In a class of 25 distinct students, how many different 3-student committees can be formed? (Order of students within the committee does not matter.)

Introduction / Context:Forming committees is choosing subsets, not arranging sequences. The correct tool is combinations C(n, k).

Given Data / Assumptions:

• n = 25 students.
• k = 3 students per committee.

Concept / Approach:Compute C(25, 3) = 25! / (3! * 22!).

Step-by-Step Solution:

Verification / Alternative check:Multiplying 2524*23 = 13,800 and dividing by 6 gives 2300 exactly.

Why Other Options Are Wrong:Nearby values (2200, 2400, 3200) reflect arithmetic slips or using permutations.

Common Pitfalls:Confusing ordered selections with unordered committee formation.

Final Answer:2300