Of the 8 students in the chess club, three will represent the club at an upcoming competition. How many different 3-person teams can be formed from the 8 students?

I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C (n, r) = n! / (n-r) ! r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have

C (8,3) = 8! / (8-3) !3!

C (8,3) = 56

There are 56 3-person teams that can be formed from the 8 students.