Hard GRE Combinations Questions Practice Questions

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. Unlike permutations, where the sequence is critical, combinations focus solely on the membership of the group. For example, selecting a committee of three people from a group of ten is a combination problem because the order in which you pick the members doesn't change the composition of the committee. On the GRE Prep journey, you will encounter the standard formula for combinations, which is expressed as:

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

In this formula, $n$ is the total number of items, and $r$ is the number of items being chosen. Hard GRE combinations questions often involve additional constraints, such as "at least" or "at most" conditions, or the requirement to select from multiple distinct groups simultaneously. To solve these, you must often calculate multiple individual combinations and then either add them (if the scenarios are mutually exclusive) or multiply them (if the selections are independent steps in a single process). Familiarizing yourself with GRE practice questions with explanations can help clarify when to use each approach.

Solved Examples

1. Example 1: Multiple Groups
   A research team must be formed by selecting 3 biologists from a pool of 6 and 2 chemists from a pool of 5. How many different teams can be formed?
   
   1. Calculate the ways to choose biologists: $\ \binom{6}{3} = \ \frac{6 \ \times 5 \ \times 4}{3 \ \times 2 \ \times 1} = 20$.
   2. Calculate the ways to choose chemists: $\ \binom{5}{2} = \ \frac{5 \ \times 4}{2 \ \times 1} = 10$.
   3. Multiply the results because both actions must occur: $20 \ \times 10 = 200$.
2. Example 2: The "At Least" Constraint
   A committee of 4 is to be chosen from 5 men and 4 women. How many committees contain at least 3 women?
   
   1. Identify the scenarios: (3 women and 1 man) OR (4 women and 0 men).
   2. Scenario A: $\ \binom{4}{3} \ \times \ \binom{5}{1} = 4 \ \times 5 = 20$.
   3. Scenario B: $\ \binom{4}{4} \ \times \ \binom{5}{0} = 1 \ \times 1 = 1$.
   4. Add the scenarios: $20 + 1 = 21$.
3. Example 3: Complementary Counting
   There are 7 distinct points on a circle. How many triangles can be formed using these points as vertices, such that a specific point P is NOT included?
   
   1. There are 7 points total. To exclude point P, we only choose from the remaining 6 points.
   2. We need 3 points to form a triangle.
   3. Calculate the combination: $\ \binom{6}{3} = \ \frac{6 \ \times 5 \ \times 4}{3 \ \times 2 \ \times 1} = 20$.

Practice Questions

1. A basket contains 8 apples and 6 oranges. In how many ways can a person select 5 fruits such that there are exactly 3 apples and 2 oranges?

2. A debating club has 10 members. A delegation of 4 members is to be sent to a convention. However, two members, Alice and Bob, refuse to attend together. How many different delegations can be formed?

3. A pizza shop offers 10 different toppings. A customer wants to order a pizza with at least 8 toppings. How many different topping combinations are possible?

4. In a group of 12 people, there are 7 lawyers and 5 doctors. A committee of 5 is to be formed such that it contains more doctors than lawyers. How many such committees are possible?

5. How many ways can a student choose 6 courses from a list of 10 if 2 specific courses are mandatory and must be included?

6. A set of 15 points is chosen in a plane such that no three points are collinear. How many distinct lines can be drawn connecting any two points?

7. A photographer needs to choose 4 photos from a collection of 6 landscape photos and 5 portrait photos. If the selection must include at least one of each type, how many combinations are possible?

8. A company is hiring 3 interns from a pool of 8 applicants. If one specific applicant, Sarah, must be included, and another applicant, Tom, must be excluded, how many ways can the interns be selected?

Answers & Explanations

1. Answer: 840. Calculate $\ \binom{8}{3} = 56$ and $\ \binom{6}{2} = 15$. Multiply them: $56 \ \times 15 = 840$.
2. Answer: 182. Total ways to choose 4 from 10 is $\ \binom{10}{4} = 210$. Ways where Alice and Bob are both included: $\ \binom{8}{2} = 28$. Subtract: $210 - 28 = 182$.
3. Answer: 56. Scenarios are 8, 9, or 10 toppings: $\ \binom{10}{8} + \ \binom{10}{9} + \ \binom{10}{10} = 45 + 10 + 1 = 56$.
4. Answer: 81. Cases: (3D, 2L), (4D, 1L), (5D, 0L).
   (3D, 2L): $\ \binom{5}{3} \ \times \ \binom{7}{2} = 10 \ \times 21 = 210$ (Wait, the question asks for more doctors than lawyers).
   $\ \binom{5}{3} \ \times \ \binom{7}{2} = 210$.
   $\ \binom{5}{4} \ \times \ \binom{7}{1} = 5 \ \times 7 = 35$.
   $\ \binom{5}{5} \ \times \ \binom{7}{0} = 1 \ \times 1 = 1$. Total: $210 + 35 + 1 = 246$.
5. Answer: 70. Since 2 are mandatory, we only need to choose 4 more from the remaining 8: $\ \binom{8}{4} = \ \frac{8 \ \times 7 \ \times 6 \ \times 5}{4 \ \times 3 \ \times 2 \ \times 1} = 70$.
6. Answer: 105. A line requires 2 points. $\ \binom{15}{2} = \ \frac{15 \ \times 14}{2} = 105$.
7. Answer: 310. Total ways to pick 4 from 11 is $\ \binom{11}{4} = 330$. Subtract cases with only landscapes: $\ \binom{6}{4} = 15$. Subtract cases with only portraits: $\ \binom{5}{4} = 5$. Result: $330 - 15 - 5 = 310$.
8. Answer: 15. Sarah is in, Tom is out. We need to choose 2 more interns from the remaining 6 applicants (8 total - Sarah - Tom). $\ \binom{6}{2} = 15$.

1. How many ways can a committee of 3 be chosen from 7 people?

• A21
• B35
• C42
• D210

Frequently Asked Questions

What is the difference between a combination and a permutation?

In a combination, the order of the items does not matter, whereas in a permutation, the sequence is significant. For instance, {A, B} and {B, A} are the same combination but two different permutations.

How do I handle "at least" questions on the GRE?

You should either sum the combinations for every scenario that meets the criteria or subtract the unwanted scenarios from the total possible combinations. Using AI-powered practice software can help you master these logic shifts.

Why do we divide by r! in the combination formula?

Dividing by $r!$ removes the redundancies created by different orderings of the same set of items. This ensures that a group is only counted once regardless of how its members were arranged.

Can nCr ever be a decimal?

No, combinations always result in a whole number because you are counting distinct ways to group physical objects. If your calculation results in a fraction, you have made an error in simplifying the factorials.

Is 0! equal to 0 or 1?

In mathematics, $0!$ is defined as 1. This convention allows the combination and permutation formulas to work consistently, such as when choosing all $n$ items from a set.