3}=220$ thì $n$ bằng
A. $12$.
B. $13$.
C. $14$.
D. $11$.
A. $12$.
B. $13$.
C. $14$.
D. $11$.
Điều kiện: $\left\{ \begin{aligned}
& n\in \mathbb{N} \\
& n\ge 3 \\
\end{aligned} \right.$
$C_{n}^{3}=\dfrac{n!}{3!.\left( n-3 \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}$ mà $C_{n}^{3}=220$ nên
$\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}=220\Leftrightarrow n\left( n-1 \right)\left( n-2 \right)=1320\Leftrightarrow {{n}^{3}}-3{{n}^{2}}+2n-1320\Leftrightarrow n=12$ (thỏa mãn).
& n\in \mathbb{N} \\
& n\ge 3 \\
\end{aligned} \right.$
$C_{n}^{3}=\dfrac{n!}{3!.\left( n-3 \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}$ mà $C_{n}^{3}=220$ nên
$\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}=220\Leftrightarrow n\left( n-1 \right)\left( n-2 \right)=1320\Leftrightarrow {{n}^{3}}-3{{n}^{2}}+2n-1320\Leftrightarrow n=12$ (thỏa mãn).
Đáp án A.