UGC NET Questions (Paper – 1)

Statements A and B apply the multiplication principle and permutation formula correctly. A recognises that each of the 3 distinct books has 2 choices of shelf, so 2 × 2 × 2 = 2³ placements. B notes that 3 distinct books can be arranged in 3! orders on one shelf. C is false because identical books reduce the number of distinct permutations to 3!/2!, not 3!, and D is false because when counts depend on previous choices, we must adjust rather than blindly multiply m and n. E is false since choosing a student–teacher pair uses m × n, not m + n. Thus only A and B are correct.

Option A:
Option A is incomplete as it includes only statement A and omits B, thereby missing a standard example of permutations with distinct objects that is also correct.

Option B:
Option B is correct because it selects exactly the statements that correctly apply counting principles while excluding C, D and E, each of which misuses permutations or the multiplication rule. It reflects the proper reasoning expected in NET quantitative aptitude.

Option C:
Option C is wrong because it adds C, which ignores the effect of identical objects and incorrectly counts permutations as if all items were distinct.

Option D:
Option D is incorrect since it includes D and still allows the false “always m × n regardless of dependence” claim, so the combination cannot be correct.

The word LEVEL has five letters in total, with L appearing twice and E appearing twice, while V appears once. The formula for permutations of n objects with repetitions is n factorial divided by the product of factorials of repeated counts. Thus, the number of distinct permutations is 5! ÷ (2! × 2!). Calculating this gives 120 ÷ 4 = 30. Hence, there are 30 different arrangements of the letters in LEVEL.

Option A:
Option A, 20, underestimates the actual number and cannot be obtained by the correct factorial expression 5! ÷ (2! × 2!). It likely comes from an incorrect guess or misapplied formula.

Option B:
Option B accurately uses the repetition-adjusted permutation formula, recognizing two repeated L’s and two repeated E’s. By evaluating 120 ÷ 4, it yields 30, which counts all unique arrangements without overcounting identical ones.

Option C:
Option C, 40, exceeds the true count and would require a different numerator or denominator in the formula. It suggests that the effect of repeated letters has not been properly accounted for.

Option D:
Option D, 60, is exactly half of 120 and may arise from dividing by only one 2! instead of both 2!. This partially correct adjustment still leads to overcounting because both L and E are repeated.

Statement A is correct for linear permutations. Statement B is true because circular permutations fix rotation, giving (n − 1)!. Statement C is true via the block method for adjacency. Statement E is true because extra constraints reduce valid arrangements. Statement D is false because in permutations, order matters. Thus A, B, C and E only are correct.

Option A:
Option A is correct because it includes all true statements and excludes D (false).

Option B:
Option B is incomplete as it omits E (constraints).

Option C:
Option C is wrong because it includes D (false) and omits A.

Option D:
Option D is incorrect because it includes D (false) and omits B (circular formula).

This option is correct because permutations count ordered arrangements of r objects chosen from n distinct objects. The standard notation for this is nPr. It takes into account both the selection and the ordering of the objects. Therefore, the correct symbol is nPr.

Option A:
nCr represents combinations, which count selections without regard to order. It does not distinguish between different arrangements of the same r objects. Since the question explicitly mentions arranging, this symbol is not appropriate.

Option B:
rPn is not the standard way of denoting arrangements of r out of n. It reverses the positions of n and r and does not correspond to any common formula in basic combinatorics. Hence, this notation is incorrect.

Option C:
The symbol nPr captures both how many objects are chosen and how they are ordered. It equals n!/(n−r)!, which counts all possible ordered selections. This matches the description of selecting and arranging r objects from n, so this option is correct.

Option D:
rCn would again reverse the roles of n and r in combination notation. It is not used to represent either permutations or combinations in standard texts. Therefore, it is not the correct symbol.

Statements A, B, D and E are all correct descriptions of the role of permutations and combinations. Permutations count ordered arrangements, combinations count unordered selections, seating arrangement problems rely heavily on order-based counting, and both concepts are frequently tested in UGC NET aptitude. Statement C is false because, for r greater than 1, nPr is larger than nCr since each combination can be arranged in several ordered ways. Thus the only option that includes all true statements while excluding the false one is the combination A, B, D and E.

Option A:
Option A is incomplete because it omits E, failing to note explicitly that both permutations and combinations are actively used in NET mathematical reasoning questions. Without E, the statement about exam relevance is missing.

Option B:
Option B is correct because it gathers A, B, D and E, which together present the conceptual distinction between order and selection and their examination usage, while rightly rejecting C, which reverses the relative sizes of permutations and combinations.

Option C:
Option C is incorrect since it leaves out A and therefore does not mention that permutations are arrangements where order matters, a central concept for the topic. Though it contains some true statements, it is not complete.

Option D:
Option D is wrong because it includes C, the false claim that nPr is always less than nCr, and thereby mixes an incorrect statement with true ones. The presence of C makes the entire combination unacceptable.

The word NET has three distinct letters: N, E and T. The number of permutations of n distinct objects taken all at a time is n!, so here it is 3!. Calculating 3! gives 3 × 2 × 1 = 6. Therefore, there are 6 possible distinct arrangements of the letters in NET.

Option A:
Option A uses the factorial formula for permutations correctly, recognizing that there are no repeated letters. By computing 3! as 6, it counts all possible orderings such as NET, NTE, ENT, ETN, TEN and TNE.

Option B:
Option B, 3, would be the number of letters, not the number of arrangements. It undercounts the permutations by ignoring the different orders in which letters can appear.

Option C:
Option C, 9, may come from incorrectly squaring the number of letters (3²), which does not match the combinatorial rule for arrangements of distinct objects.

Option D:
Option D, 12, overestimates the possible arrangements and could arise from misapplying permutation formulas that involve repetition or additional positions.

A and B correctly give the formulas for permutations and combinations without repetition. C is true because allowing repetition means each of the r positions can be filled in n ways, giving nʳ arrangements. E is also correct since identifying whether order matters is essential to selecting the right counting approach. D is false because without repetition, r-length arrangements from n objects are counted by nPr, not rⁿ. Thus A, B, C and E only are correct.

Option A:
Option A misses E and therefore does not state the important strategic step of checking whether order is significant. Although it contains correct formulas, it is not complete as an answer.

Option B:
Option B omits A and so fails to mention the basic permutation formula, even though it includes other true statements. Leaving out A makes the set incomplete for the topic.

Option C:
Option C is correct because it collects all the true statements while excluding D, which misapplies exponent notation. It reflects the range of arrangements encountered in NET questions.

Option D:
Option D excludes B and so ignores the combination formula, and it does not correct the error in D. Consequently, it does not represent the full correct set.

B is correct: permutations count ordered arrangements. C is correct because nPr = nCr × r! and r! ≥ 1. D is correct since choosing 0 or all items yields one way. E is correct since choosing 1 item from n gives n ways. A is false because combinations ignore order. Hence B, C, D and E only are correct.

Option A:
Option A is incomplete because it omits D and E, which are standard special cases of combinations.

Option B:
Option B is incorrect because it omits E (true).

Option C:
Option C is incorrect because it omits C (true) even though B, D and E are true; it does not list all correct statements.

Option D:
Option D is correct because it includes all true statements and excludes A, which reverses the definition of combinations.

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.