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.