This question tests counting with a restriction on adjacency. Order matters since we're arranging books, so we use permutations. Total arrangements without restriction: 5! = 120. To find arrangements where A and B are not adjacent, we subtract arrangements where they are adjacent. When A and B are adjacent, treat them as a single unit: we have 4 units to arrange in 4! = 24 ways, and A and B can be ordered within their unit in 2! = 2 ways, giving 24×2 = 48 arrangements with A and B adjacent. Therefore, arrangements where A and B are not adjacent: 120 - 48 = 72. A common error is forgetting to multiply by 2 when A and B are treated as a unit, giving the incorrect answer of 96.

This question tests permutations with repeated elements. Order matters since we're arranging letters to form strings, but we must account for identical letters. BANANA has 6 letters: B(1), A(3), N(2). If all letters were distinct, we'd have 6! = 720 arrangements. However, we must divide by the factorial of each letter's frequency to avoid overcounting identical arrangements. The number of distinct arrangements is 6!/(1!×3!×2!) = 720/(1×6×2) = 720/12 = 60. A common mistake is forgetting to divide by the factorials of repeated letters, leading to the incorrect answer of 720.

This question tests counting with the stars and bars method combined with constraints. Order doesn't matter for fruit purchases, making this a combinations problem. We need exactly 6 pieces total with at least 1 of each type. First, we allocate 1 piece to each of the 4 types, using 4 pieces. This leaves 2 pieces to distribute freely among the 4 types. Using stars and bars, distributing 2 identical items into 4 distinct categories gives us C(2+4-1, 4-1) = C(5,3) = 10 ways. The constraint of at least one of each type is handled by the initial allocation. A common mistake would be trying to use C(6+4-1, 4-1) = C(9,3) = 84, which doesn't account for the minimum requirement.

This question tests permutations with assignment restrictions. Order matters since tasks are distinct and assigned to specific employees. Without restrictions: 4! = 24 ways to assign 4 tasks to 4 employees. With the restriction that Employee X cannot do Task 1 or Task 2, X has only 2 choices (Task 3 or Task 4). Once X is assigned, the remaining 3 employees can be assigned to the remaining 3 tasks in 3! = 6 ways. Total valid assignments: 2 × 6 = 12. Choice C correctly identifies this count. Choice E (24) represents the unrestricted case, failing to account for X's limitations.

This question tests counting permutations with a restriction on adjacency. Order matters since we're arranging students in specific seats. We can use complementary counting: find the total arrangements without restrictions, then subtract arrangements where X and Y sit together. Total arrangements of 5 students is 5! = 120. To count arrangements where X and Y are adjacent, treat them as a single unit that can be arranged in 2 ways (XY or YX), and this unit plus the 3 other students gives us 4 units to arrange in 4! ways. So there are 2 × 4! = 2 × 24 = 48 arrangements where X and Y are adjacent. Therefore, arrangements where they're not adjacent is 120 - 48 = 72. A common error is forgetting to account for the two ways X and Y can be ordered within their adjacent pair.

This question tests counting with inclusion-exclusion principle. Order matters for strings, and we can use any of 3 letters in each of 8 positions. We want strings with at least one A and at least one B, which we can find using inclusion-exclusion: total strings minus those missing A or B. Total 8-character strings using {A,B,C} is $3^8$. Strings without A use only {B,C}, giving $2^8$ possibilities. Strings without B also give $2^8$ possibilities. Strings without both A and B use only C, giving $1^8$ = 1 possibility. By inclusion-exclusion: $3^8$ - $2×2^8$ + 1. The +1 corrects for double-subtracting strings with only C. A common error is forgetting this correction term or applying inclusion-exclusion incorrectly.

This question tests counting with order and distinctness constraints. Order matters for passwords, and we need distinct characters within each category. For 3 distinct letters from 26: first letter has 26 choices, second has 25, third has 24, giving 26×25×24 = 15,600 ways. For 2 distinct digits from 10: first digit has 10 choices, second has 9, giving 10×9 = 90 ways. Total passwords = 15,600 × 90 = 1,404,000. The key insight is recognizing this as a permutation problem (order matters) with no repetition allowed. A common error is using combinations instead of permutations, as in answer B, which incorrectly treats the password as an unordered selection.