Answer: 455

This is a combination problem where order doesn't matter. The formula for combinations is: $$_nC_r = \frac{n!}{(n-r)!r!}$$ Where: - n = 15 (total students) - r = 3 (students per group) $$_{15}C_3 = \frac{15!}{(15-3)!3!} = \frac{15!}{12!3!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455$$ Therefore, there are 455 different groups of 3 students possible from 15 students.

Author: Nikitesh Somanthe