Answer: 455

## Explanation This is a combination problem where we need to select groups of 3 students from 15, and the order of selection doesn't matter. ### Formula Used We use the combination formula: $$\binom{n}{r} = \frac{n!}{(n-r)!r!}$$ ### Calculation - n = 15 (total students) - r = 3 (students per group) $$\binom{15}{3} = \frac{15!}{(15-3)! \cdot 3!} = \frac{15!}{12! \cdot 3!}$$ $$= \frac{15 \times 14 \times 13 \times 12!}{12! \cdot 3 \times 2 \times 1} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455$$ ### Why Combination Instead of Permutation? - **Combination** is used when order doesn't matter (selecting groups where {Student A, Student B, Student C} is the same as {Student B, Student A, Student C}) - **Permutation** would be used if the order mattered (like selecting president, vice-president, secretary) Therefore, there are **455** different possible groups of 3 students that can be formed from 15 students.

Author: Tanishq Prabhu