Lời giải chi tiết:

\(C_{n + 4}^{n + 1} - C_{n + 3}^n = 7\left( {n + 3} \right)\)\(\left( {n \in N} \right)\)

Có tính chất: \(C_n^k = C_n^{n - k}\)\( \Rightarrow \left\{ \begin{array}{l}C_{n + 4}^{n + 1} = C_{n + 4}^3\\C_{n + 3}^n = C_{n + 3}^3\end{array} \right.\)

\( \Leftrightarrow C_{n + 4}^3 - C_{n + 3}^3 = 7\left( {n + 3} \right)\)

\( \Leftrightarrow \dfrac{{\left( {n + 4} \right)!}}{{3!.\left( {n + 1} \right)!}} - \dfrac{{\left( {n + 3} \right)!}}{{3!.n!}} = 7\left( {n + 3} \right)\)

\( \Leftrightarrow \dfrac{{\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}}{{3!}} - \dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{{3!}} = 7\left( {n + 3} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}n + 3 = 0 \Leftrightarrow n =  - 3\,\,(Loai)\\\dfrac{{\left( {n + 2} \right)\left( {n + 4} \right)}}{{3!}} - \dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} = 7\end{array} \right.\)

\( \Leftrightarrow \left( {n + 2} \right)\left( {n + 4} \right) - \left( {n + 1} \right)\left( {n + 2} \right) = 42\)

\( \Leftrightarrow 3n + 6 = 42\)

\( \Leftrightarrow n = 12\)

Chọn D.