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.
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