Đáp án + Giải thích các bước giải:

Ta có:

$\textbf{A} = \dfrac{\textbf{A}^6_n + \textbf{A}^5_n}{\textbf{A}^4_n}$

$\phantom{\textbf{A}} = \dfrac{\frac{n!}{(n - 6)!} + \frac{n!}{(n -5 )!}}{\frac{n!}{(n - 4)!}}$ 

$\phantom{\textbf{A}} = \dfrac{\frac{n!}{(n - 4)!}(n - 4)(n - 5) + \frac{n!}{(n -4)!}(n -4 )}{\frac{n!}{(n - 4)!}}$ 

$\phantom{\textbf{A}} = (n - 4)(n - 5) + (n - 4)$

$\phantom{\textbf{A}} = (n - 4)^2$

$\textbf{C}^{n - 3}_n = 1140$

$\Leftrightarrow \dfrac{n!}{(n - 3)! \cdot 3!} = 1140$

$\Leftrightarrow \dfrac{n!}{(n -3 )!} = 6840$

$\Leftrightarrow n(n - 1)(n -2 ) = 6840$

$\Leftrightarrow n^3 - 3n^2 + 2n -6840 = 0$

$\Leftrightarrow n^3 - 20n^2 + 17n^2 - 340n + 342n - 6840 = 0$

$\Leftrightarrow n^2(n - 20) + 17(n - 20) + 342(n - 20) = 0$

$\Leftrightarrow \left(n^2 + 17n + 342\right)(n - 20) = 0$

Do $n^2 + 17n + 342= \left(n + \dfrac{17}{2}\right)^2 + \dfrac{1079}{4} > 0, \forall n$

$\Rightarrow n - 20 = 0$

$\Leftrightarrow n = 20$

$\Rightarrow \textbf{A} = (20 - 4)^2 = 256$