GRE Combinations Questions Practice Questions with Answers

Concept Explanation

Combinations represent the number of ways to select a subset of items from a larger set where the order of selection does not matter. In the context of the Quantitative Reasoning section of the exam, combinations differ from permutations because the sequence in which you pick the items is irrelevant; picking Alice and Bob for a committee is identical to picking Bob and Alice. This mathematical concept is fundamental for tackling probability and counting problems on the GRE Prep journey.

The standard formula for calculating combinations is expressed as "n choose r," written as $C(n, r)$ or $\binom{n}{r}$. The formula is defined as:

$$C(n, r) = \frac{n!}{r!(n - r)!}$$

Where:

• n is the total number of items in the set.
• r is the number of items being chosen.
• ! denotes the factorial, which is the product of all positive integers up to that number (e.g., $4! = 4 \times 3 \times 2 \times 1 = 24$).

A helpful tip for GRE takers is to recognize the symmetry property: $C(n, r) = C(n, n-r)$. For example, choosing 8 people out of 10 to go on a trip is the same as choosing which 2 people stay home. This often simplifies calculations significantly. For those looking to sharpen their skills across different standardized tests, understanding these principles is as vital as practicing with USMLE Practice Questions with Answers or other high-stakes math assessments.

Solved Examples

1. Example 1: Basic Selection
   A basketball coach must choose 5 starting players from a roster of 12. How many different starting lineups can the coach create?
   1. Identify the total (n = 12) and the selection size (r = 5).
   2. Apply the formula: $\frac{12!}{5!(12-5)!} = \frac{12!}{5!7!}$.
   3. Expand the larger factorial and cancel: $\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$.
   4. Simplify: $12 \div (4 \times 3) = 1$, and $10 \div (5 \times 2) = 1$.
   5. Multiply the remaining terms: $11 \times 9 \times 8 = 792$.
2. Example 2: Group Constraints
   A committee of 3 must be formed from 5 men and 4 women. If the committee must contain exactly 2 women, how many ways can it be formed?
   1. Select 2 women from 4: $C(4, 2) = \frac{4 \times 3}{2 \times 1} = 6$.
   2. Select 1 man from 5 (to complete the 3-person committee): $C(5, 1) = 5$.
   3. Multiply the independent selections: $6 \times 5 = 30$.
3. Example 3: Complementary Counting
   In a group of 6 friends, how many ways can at least one person be chosen to go to the store?
   1. Calculate the total possible subsets, which is $2^n$. Here, $2^6 = 64$.
   2. Subtract the case where "zero" people are chosen (the empty set): $C(6, 0) = 1$.
   3. Final calculation: $64 - 1 = 63$.

Practice Questions

1. A bookstore has 8 different mystery novels. If a customer wants to buy 3 of them, how many different combinations of novels can the customer choose?

2. A research team of 4 members is to be selected from a pool of 10 scientists. How many different teams are possible?

3. A pizza parlor offers 7 different toppings. How many different 2-topping pizzas can be made? (Note: The order of toppings does not matter.)

4. A box contains 5 red marbles and 6 blue marbles. How many ways can a person select 2 red marbles and 2 blue marbles?

5. From a group of 10 students, 3 are to be chosen to attend a conference. However, 2 of the students are best friends and will only go if they both go together. How many ways can the group be selected?

6. A photographer needs to choose 4 photos from a set of 9 to display in a gallery. How many different sets of 4 photos can be chosen?

7. A restaurant menu offers 5 appetizers and 8 main courses. A diner wants to choose 2 appetizers and 1 main course. How many different meal combinations are possible?

8. How many different triangles can be formed by joining any three vertices of a regular hexagon?

9. A deck of cards has 5 unique cards. If you are dealt 3 cards, how many different hands are possible?

10. A quality control inspector must pick 3 lightbulbs from a batch of 15 to test. If the batch contains 2 defective bulbs, how many ways can the inspector pick a sample that contains exactly 1 defective bulb?

Answers & Explanations

1. Answer: 56
   Using the formula $C(8, 3)$, we calculate $\frac{8 \times 7 \times 6}{3 \times 2 \times 1}$. The 6 in the denominator cancels the 6 in the numerator, leaving $8 \times 7 = 56$.
2. Answer: 210
   Calculate $C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$. Simplifying gives $10 \times 3 \times 7 = 210$.
3. Answer: 21
   Calculate $C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21$.
4. Answer: 150
   First, choose the red marbles: $C(5, 2) = 10$. Next, choose the blue marbles: $C(6, 2) = 15$. Multiply them: $10 \times 15 = 150$.
5. Answer: 28
   Case 1: The two friends go. We need 1 more student from the remaining 8. $C(8, 1) = 8$. Case 2: The two friends do not go. We need 3 students from the remaining 8. $C(8, 3) = 56$. Wait, the prompt implies they go together or not at all. Summing these gives 64. Correction: If they only go if they both go, we check the case where they both go (8 ways) and the case where neither goes (56 ways). Total = 64. However, if the question implies they can't go separately, the logic holds. Let's re-verify: $C(8,1) + C(8,3) = 8 + 56 = 64$.
6. Answer: 126
   Calculate $C(9, 4) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
7. Answer: 80
   Appetizers: $C(5, 2) = 10$. Main course: $C(8, 1) = 8$. Total: $10 \times 8 = 80$.
8. Answer: 20
   A hexagon has 6 vertices. A triangle is formed by choosing any 3 vertices. $C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
9. Answer: 10
   Calculate $C(5, 3)$, which is the same as $C(5, 2)$. $\frac{5 \times 4}{2 \times 1} = 10$.
10. Answer: 156
    Choose 1 defective bulb from 2: $C(2, 1) = 2$. Choose 2 non-defective bulbs from 13: $C(13, 2) = \frac{13 \times 12}{2} = 78$. Total: $2 \times 78 = 156$.

1. Which of the following best describes the main difference between a combination and a permutation?

• ACombinations require a larger set of numbers.
• BOrder matters in permutations but not in combinations.
• CPermutations are only used for prime numbers.
• DCombinations involve multiplying by the number of items.

Frequently Asked Questions

When should I use combinations instead of permutations on the GRE?

Use combinations when the order of the items selected does not change the outcome, such as picking a team or a set of fruit. If the items have specific roles or positions, like President and Vice President, use permutations instead.

Does the GRE provide the combination formula during the test?

No, the GRE does not provide a formula sheet for the Quantitative Reasoning section. You must memorize the combination and permutation formulas and understand how to apply them to various word problems.

How can I quickly calculate factorials for large numbers?

You rarely need to calculate the full value of a large factorial; instead, write out the terms and cancel the common factors in the numerator and denominator. This "cancellation method" is much faster and reduces the risk of arithmetic errors.

What is the relationship between probability and combinations?

Combinations are often used to find the total number of successful outcomes and the total number of possible outcomes in probability problems. For instance, the probability of picking a specific group is 1 divided by the total number of combinations possible.

Are combinations tested frequently on the GRE?

While not every test includes high-level combinatorics, basic counting principles and combinations appear regularly. Mastering these ensures you are prepared for Medium and Hard difficulty questions that often appear in the second math section.