Choosing 2 from 5 students – UGC NET Question

The number of ways to choose r objects from n distinct objects is given by the combination formula nCr = n!/[r!(n−r)!]. Here n = 5 and r = 2, so 5C2 = 5!/[2!3!] = (5×4)/(2×1) = 10. Therefore, there are 10 different pairs of students that can be chosen from 5.

Option A:
A value of 5 would arise if we incorrectly assumed that each student could only form one pair or confused combinations with simple counting. However, with 5 students, there are more than 5 distinct pairs. So, this option is an underestimate and is not correct.

Option B:
A value of 8 does not match the formula 5C2 and suggests a miscalculation. No standard counting principle or correct formula would yield 8 for this problem. Therefore, this option is incorrect.

Option C:
A value of 9 is also inconsistent with the formula for combinations. It lies between nearby values but does not have any combinatorial justification here. Since 5C2 has been clearly computed as 10, 9 cannot be the right answer.

Option D:
10 is correct because it results directly from applying the combination formula and correctly accounts for all unordered pairs of students. It avoids double counting pairings like (A,B) and (B,A), which are considered the same pair in combinations. This understanding is essential in many selection problems in aptitude tests.