Choosing 3 males from 6 males and 2 females from 4 females to form a committee of five, we have
\(\Rightarrow{ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\)
\(=\frac{6 !}{(6-3) \mid 3 !} \times \frac{4 !}{(4-2) \mid 2 !}\)
\(=\frac{6 \times 5 \times 4 \times 3 !}{3 ! 3 !} \times \frac{4 \times 3 \times 2 !}{2 ! 2 !}\)
\(=\frac{6 \times 5 \times 4}{3 \times 2} \times \frac{4 \times 3}{2 \times 1}\)
\(5 \times 4 \times 2 \times 3=120\) ways