UGC NET Questions (Paper – 1)

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.

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.

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.

There are 4 aces in a standard 52-card deck. The probability that the first card is an ace is 4/52, since any of the 4 aces could appear among the 52 cards. After one ace has been drawn, 3 aces remain in a reduced deck of 51 cards, so the probability that the second card is also an ace is 3/51. Because the draws are without replacement and the events are dependent, we multiply the probabilities: (4/52) × (3/51) = 12/2652, which simplifies to 1/221. Therefore, the correct probability that both drawn cards are aces is 1/221.

Option A:
Option A (1/13) usually comes from considering only the probability that a single card is an ace (4/52 = 1/13) and ignoring that we need two consecutive aces. It does not incorporate the second dependent draw, so it does not represent the joint probability of both events occurring. Because it treats the problem as if only one ace were required, it significantly overestimates the true probability. Hence, 1/13 is not correct for “both cards are aces.”

Option B:
Option B (1/169) typically arises from incorrectly assuming that the two draws are independent with replacement and computing (1/13) × (1/13). However, in this question the cards are drawn without replacement, which means the second probability changes after the first card is drawn. Using 1/169 ignores the reduced deck on the second draw and does not reflect the correct dependent-event structure. Therefore, it underestimates the correct probability.

Option C:
Option C (2/221) looks close to the correct answer but is exactly double 1/221, and there is no step in the correct derivation that would justify multiplying the final probability by 2. This kind of error often happens when students miscount favourable outcomes or mistakenly think order does not matter when they have already accounted for it. Since the exact product (4/52) × (3/51) simplifies to 1/221, any multiple of that value, including 2/221, is incorrect.

Option D:
Option D (1/221) matches the precise result from multiplying the probability of the first ace and the second ace without replacement: (4/52) × (3/51). The numerator 12 counts the ordered ways to pick two aces in succession, and the denominator 2652 counts all ordered pairs of cards, which simplifies to 1/221. This value correctly reflects both the limited number of aces and the shrinking deck size after the first draw. Hence, it is the only option fully consistent with the rules of probability for dependent events.

When order does not matter, the number of ways to choose r objects from n is given by the combination formula nCr. Here n is 5 and r is 2, so the number of committees is 5C2. Using the formula, 5C2 = 5! ÷ (2! × 3!) = (5 × 4) ÷ (2 × 1) = 10. Therefore, there are 10 distinct ways to select a pair of students from the group.

Option A:
Option A, 5, corresponds to the number of students rather than the number of pairs that can be formed. It ignores the multiple distinct combinations that are possible when selecting two members.

Option B:
Option B, 8, does not match the combinatorial calculation for 5C2 and appears to be an arbitrary number without basis in the formula.

Option C:
Option C, 9, is close to 10 but still does not equal the true combination count as derived from factorial operations, showing a small but important error.

Option D:
Option D correctly employs the combination formula and simplifies it accurately. It recognizes that for each choice of two students, the order is irrelevant, leading to exactly 10 possible committees.

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.

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.