In a class of 16 students, there are 6 male students and 10 female students. In how many ways can you select a group of 5 students from this class such that 3 of the 5 students are males and the other two are females?

3 male students we can select by $C_6^3 = \frac{{6!}}{{3!(6 - 3)!}} = \frac{{6!}}{{3!3!}} = \frac{{4 \cdot 5 \cdot 6}}{6} = 20$ ways.

Other 2 female students we can select by $C_{10}^2 = \frac{{10!}}{{2!(10 - 2)!}} = \frac{{10!}}{{2!8!}} = \frac{{9 \cdot 10}}{2} = 45$ ways.

then the total number of ways is 20*45=900.

answer: 900 ways