GRE Math: Permutations and Combinations Practice Questions & Tips to solve

Permutations and combinations are fundamental concepts in probability and counting that appear in the Quantitative Reasoning section of the GRE. A permutation is an arrangement of items in a specific order. In permutations, the order of the items matters. A combination is a selection of items where the order does not matter. Permutation and Combination comes under the data analysis section. While they are a small part of the overall exam, they can make a big difference in your score, especially if you are aiming for a high percentile.

The GRE typically includes 1-2 permutation and combination questions per Quantitative section. The difficulty level varies, but they are generally considered medium to hard questions compared to the rest of the Quant section. If you're wondering how to get better at permutations and combinations, check out this article to learn the differences between them, along with practice questions, formulas, solutions and tips to ace GRE.

What are Permutations and Combinations?

Permutation: A permutation is an arrangement of items in a specific order. In permutations, the order of the items matters. For example, if you have three letters A, B, and C, the permutations would be ABC, ACB, BAC, BCA, CAB, and CBA. The formula to calculate permutations is given by:

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

where n is the total number of items, and k is the number of items to arrange.

Combination: A combination is a selection of items where the order does not matter. For example, if you have three letters A, B, and C, the combinations of two letters would be AB, AC, and BC. Here, AB and BA are considered the same combination. The formula to calculate combinations is given by:

C(n,k) = n!/k!(n-k)!

where n is the total number of items, and k is the number of items to select.

Look at the differences in the table below:

GRE Permutation and Combination Practice Questions with Solutions

Check out the provided gre permutation and combination practice problems with solutions for better understanding:

Example Questions: Permutation/Combination

1. How many 3-digit numbers can be formed using the digits 1, 2, and 3 if repetition of digits is not allowed?

Possible Answers:

1. 5

2. 6

3. 8

4. 4

Correct Answer: 6

Explanation:

Since the problem asks for a specific order of digits (i.e., forming a sequence), we are dealing with permutations.

We need to create a 3-digit number, which means we need to arrange three digits in a specific order.

Since we are forming a 3-digit number, the order in which the digits are arranged matters. For example, 123 is different from 231.

Since repetition is not allowed, once a digit is used in one position, it cannot be used again in another position.

Given these conditions, this is a permutation problem because we are arranging a specific number of digits (3) in a specific order from a larger set of available digits (3).

The formula is P (n, k) = n!/(n-k)!

However, since we're using all 3 digits to form a 3-digit number, we can directly use the factorial formula n! for the full arrangement

P(3,3) = 3! = 3×2×1=6

The total number of 3-digit numbers that can be formed using the digits 1, 2, and 3 without repetition is 6.

2. In how many ways can you choose 3 scoops of ice cream from 5 different flavors, if you can choose more than one scoop of the same flavors?

Possible Answers:

1. 34

2. 38

3. 35

4. 42

Correct Answer: 35

Explanation:

Since the order in which we choose the items does not matter, we can crack this using combinations.

We have 5 different flavors of ice cream.

We need to choose 3 scoops.

We can choose more than one scoop of the same flavor.

The order in which we choose the scoops doesn't matter. For example, choosing chocolate, vanilla, and strawberry is the same as choosing strawberry, chocolate, and vanilla.

Since more than one scoop of the same flavor can be chosen, repetition is allowed.

Given these conditions, this is a combination problem with repetition ( combinations with replacement).

The formula for combinations with replacement (or repetition) is:

C(n+r-1,r) = (n+r-1)!/r!(n-1)!

When repetition is allowed, the formula for combinations with replacement is ((n+r-1)/r)

Here, n=5(Flavors) and r=3 (scoops), 

C(5+3−1,3) =C(7,3)=  7!/3!(7-3)! =  7!/3!4!

so: 5040/624    =  5040/144  = 35

The total number of ways to choose 3 scoops of ice cream from 5 flavors, with repetition allowed, is 35.

3. Quantity A: The number of possible combinations if four unique choices are made from eight possible options.

Quantity B: The number of possible permutations if two unique choices are made from seven possible options.

Possible Answers:

1. The relationship cannot be established.

2. The two quantities are equal.

3. Quantity A is greater.

4. Quantity B is greater.

Correct Answer: Quantity A is greater.

Explanation:

Combination (Quantity A):

Since the order in which we choose the items does not matter, we use the combination formula.

The formula is C (n, k) = n!/(n-k)!k!

C (8,4) = 8!/(8-4)!4!  = 8!/(4!×4!) = (8×7×6×5)/(4×3×2×1) = 70

Permutation (Quantity B):

Since the order in which the items are selected matters, we use the permutation formula.

The formula is P (n, k) = n!/(n-k)!

P (7,2) = 7!/(7-2)!  = 7!/5! = 7 ×6 = 42

In this case, Quantity A is greater.

4. How many 5-digit sequences can be formed using the digits 1, 2, 3, 4, and 5 if the sequence must start with an odd number?

Possible Answers:

1. 69

2. 82

3. 79

4. 72

Correct Answer: 72

Explanation:

Since the problem asks for a sequence, the order of the digits matters, so we are dealing with a permutation problem.

Determine the choices for the 1st digit

The first digit must be an odd number.The odd digits available are (1, 3, and 5).

Since the first digit must be one of these three odd digits, we have 3 choices for the first digit.

Arrange the Remaining Digits

After selecting the first digit, we are left with 4 digits.

• Without Replacement: Each digit can only be used once, so once the first digit is chosen, it is not available for the remaining positions. This means that for the remaining 4 digits, we need to determine how many ways we can arrange them.

• Formula: The number of ways to arrange n unique items is given by n! where n! is the product of all positive integers from 1 to n.

The remaining 4 digits can be arranged in 4! ways. 4! = 4×3×2×1 = 24

There are 24 different ways to arrange the remaining 4 digits.

Total Number of Sequences: To find the total number of 5-digit sequences, we multiply the number of choices for the first digit by the number of arrangements of the remaining digits.

Total sequences = 3×4! =3× (4×3×2×1) =3×24=72

The total number of 5-digit sequences that can be formed under the given conditions is 72.

5.   Quantity A: The number of possible combinations if five unique choices are made from twelve possible options.

Quantity B: The number of possible permutations if three unique choices are made from twelve possible options.

Possible Answers:

1. The relationship cannot be established.

2. The two quantities are equal.

3. Quantity A is greater.

4. Quantity B is greater.

Correct Answer:  Quantity B is greater.

Explanation:

Combination (Quantity A):

When the order does not matter, we use combinations. This means that selecting five items in any order is considered the same (e.g., selecting A, B, C, D, E is the same as selecting E, D, C, B, A).

Formula: C(n,k )= n!/(n-k)!k!

C (12,5) = 12!/(12-5)!5!  = 12!/(7!×5!) = (12×11×10×9×8)/(5×4×3×2×1) = 792

Permutation (Quantity B):

When the order does matter, we use permutations. This means that selecting three items in different orders counts as different permutations (e.g., selecting A, B, C is different from selecting C, B, A).

Formula P (n, k) = n!/(n-k)!

Quantity B:

P (12,3) = 12!/(12-3)!  = 12!/9! = 12 ×11×10 = 1320

In this case, Quantity B is greater.

6. How many ways can you choose 3 letters from the word "APPLE" with repetition allowed?

Possible Answers:

1. 70

2. 72

3. 35

4. 40

Correct Answer: 35

Explanation:

Since the problem asks for the number of ways to choose letters and the order in which we select the letters does not matter, we are dealing with combinations.

The problem allows for repetition of letters, meaning we can choose the same letter more than once.

The formula for combinations with replacement (or repetition) is:

C(n+r-1,r) = (n+r-1)!/r!(n-1)!

where n= 5 (distinct letters in APPLE: A, P, L, E) and r=3.

C(5+3−1,3) =C(7,3)=  7!/3!(7-3)! =  7!/3!4!

so: 210/6    =  35

The number of ways to choose 4 letters from "APPLE" with repetition allowed is 35

7. Sophia is selecting books from a shelf. If there are 45 unique books on the shelf, a selection of how many books from the options below will give the minimum number of potential book combinations?

Possible Answers:

1. 23

2. 22

3. 19

4. 20

Correct Answer: 19

Explanation:

Sophia is selecting books from a shelf of 45 unique books.

The goal is to find the selection size kkk that minimizes the number of possible combinations.

Combinations are used because the order of selecting the books does not matter.

Since the problem asks for the number of ways to select books where the order doesn't matter, we need to use the combinations formula.

Once a book is selected, it cannot be selected again(without replacement), meaning each selection is unique.

The formula is C (n, k) = n!/(n-k)!k!

n is the total number of items to choose from (in this case, 45 books).

k is the number of items to be selected.

The number of combinations C(n,k) is maximized when k is close to n/2. This is because the combination formula involves both k! and (n−k)!, and these factors are minimized when k is near the middle of the range (around n/2).

For an even n:

 k_max = n/2

For an odd n: 

k_max1,2 = (n±1)/2

Given n=45 (an odd number):

k_max1 = (45+1)/2 = 23

k_max2 = (45-1)/2 = 22

Evaluating the options:

|Option( k) | |k -  k_max1| | | k - k_max2| |

| 23 |                 | 0  |                  | 1 | 

| 22 |                 | 1  |                  | 0 |

| 21 |                 | 2  |                  | 1 | 

| 20 |                 | 3 |                   | 2 | 

| 19 |                 | 4 |                   | 3 |

k=19 is the farthest from both k_max1 = 23 and k_max2 =22

Thus, 19 gives the smallest number of potential combinations.

8. Tom is getting ready for a week-long hiking trip. He plans to pack 6 different jackets and 2 pairs of gloves. If he has 12 jackets to choose from, along with 8 different kinds of gloves, how many ways can he pack his items?

Possible Answers:

1. 3360

2. 665280

3. 39866880

4. 37255680

Correct Answer: 37255680

Explanation: Since the order in which Tom packs his jackets and gloves matters, we are dealing with permutations.

The formula for permutations P (n, k) is given by:

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

Step-by-step solution:

For the jackets:

P (12,6) = 12!/(12-6)!  = 12!/6! = 12 ×11×10 ×987 = 665280

For the gloves:

  P (8,2) = 8!/(8-2)!  = 8!/6! = 8 ×7 = 56

Total combinations: 665280 × 56 = 37255680

9. How many 5-digit sequences can be formed using the digits 1, 2, 3, and 4 without repetition?

Possible Answers:

1. 16

2. 24

3. 120

4. 8

Correct Answer: 120

Explanation: 

Since the problem asks for a specific order of digits (i.e., forming a sequence), we are dealing with permutations.

The problem specifies that digits cannot be repeated, meaning once a digit is used, it cannot be used again in the sequence.

Formula P (n, k) = n!/(n-k)!

n=5 n is the total number of items to choose from (in this case, the 4 digits: 1, 2, 3, 4)

k=4 k is the number of items to be arranged (in this case, we are arranging 4 digits).

P (5,4) =5! =54×3×2×1=120

The number of 5-digit sequences that can be formed using the digits 1, 2, 3, and 4 without repetition is 120.

10. A new clothing store offers a promotion where you can choose one of three shirts, one of five pairs of pants, and four of twelve accessories. How many possible outfit combinations are there?

Possible Answers:

1. 7425

2. 5280

3. 31680

4. 240

Correct Answer: 7425

Explanation: 

We need to determine the total number of possible outfit combinations, which includes choosing one shirt, one pair of pants, and three accessories.

The process involves calculating the combinations for accessories and then multiplying by the number of options for shirts and pants.

Step-by-step solution:

1. Calculate the number of ways to choose the accessories:

Since order does not matter for choosing accessories, we use the combinations formula: 

C (n, k) = n!/(n-k)!k!

Plugging in the values, where n=12 (total accessories) and r=3(chosen accessories):  C (12,4) = 12!/(12-4)!4! = 12!/8!4!  = (12×11×109)/(4×3×2)  = 495

2. Calculate the total number of combinations for shirts and pants:

Number of ways to choose one shirt from three options: 3

Number of ways to choose one pair of pants from six options: 5

3. Multiply the results:

Total combinations = (Number of shirts) × (Number of pants) × (Number of accessory combinations) = 3×5×495=7425

So, the correct number of possible outfit combinations is 7425.

Top 6 Strategies to Crack Permutations and Combinations Questions

Let’s look at the tips to ace GRE Permutations and Combinations (P&C) below:

• Verify that the question is about P&C: Ensure the question involves permutations and combinations by looking for keywords like "event," "outcome," "random selection," and "chance." Learn to distinguish between permutation and combination problems. This will help you apply the correct formula quickly.
• Remember Formulas: Memorizing key formulas saves time during the exam. Focus on probability, odds in favor, and odds against formulas.
• Focus on Language: Understand the problem's language to determine whether to use combinations (forming groups) or permutations (finding order). Look for clues like "first," "second," "third," or "how many different ways."
• Practice Regularly: Consistent practice helps familiarize yourself with the question style and format. Focus on medium to high difficulty questions to prepare for the toughest problems.
• Look for Additional Resources: Use prep books, online forums, tutors, and GRE courses for additional insights and strategies.
• Evaluate Mistakes: Document and review mistakes from practice tests. This will help you identify areas that need improvement and prevent repeating errors.
• Mock Tests : Take full-length practice tests under timed conditions to simulate the exam environment, which will help you manage time effectively during the actual test
• Use Shortcuts : For permutations, remember to use n! for arrangements without repetition and  n!r! for arrangements with repetition. For combinations, use nCr for selections without repetition    and and r+n−1Cr for selections with repetition

Conclusion

We hope the practice problems with solutions and tips help you ace the GRE exam. You can master these questions and improve your GRE score with focused practice and thorough preparation. You can take guidance from experts like Kanan International, we're here to help you achieve the best GRE score and support you throughout your study abroad journey.

Frequently Asked Questions (FAQs)

1. Is permutation and combination important for GRE?

Permutation and Combination (P&C) is a crucial chapter in GRE Math. Knowing how to answer a few permutation and combination questions can significantly improve your score on the GRE Quantitative exam. Answering P&C questions correctly could also determine your chances of getting admission at top universities.You can achieve your ideal GRE Quant score of 170 by mastering the fundamentals of permutations and combinations.

2. Is it easy to get a 170 on the GRE?

Achieving a perfect 170 on the GRE Quant section is challenging, as it places you in the 97th percentile, meaning only 3 out of every 100 students reach this score. The average GRE Quant score is around 152. However, with consistent practice and effective strategies, you can attain this high score.

3. What is the major difference between permutation and combination?

The difference between permutation and combination is that the order of the members is taken into consideration in permutation, whereas it is irrelevant in combination.

4. How to identify if it is permutation or combination?

For example, if you're choosing five people to be in a group where everyone has the same role, it's a combination because the order doesn't matter. But if you're picking five people and their role depends on the order you pick them, it's a permutation because the order matters.

5. How can I tackle complex permutation and combination questions effectively?

To tackle complex permutation and combination questions effectively, break the problems down into smaller manageable parts and apply the formulas step by step. Regular practice and a good understanding of advanced concepts are also essential for mastering these questions.