UGC NET Questions (Paper – 1)

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.

This option is correct because combinations count the number of ways to select r objects from n distinct objects when order does not matter. The standard notation for this is nCr. It is also called the binomial coefficient and is given by n!/[r!(n−r)!]. Therefore, nCr represents the required count.

Option A:
nPr is used for permutations, where order of arrangement matters. It counts more cases than combinations because each selection can be arranged in many orders. Since the question specifies "without regard to order," nPr is not suitable.

Option B:
rCn is not the conventional way of writing combinations of r from n. It reverses the parameters and is not used in standard notation. This makes the option incorrect.

Option C:
rPn again refers to permutations but with reversed parameters compared to typical notation. It also considers ordering, which the question explicitly ignores. Thus, this symbol does not answer the question correctly.

Option D:
nCr correctly denotes the number of combinations of n objects taken r at a time. It ignores order and is widely used in probability and counting problems. This precisely matches the description given, so this option is correct.

Statements A and B correctly distinguish permutations and combinations in terms of the role of order. D is true because permutations can be obtained from combinations by arranging the chosen r objects in r! ways. E is also correct since there is only one way to choose all n objects from n. Statement C is false because nPr generally exceeds nCr when 0 < r < n, so permutations and combinations are not equal in count. Thus the correct combination is A, B, D and E only. Option A: Option A is incomplete as it omits D and E, and hence does not show the quantitative relationship between permutations and combinations or the special case when all objects are chosen. It therefore does not capture the full set of correct statements. Option B: Option B is incorrect because it leaves out E and includes only A, B and D, so it does not mention the important special case nCr = 1 when r = n. This omission makes the combination incomplete. Option C: Option C is correct because it gathers the true distinctions and the exact formula relating nPr and nCr and excludes C, which wrongly claims equality of counts. It therefore matches the combinatorial facts emphasised in NET aptitude. Option D: Option D is wrong as it drops A and includes B, D and E only, failing to state explicitly that permutations take order into account while still not rejecting C. As a result, it does not provide the precise set of correct statements requested.

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.

The word BANANA has 6 letters, with A repeated 3 times, N repeated 2 times and B appearing once. The formula for permutations of n objects with repeated groups is n! divided by the product of factorials of the counts of repeated items. Here this gives 6! ÷ (3! × 2!). Evaluating this gives 720 ÷ (6 × 2) = 720 ÷ 12 = 60. So there are 60 distinct arrangements of the letters of BANANA.

Option A:
Option A, 30, underestimates the total arrangements and would correspond to a different combination of repeats or a smaller word. It does not follow the factorial formula for this word.

Option B:
Option B, 36, arises from miscalculating either 6! or one of the factorials in the denominator. It does not equal 720 ÷ 12 and so fails to match the correct combinatorial reasoning.

Option C:
Option C, 48, again does not match the computed result and suggests partial adjustment for repetition without applying the full divisor 3! × 2!.

Option D:
Option D carefully applies the repetition-adjusted permutation formula. It fully accounts for indistinguishable arrangements caused by repeated letters and yields the correct count of 60.

If one particular book must always be included, we first include that book, leaving 6 other books to choose from. We then need to select 2 additional books from the remaining 6. The number of ways to do this is 6C2, which equals (6 × 5) ÷ (2 × 1) = 15. Hence, there are 15 different selections of 3 books that always contain the specified book.

Option A:
Option A, 10, corresponds to 5C2 and would be correct only if there were 5 remaining books instead of 6, so it does not match the actual numbers given.

Option B:
Option B correctly applies the combination formula with n = 6 and r = 2, reflecting the need to pick the remaining books after fixing the one mandatory book.

Option C:
Option C, 20, might come from using 6P2 or another permutation-based miscalculation, but we are counting combinations where order does not matter.

Option D:
Option D, 35, is 7C3, which counts all combinations of 3 books without the restriction that the particular book must always be included.

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.