UGCNET Questions

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 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.

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.