Permutations of 5 taken 3 at a time – UGC NET Question

The number of permutations of n distinct objects taken r at a time is given by nPr = n!/(n−r)!. Here n = 5 and r = 3, so 5P3 = 5!/(5−3)! = 5!/2! = (5×4×3×2×1)/(2×1) = 5×4×3 = 60. Thus, there are 60 different ordered arrangements of 3 objects chosen from 5.

Option A:
A value of 10 corresponds to the combination 5C2 or 5C3, which counts unordered selections, not ordered arrangements. Since permutations consider order, 10 is too small and does not account for all possible orderings of 3 objects. Therefore, this option is incorrect.

Option B:
60 is correct because it counts each distinct ordering separately. For example, selecting A, B and C yields six permutations: ABC, ACB, BAC, BCA, CAB and CBA, and similar counts apply for other triplets. Multiplying the number of choices for each position, 5×4×3, naturally arrives at 60.

Option C:
A value of 20 does not match any standard permutation or combination formula for the given n and r. It likely arises from a miscalculation or misunderstanding of order versus selection. Hence, this option cannot be supported.

Option D:
120 equals 5!, which counts permutations of all 5 objects taken at a time, not just 3 out of 5. The question restricts us to arrangements of length 3, so 120 overcounts and is not correct here.