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.

When you encounter arrangement problems with grouping restrictions, think of it as a two-step process: first arrange the groups, then arrange items within each group.

Since books of the same subject must stay together, treat each subject as a single unit. You have 3 units (history, science, math) that can be arranged in $$3! = 6$$ ways.

Next, arrange books within each subject group:

• 4 history books can be arranged in $$4! = 24$$ ways
• 3 science books can be arranged in $$3! = 6$$ ways
• 2 math books can be arranged in $$2! = 2$$ ways

Using the multiplication principle, the total arrangements are: $$3! \times 4! \times 3! \times 2! = 6 \times 24 \times 6 \times 2 = 1728$$

Answer choice A (288) represents a common error where students forget to arrange the subject groups themselves, calculating only $$4! \times 3! \times 2! = 288$$. Choice B (864) occurs when students incorrectly calculate one of the factorials or miss a step in the multiplication. Choice D (362,880) is what you'd get if you ignored the grouping restriction entirely and arranged all 9 books freely: $$9! = 362,880$$.

The correct answer is C (1728).

Remember this pattern: for grouped arrangement problems, multiply the arrangements of the groups by the arrangements within each group. Always account for both levels of organization—the macro arrangement and the micro arrangements.

When you encounter "at least" problems in combinatorics, break them down systematically by considering all valid cases separately. This question asks for selections with "at least 2 SUVs," meaning exactly 2, 3, or 4 SUVs.

Let's calculate each case using combinations. With 6 sedans and 5 SUVs available:

Case 1: Exactly 2 SUVs (and 2 sedans)

$$\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$$

Case 2: Exactly 3 SUVs (and 1 sedan)

$$\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$$

Case 3: Exactly 4 SUVs (and 0 sedans)

$$\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$$

Total: $$150 + 60 + 5 = 215$$

Choice A (115) represents a common error where students only calculate one case, likely the 2 SUVs + 2 sedans scenario but make computational mistakes. Choice B (150) is exactly the first case alone—students who miss that "at least 2" means multiple scenarios. Choice D (330) is the total number of ways to choose any 4 vehicles from 11, ignoring the SUV constraint entirely: $$\binom{11}{4} = 330$$.

Strategy tip: For "at least/at most" combination problems, always list out each valid case explicitly before calculating. Alternatively, you could use the complement method: total selections minus selections with 0 or 1 SUVs. Both approaches should yield the same answer and serve as a good check.

When you encounter overlapping sets problems like this, you're dealing with the inclusion-exclusion principle. The key insight is that when you add up individual group sizes, you're double-counting students in multiple subjects, so you need to systematically account for all overlaps.

Let's use the inclusion-exclusion formula. If we let C = Chemistry, P = Physics, and B = Biology, and x = students in all three subjects, we can write: $$|C \cup P \cup B| = |C| + |P| + |B| - |C \cap P| - |P \cap B| - |C \cap B| + |C \cap P \cap B|$$

Substituting our values: $$80 = 45 + 35 + 20 - 15 - 10 - 8 + x$$

Simplifying: $$80 = 100 - 33 + x = 67 + x$$

Therefore: $$x = 80 - 67 = 13$$

Now let's examine why the wrong answers are incorrect. Choice (A) 3 would mean we're subtracting too much from our total, suggesting insufficient overlap between all three subjects given the pairwise overlaps. Choice (B) 7 similarly underestimates the necessary triple overlap. Choice (D) 20 equals the total Biology enrollment, which would mean every Biology student is also in Chemistry and Physics—an impossibly high overlap given the constraints.

The correct answer is (C) 13.

Strategy tip: For overlapping sets problems, always write out the inclusion-exclusion formula first, then substitute known values. Remember that pairwise intersections include the triple intersection, so the formula automatically accounts for this relationship. Practice recognizing when the given totals require calculating an unknown intersection.

When you encounter combinatorics problems with restrictions, break them down by cases that satisfy the constraint. This problem asks for teams with "at most 2 designers," meaning 0, 1, or 2 designers allowed.

Calculate each valid case separately using combinations:

Case 1: 0 designers, 5 developers

Choose 5 from 7 developers: $$\binom{7}{5} = 21$$

Case 2: 1 designer, 4 developers

Choose 1 from 5 designers AND 4 from 7 developers: $$\binom{5}{1} \times \binom{7}{4} = 5 \times 35 = 175$$

Case 3: 2 designers, 3 developers

Choose 2 from 5 designers AND 3 from 7 developers: $$\binom{5}{2} \times \binom{7}{3} = 10 \times 35 = 350$$

Total valid teams: $$21 + 175 + 350 = 546$$

Choice A (246) likely comes from miscalculating one of the combination formulas or forgetting a case entirely. Choice B (350) represents only Case 3—students might mistakenly think "at most 2" means "exactly 2." Choice D (792) appears to be the total possible 5-person teams without any restrictions: $$\binom{12}{5} = 792$$.

The correct answer is C (546).

Strategy tip: For "at most" or "at least" problems, always list out all valid cases explicitly before calculating. Double-check that you're using the multiplication principle correctly when selecting from multiple groups simultaneously, and remember that "at most X" includes all values from 0 up to X.

This is a combinations with repetition problem, where you need to count selections when items can be repeated and order doesn't matter. When you see questions about selecting items from categories with unlimited supply, think "stars and bars" - a classic combinatorics technique.

You're choosing 6 cards from 4 types where repetition is allowed. This is equivalent to finding the number of ways to distribute 6 identical items into 4 distinct categories. The formula is $$\binom{n+k-1}{k}$$ where $$n$$ is the number of types (4) and $$k$$ is the number of items selected (6).

So we calculate $$\binom{4+6-1}{6} = \binom{9}{6} = \frac{9!}{6!(9-6)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$

Choice A (15) represents $$\binom{6}{2}$$, which might come from incorrectly applying the standard combinations formula. Choice C (126) equals $$\binom{9}{3}$$, which you'd get by mistakenly using $$\binom{9}{k-1}$$ instead of the correct formula. Choice D (4096) equals $$4^6$$, which would be correct if order mattered and you were counting arrangements rather than selections.

The key trap here is recognizing that this isn't ordinary combinations (since repetition is allowed) nor permutations (since order doesn't matter for a "selection"). When you see unlimited supply problems, always consider whether the stars and bars method applies - it's frequently tested on the GRE quantitative section.

When you encounter counting problems with restrictions, break them down systematically by identifying the constraints first. This question involves forming 4-digit numbers with two key restrictions: no repeated digits and the number must be even.

For a number to be even, it must end in 0, 2, or 4. Let's count by cases based on the last digit:

Case 1: Ends in 0

• Last position: 1 choice (0)
• First position: 4 choices (1, 2, 3, 4 — can't be 0 since it's used, and can't start with 0 anyway)
• Second position: 3 remaining choices
• Third position: 2 remaining choices
• Total: $$4 \times 3 \times 2 \times 1 = 24$$

Case 2: Ends in 2

• Last position: 1 choice (2)
• First position: 3 choices (1, 3, 4 — can't be 0, and 2 is used)
• Second position: 3 remaining choices (including 0 now)
• Third position: 2 remaining choices
• Total: $$3 \times 3 \times 2 \times 1 = 18$$

Case 3: Ends in 4

• Same logic as Case 2: $$3 \times 3 \times 2 \times 1 = 18$$

Total even 4-digit numbers: $$24 + 18 + 18 = 60$$

Choice (A) 36 likely miscounts by forgetting one of the even-ending cases. Choice (B) 48 might incorrectly handle the restriction about not starting with 0. Choice (D) 96 probably ignores the "even" requirement entirely.

Strategy tip: In counting problems with multiple constraints, organize by the most restrictive condition first (here, the even requirement), then apply remaining restrictions systematically to avoid missing cases.