UGCNET Questions

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.

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.