Bài 104170

Ta có:

$C^7_n=2.C^3_n \Leftrightarrow \frac{n!}{7!(n-7)!}=2\frac{n!}{3!(n-3)!} \Leftrightarrow (n-6)(n-5)(n-4)(n-3)=1680$

$\Leftrightarrow (n-6)(n-5)(n-4)(n-3)=1680$

$\Leftrightarrow (n^2-9n+18)(n^2-9n+20)=1680$

Đặt $t=n^2-9n+18$

Suy ra: $t(t+2)=1680 \Leftrightarrow t^2-2t-1680=0 \Leftrightarrow \left[ \begin{array}{l}t = 40\\t = -42\end{array} \right.$

* Khi: $t=n^2-9n+18=40 \Leftrightarrow n^2-9n-22=0 \Leftrightarrow \left[ \begin{array}{l}n = 11\\n = -2\end{array} \right.\Rightarrow n=11$

* Khi: $t=n^2-9n+18=-42 \Leftrightarrow n^2-9n+60=0 \Leftrightarrow n \in \varnothing$

* Kết luận: $n=11$.