\({7 \choose 2}\) ways to choose 2 freshmen for the committee,
\({9 \choose 3}\) ways to choose 3 sophomores for the committee,
\({7 \choose 4}\) ways to choose 4 juniors for the committee, and
\({7 \choose 5}\) ways to choose 5 seniors for the committee. So by the generalized basic principle of counting, there are a total of
\begin{equation*}
{7 \choose 2} \cdot {9 \choose 3} \cdot {7 \choose 4} \cdot {7 \choose 5}
= \frac{7 !}{2!5 !} \cdot \frac{9 !}{3! 6 !} \cdot \frac{7 !}{4! 3 !} \cdot \frac{7 !}{5! 2 !} = 1296540
\end{equation*}
different possible committees.