GRE Quiz: Counting Combinatorics

What this quiz covers

This quiz focuses on Counting Combinatorics, giving you a quick way to practice the rules, question types, and explanations that matter most for GRE.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 24
2. 48
3. 96
4. 72 (correct answer)
5. 120

Explanation: 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.

Question 2

1. 720
2. 120
3. 60 (correct answer)
4. 90
5. 30

Explanation: 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.

Question 3

1. 20
2. 15
3. 12
4. 10 (correct answer)
5. 6

Explanation: 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.

Question 4

1. 8
2. 10
3. 12 (correct answer)
4. 16
5. 24

Explanation: 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.

Question 5

1. 3
2. 7
3. 13 (correct answer)
4. 20

Explanation: 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.

Question 6

1. 246
2. 350
3. 546 (correct answer)
4. 792

Explanation: 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.

Question 7

1. 48
2. 96
3. 120
4. 72 (correct answer)
5. 24

Explanation: 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.

Question 8

1. 3^8 - 2\cdot 2^7 + 1
2. 3^8 - 2^8
3. 3^8 - 2\cdot 2^8
4. 3^8 - 2\cdot 2^8 + 1 (correct answer)
5. 2^8 - 1

Explanation: 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.

Question 9

1. $26^3\cdot 10^2=1,757,600$
2. $\binom{26}{3}\binom{10}{2}=32,500$
3. $26\cdot25\cdot24\cdot10\cdot9=1,404,000$ (correct answer)
4. $\binom{26}{3}\cdot(10\cdot9)=234,000$
5. $(26\cdot25\cdot24)\cdot\binom{10}{2}=780,000$

Explanation: 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.

Question 10

1. 288
2. 864
3. 1728 (correct answer)
4. 362,880

Explanation: 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

Question 11

1. 115
2. 150
3. 215 (correct answer)
4. 330

Explanation: 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.

Question 12

1. 15
2. 84 (correct answer)
3. 126
4. 4096

Explanation: 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.

Question 13

1. 36
2. 48
3. 60 (correct answer)
4. 96

Explanation: 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: $4 \times 3 \times 2 \times 1 = 24$

• 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: $3 \times 3 \times 2 \times 1 = 18$

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

Question 14

1. 35
2. 70
3. 105 (correct answer)
4. 126

Explanation: This is a combinations problem with restrictions, which requires careful application of the complement principle. When you encounter "refuse to work together" scenarios, the most efficient approach is to calculate the total unrestricted combinations, then subtract the forbidden cases. First, find the total number of ways to select 4 people from 9 without restrictions: $\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ Next, calculate how many committees would include both Alice and Bob. If both are on the committee, you need to select 2 more people from the remaining 7: $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$ Therefore, the number of valid committees is: $126 - 21 = 105$ Choice A (35) represents a common error of trying to count valid committees directly without proper case analysis. Choice B (70) might result from incorrectly calculating $\binom{7}{4} + \binom{7}{4}$, attempting to count committees with Alice but not Bob, plus committees with Bob but not Alice, but missing committees with neither. Choice D (126) is the trap of forgetting about the restriction entirely—this would be correct if Alice and Bob had no objection to serving together. Strategy tip: For "refuse to work together" problems, always use the complement: Total possibilities minus forbidden arrangements. This approach is more reliable than trying to count valid cases directly, which often leads to overcounting or missing scenarios.

Question 15

1. 6
2. 144 (correct answer)
3. 576
4. 5040

Explanation: When you encounter circular arrangement problems with restrictions, break them down into manageable steps using the constraint as your guide. Since no two women can sit adjacent, you must first place the 4 men around the circular table, which creates spaces between them for the women. For circular arrangements, we fix one person's position to avoid counting rotations as different arrangements, so arrange the 4 men in $(4-1)! = 3! = 6$ ways. Once the men are seated, there are exactly 4 spaces between them (since it's circular) where the 4 women must sit. This automatically ensures no two women are adjacent. The 4 women can be arranged in these 4 designated spaces in $4! = 24$ ways. Therefore, the total arrangements = $6 \times 24 = 144$. Choice A (6) represents only the arrangements of the men, ignoring the women entirely. Choice C (576) likely comes from incorrectly calculating $4! \times 4!$ without accounting for the circular nature, treating it as a linear arrangement. Choice D (5040) equals $7!$, which might result from mistakenly calculating arrangements of 7 people instead of 8, or from other computational errors that ignore both the circular constraint and the adjacency restriction. Strategy tip: For restricted circular seating problems, always place the unrestricted group first to create the framework, then arrange the restricted group within the allowable positions. Remember that circular arrangements of $n$ objects have $(n-1)!$ arrangements, not $n!$.

Question 16

1. 120
2. 3360 (correct answer)
3. 6720
4. 40,320

Explanation: When you encounter a problem about arranging letters in a word, you're dealing with permutations with repetition. The key insight is that identical letters create fewer distinct arrangements than if all letters were different. The word "ENGINEER" has 8 letters total: E appears 3 times, N appears 2 times, and G, I, R each appear once. To find distinct arrangements, you start with the total possible arrangements if all letters were different (8!), then divide by the factorial of each repeated letter's frequency to account for identical permutations. The formula is: $\frac{8!}{3! \times 2! \times 1! \times 1! \times 1!} = \frac{40,320}{6 \times 2 \times 1 \times 1 \times 1} = \frac{40,320}{12} = 3,360$ Choice A (120) represents a significant undercounting, likely from incorrectly applying a much smaller factorial or misidentifying the problem type entirely. Choice C (6,720) suggests dividing by only one of the repetition factors—perhaps accounting for the repeated E's but forgetting about the repeated N's, giving $\frac{40,320}{6} = 6,720$. Choice D (40,320) is simply 8!, which ignores repetitions entirely and treats all letters as distinct. Remember: whenever letters repeat in permutation problems, you must divide by the factorial of each letter's frequency. Double-check that you've identified all repeated elements—missing even one repetition factor will lead to an incorrect (and likely available) answer choice.

Question 17

1. 56
2. 168
3. 336 (correct answer)
4. 512

Explanation: This problem tests your understanding of multi-step counting with combinations and permutations. When you see a selection process with multiple stages, break it down step by step and identify whether order matters at each stage. The process has two distinct steps. First, you need to choose 3 people from 8 candidates for the committee. Since the order doesn't matter for committee membership, this is a combination: $C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways. Second, from these 3 committee members, you must select a chairperson and a secretary. Since these are different roles, order matters here—this is a permutation. You have 3 choices for chairperson, then 2 remaining choices for secretary, giving you $3 \times 2 = 6$ ways to assign these positions. Using the multiplication principle, the total number of ways is $56 \times 6 = 336$. Choice A (56) represents only the first step—selecting the committee but forgetting to assign the leadership roles. Choice B (168) likely comes from incorrectly calculating the permutation step as $C(3,2) = 3$ instead of $P(3,2) = 6$, then multiplying $56 \times 3$. Choice D (512) might result from treating the entire process as if order matters throughout, such as calculating $P(8,3) \times 2$. Remember: when facing multi-step counting problems, identify each stage separately and determine whether order matters at that particular step. Then multiply the results together.