$3 \times10$

The first student can be chosen in 30 ways, the second in 29 ways and the third on 28 ways. Thus we have $30 \times 29 \times 28$ ways.

However, each team was counted several times: the same trio of students can be chosen in different ways.

For instance, choosing student A first, then student B, and finally, C is the same as choosing C first, then A, and then B. Since the number of permutations of 3 elements is 3!, each team was counted exactly 3!=6 times.

Therefore the answer is $\frac {30 \times 29 \times 28}{3!}$ .

= $\boxed{4060}.$