GMAT Permutation & Combination Practice Questions with Answers

Within the GMAT Quantitative section, Permutation and Combination serve as core topics in combinatorics and probability. Grasping these two concepts is imperative because they revolve around how items can be selected or ordered, which frequently appears across different problem types on the GMAT.

• Permutation: This denotes the ordering of objects in a specific sequence. For instance, arranging the letters A, B, and C in different sequences yields permutations like ABC, ACB, BAC, and so forth.
• Combination: This denotes the picking of objects without regard to their sequence. For instance, choosing two letters from the set containing A, B, and C produces the combinations AB, AC, and BC.

Both concepts are essential for comprehending data structures, and mastering them is vital for solving GMAT questions that involve arrangement and selection. Ashiraj Prep provides comprehensive materials to assist you in mastering the Quantitative portion of the GMAT, covering both Permutation and Combination.

To resolve GMAT problems concerning Permutation and Combination, it is essential to understand the following fundamental formulas:

Permutation Formula: Pn,r=n!n−r!

Where:

• n = total number of objects
• r = number of objects to arrange
• ! denotes factorial (e.g., 5!=54321)

Example:If you want to arrange 3 objects out of 5, you would calculate:

P5,3=5!5−3!=5!2!=5431=60

So, there are 60 ways to arrange 3 objects from 5.

Combination Formula: Cn,r=n!r!n−r!

Where:

• n = total number of objects
• r = number of objects to select
• ! denotes factorial

Example:If you want to select 3 objects from a set of 5, you would calculate: C5,3=5!3!5−3!=543321=10So, there are 10 ways to select 3 objects from 5.

Here is a collection of 15 GMAT-style questions covering Permutation and Combination along with detailed solutions:

Question 1: How many ways can 5 people be arranged in a row?

Solution:This is a basic permutation problem. We are arranging all 5 people in a specific order. Using the permutation formula:

P5,5=5!5−5!=5!=54321=120

So, the number of ways to arrange 5 people in a row is 120.

Question 2: In how many ways can a committee of 3 people be selected from a group of 10 people?

Solution:This is a combination problem because the order doesn’t matter. We use the combination formula:

C10,3=10!3!10−3!=1098321=120

So, there are 120 ways to select 3 people from 10.

Question 3: How many 4-letter words can be formed using the letters A, B, C, D, and E?

Solution:This is a permutation problem where we are arranging 4 letters from a set of 5. Using the permutation formula:

P5,4=5!5−4!=5!=120

So, 120 distinct 4-letter words can be formed.

Question 4: How many different 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition?

Solution:This is another permutation problem where we are selecting 3 digits from 6 and arranging them. Using the permutation formula:

P6,3=6!6−3!=654=120

So, there are 120 possible 3-digit numbers.

Question 5: How many ways can 4 men and 3 women be arranged in a row such that all the men sit together?

Solution:Treat all 4 men as one unit. So, we have 5 units (the 4 men as one unit + 3 women). The number of ways to arrange these 5 units is:

P5,5=5!=120

Within the men’s unit, the 4 men can be arranged in 4!=24 ways.Thus, the total number of arrangements is:

5!4!=12024=2880

So, there are 2880 ways to arrange them.

Question 6: How many ways can you select 2 red balls from 5 red balls and 3 blue balls?

Solution:This is a combination problem because we are selecting 2 red balls from 5 red balls. Using the combination formula:

C5,2=5!2!5−2!=5421=10

So, there are 10 ways to select 2 red balls.

Question 7: How many ways can 4 different books be arranged on a shelf?

Solution:This is a simple permutation problem where we are arranging 4 distinct objects (books) on a shelf. The number of ways to arrange 4 books is:

P4,4=4!=4321=24

So, the number of ways to arrange 4 different books is 24.

Question 8: How many ways can a 4-digit number be formed using the digits 1, 2, 3, 4, 5, 6 without repetition?

Solution:This is a permutation problem where we need to arrange 4 digits out of 6. The number of ways to arrange the digits is:

P6,4=6!6−4!=6543=360

So, 360 different 4-digit numbers can be formed.

Question 9: In how many ways can 3 women be arranged in a line if they cannot sit next to each other?

Solution:First, calculate the total number of ways to arrange all 3 women and 4 men without any restrictions:

P7,7=7!=5040

Now, calculate the number of ways in which all 3 women sit together. Treat the 3 women as a single unit, which gives us 5 units (4 men + 1 women’s unit). The number of ways to arrange these 5 units is:

P5,5=5!=120

Within the women’s unit, the 3 women can be arranged in 3!=6 ways.Thus, the number of ways in which all 3 women sit together is:

1206=720

Finally, subtract this from the total number of unrestricted arrangements:

5040−720=4320

So, there are 4320 ways in which 3 women can be arranged such that they are not sitting next to each other.

Question 10: How many ways can you form a team of 2 men and 2 women from a group of 5 men and 6 women?

Answer: This involves a combination problem where 2 men and 2 women are chosen. The combinations are computed individually:

• Number of ways to pick 2 men out of 5:

C5,2=5!2!5−2!=10

• Number of ways to pick 2 women out of 6:

C6,2=6!2!6−2!=15

Hence, the overall number of ways to create the team is:

1015=150

Thus, there are 150 ways to put together a team of 2 men and 2 women.

Question 11: In how many ways can 3 vowels be picked from the 5 vowels A, E, I, O, U?

Answer: This is a combination problem where 3 vowels are selected from a group of 5. Applying the combination formula:

C5,3=5!3!5−3!=543321=10

Therefore, there are 10 ways to choose 3 vowels.

Question 12: In how many ways can 4 objects be arranged in a circle?

Answer: When items are arranged in a circular formation, the quantity of arrangements is expressed by:

Pn,n=n−1!1=n−1!

For 4 items:

P4,4=3!=6

So, there are 6 ways to arrange 4 items in a circle.

Question 13: In how many different ways is it possible to choose 2 books from a set of 5 and arrange them on a bookshelf?

Answer: This involves a combination followed by a permutation. Initially, 2 books are selected from 5 using the combination equation:

C5,2=5!2!5−2!=10

Next, the 2 chosen books are arranged:

2!=2

Consequently, the combined number of ways to pick and arrange the books is:

102=20

As a result, there are 20 ways.

Question 14: How many ways exist to deal 5 cards out of a standard 52-card deck?

Answer: This is a combination scenario where 5 cards are selected from 52. Utilizing the combination equation:

C52,5=52!5!52−5!=525150494854321=2598960

Thus, there are 2,598,960 ways to deal 5 cards from a deck of 52.

Question 15: How many ways are there to organize the letters of the word "LEVEL"?

Answer: This is a permutation problem featuring duplicate items. The quantity of unique arrangements is given by:

n!k1!k2!

Where n represents the total number of items, and k1,k2, denote the frequencies of recurring items. For "LEVEL":

P5,5=5!2!2!=5432122=30

Therefore, 30 distinct ways exist to order the letters of the word "LEVEL".

Here are the refined strategies to tackle GMAT Permutation and Combination problems proficiently:

1. Comprehend the Problem Category

Start by identifying if the question deals with ordering items (permutation) or picking items regardless of sequence (combination).

2. Recognize Essential Components:

• For Permutations: Concentrate on the overall quantity of items (n) and how many are being arranged (r).
• For Combinations: Concentrate on the overall quantity of items (n) and how many are being selected (r) irrespective of order.

3. Apply the Appropriate Formula:

• Permutation Equation: Pn,r=n!n−r!
• Combination Equation: Cn,r=n!r!n−r!

4. Spot Special Conditions:

• Look out for unique conditions, such as identical objects or circular placements, which may necessitate equation adjustments:
• Repeated objects: Apply the formula that accounts for duplicates.
• Circular layouts: Use n−1! in place of n! for circular configurations.

5. Compute Factorials: Factorials form a vital part of both formulas, so practice calculating n! and n−r! swiftly and precisely.

6. Simplify the Expression: Following formula application, reduce the factorial expressions to streamline and accelerate calculations.

7. Double-Check Your Solution: Verify whether the problem demands ordered arrangements (permutations) or unordered selections (combinations) and ensure the final result fits the problem context.