Understanding GRE Counting Methods: Concepts, Examples & Tips

Counting methods in the GRE are used to calculate the number of elements in a finite set of objects. Sets containing a small number of objects/elements can be listed out and counted easily. For sets that are too large and contain objects arranged in a pattern or systematic way, special counting techniques are used to determine the total without listing each object individually. Counting questions are an integral part of the GRE Data Analysis topic that tests the candidate’s critical thinking and problem solving skills, among other things. Read on to explore various GRE Counting Methods and learn how to effectively apply them to solve counting problems.

Key Counting Concepts on the GRE

Counting is a vital part of the GRE Quantitative Reasoning section that falls under the broader topic of Data Analysis. Let’s take a look at the key counting concepts that are used to solve problems in this particular section.

Sets

A set is a collection of distinct objects with no particular order. The objects of a set are called elements or members.

Properties of sets:

• Sets with finite or quantifiable members are listed with curly brackets. 
• Example: {2,4,6,8,10} are members of a set of even numbers, with members that are finite.
• Sets that do not have finite members are infinite sets. Infinite sets denoted by ellipsis (. . . .) at the end indicating that set goes on indefinitely.
• Example: {1,2,3,4,5, . . . } are an infinite set of all natural numbers.
• Empty sets are sets that contain no elements at all. Empty sets are denoted by Ø or {}. It signifies that there are no members or elements in the set. 
• When a set contains members that are repetitive, those are not counted as additional elements. 
• Example: Sets { 1,3,4,2,3 } and { 1,3,4,2 } are the same containing 4 elements.
• The number of elements in set S is denoted by |S|. Example: If S = {1,0,6,-2,5}, then |S| = 5. 
• Sets can also be formed by other sets. 
• The intersection of two sets, S and T, is denoted by S ∩ T, and it consists of all elements that are in both S and T.
• The union of two sets, S and T, is denoted by S ∪ T, and it consists of all elements that are in either S or T.
• Disjoint sets, also called mutually exclusive sets, have no elements in common.
• A subset of the set S is any set whose members are contained in the set S. For example, the empty set is a subset of every set.

Lists

A list is similar to a finite set consisting of an ordered set of elements. 

Properties of lists:

• Members of a list are always ordered. Changing the order of members creates a completely different list. 
• Example: the list { 1, 2, 3, 4 } and { 4, 2, 3, 1 } are different.
• Members of a list can be repeated more than once.
• Lists have a fixed length with a specific number of elements.
• The placement of elements affects the list’s identity.

Inclusion - Exclusion Principle

The Inclusion-Exclusion principle states that the number of elements in the union is equal to the sum of their individual elements, minus the number of elements in their intersection.

Example: If sets A and B intersect, then

| A ∪ B | = |A| + |B| - |A ∩ B|

If sets A and B are disjoint, then

| A ∪ B | = |A| +|B|

because | A ∪ B | = 0.

Multiplication Principle

The multiplication principle is a way to find the total number of combinations by multiplying choices together. Assume that the two choices are to be made sequentially, and the second choice is independent of the first choice. 

There are k different possibilities for the first choice. 

There are m different possibilities for the second choice. 

As per the multiplication method, there are km different possibilities for making both choices.

Example: 

If a coin is tossed two times, how many possible outcomes will there be?

On each toss, there will be 2 possible outcomes: head or tail

The 1st toss has 2 outcomes, and the 2nd toss has 2 outcomes.

There are 4 possible outcomes: 2*2 = 4

If the coin is tossed four times, there will be 2*2*2*2 = 16 possible outcomes.

Factorials

A factorial is a product of all positive integers from 1 to n. They are used to determine the possible number of ways to arrange items. It is denoted by n!. 

n! = n*(n - 1)*(n - 2)*. . . *3*2*1 

Example: 5! = 5*4*3*2*1 = 120.

This means that there are 120 ways to arrange 5 distinct objects.

Permutation and Combination

Permutation: Permutation refers to the arrangement of objects in a specific order, where the sequence of elements is important. 

To find the number of ways to arrange r objects from a total of n objects, the formula is:

P( n, r ) = n! / ( n - r )!

Where,

• n = Total number of objects

• r = Number of objects being arranged

• n! = Factorial of n, which is the product of all positive integers up to n.

• ( n - r )! = Factorial of the difference between n and r. 

Example: If you have 5 different books and want to arrange 3 of them on a shelf, what would the permutation calculation be?

P(5,3) = 5! / (5-3)! = 5*4*3*2*1 / 2* 1 = 60

Hence, there are 60 ways to arrange 3 books out of 5 when the order matters. 

Combination: A combination is a selection of items from a larger set where the order of objects does not matter. 

To calculate how many ways you can choose r objects from n without considering the order, the formula is:

C(n, r) = n! / r!*(n - r)!

Example: If you have  10 students and need to choose 3 to form a committee and the order of selection does not matter:

C(10, 3) = 10! / 3!*(10 - 3)! = 10! / 3!*7! = 10*9*8 / 3*2*1 = 120

Hence, there are 120 ways to choose 3 students out of 10 when the order does not matter. 

Related Blog: GRE Permutation and Combination

Practice Questions on GRE Counting Methods

Here are some practice questions on GRE counting to help you test your knowledge and get prepared for the exam. 

Q1. Set A contains all prime numbers less than 10. List the elements of set A.

Solution: Prime numbers less than 10 are 2, 3, 5, and 7. 

Thus, the set A is: A = {2, 3, 5, 7}. 

Q2. Given two sets A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find the union of sets A and B. 

Solution: The union of two sets is a set containing all distinct elements from both sets.  Therefore, A ∪ B = {1, 2, 3, 4, 5, 6}. 

Q3. You have 3 books: Maths, Science, and History. How many unique lists can you create by arranging these 3 books?

Solution: As the order of the books matter, the number of unique lists is:

3! = 3*2*1 = 6

Unique lists are: [Math, Science, History], [Math, History, Science], [Science, Math, History], [Science, History, Math], [History, Math, Science], [History, Science, Math] 

Q4. There are 4 students standing in line: A, B, C, and D. If the order in which they are standing matters, how many unique arrangements can be made?

Solution: As the order matters, the number of unique arrangements is:

4! = 4*3*2*1 = 24

Q5. In a class, 30 students play basketball, and 20 play soccer. If 10 students play both sports, how many students play either basketball or soccer?

Solution: Using the inclusion-exclusion principle:

∣A∪B∣=∣A∣+∣B∣−∣A∩B∣= 30+20−10 = 40

Hence, 40 students play either basketball or soccer.

Q6. If you have 3 shirts and 4 pairs of pants, how many different outfits can you create?

Solution: Using the multiplication principle:

3*4 = 12

There are 12 possible outfit combinations. 

Q7. How many ways can you arrange 5 different books on a shelf?

Solution: To find the number of arrangements, calculate the factorial of 5:

5! = 5*4*3*2*1 = 120

There are 120 ways to arrange the books.

Q8. You have 5 letters (A, B, C, D, E) and want to arrange 3 of them. How many different arrangements are possible?

Solution: Using the permutation formula:

P(5,3) = 5! / (5 - 3)! = 5! / 2! = 5*4*3 / 1 = 60

There are 60 different arrangements.

Q9. You have 7 different fruits and want to choose 3 for a fruit basket. How many combinations are possible?

Solution: Using the combination formula:

C(7,3) = 7! / 3!*(7-3)! = 7*6*5 / 3*2*1 = 35

There are 35 ways to choose 3 fruits.

Q10. From a group of 10 students, how many ways can you select 4 to form a committee?

Solution: Using the combination formula: 

C(10,4) = 10! / 4!*(10-4)! = 10*9*8*7 / 4*3*2*1 = 210

There are 210 ways to form the committee.

How to avoid common mistakes in GRE Counting?

Students often find the GRE counting questions quite tricky, making it difficult to answer the questions correctly. Some of the frequent mistakes and tips on avoiding them are given below to help you tackle GRE counting problems effectively.

• Misinterpreting the Question: Some questions have subtle wording that could be easily misinterpreted. Students should read each question carefully and identify key terms to understand the question as intended.

• Using Incorrect Formulas: Students often mix up formulas, in particular for permutations and combinations. It’s important to remember the key differences between concepts and apply the right formulas.

• Double-Counting: While counting indistinguishable items or overlapping sets, students sometimes tend to double-count. To avoid such mistakes, it’s best to carefully analyze and implement the respective counting strategies.

• Ignoring Restrictions in the Question: Some questions have certain restrictions that need to be applied. It could be placement or exclusion of certain elements. Take note of such restrictions and ensure you implement them. 

Mastering the basics of GRE counting methods is crucial for handling challenging GRE questions. This post has thoroughly explored the fundamental concepts, complete with examples, practice problems, and expert tips to help you succeed.