[ Combinations and Permutations ]

a) From a class of 30 students, two students need to be selected to participate in a mathematics olympiad. In how many ways can this be done?

b) In how many ways can a team of three students be selected from the same class?
a) The first student can be chosen in 30 ways, the second, independently of the choice of the first student, in 29 ways. In this case, each pair is counted twice. Therefore, the total number of ways is $30 \cdot 29: 2$.

b) Similarly, we get $30 \cdot 29 \cdot 28$ variants for the sequential selection of three students. In this case, each team is counted $3!=6$ times. Therefore, the number of ways to choose a team is $\frac{30 \cdot 29 \cdot 28}{3!}$.

## Answer

a) $C_{30}^{2}=435$ ways; b) $C_{30}^{3}=4060$ ways.