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12+ Permutation Questions for CAT with SOLUTIONS

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Permutation Questions for the CAT are part of the Quantitative Aptitude section in the CAT exam. Through Permutation questions, aspirants are tested to arrange objects in order. The difficulty level of the Time & Work questions can be easy to moderate.

Tips to improve Permutation Questions for CAT?

  • Inclusion-Exclusion Principle
  • Using Symmetry can make your calculations more simple
  • You can use Multiset Permutations if the question is repeating.
     

Permutation Questions Tricks for the CAT Exam

  • Understand what's the question is about. Identify the problem is about permutation or a subset.
  • Break the problem into subparts.
  • If some problems are identical adjust permutation by dividing by the factorials of frequencies.

Permutation Questions Concepts for the CAT Exam

Permutations of n distinct object: p (n) = n!.

Permutations of n objects taken r at a time : P (n,r) = n!/(n-r)!.

Permutation with repetition: For n objects where some objects are identical, the number of permutations is n! upon n1xn2x....xnk where n1, n2, ...,nk are the identical objects frequencies.

What are some CAT level Permutation Practice Questions?

Q1. In how many different ways can the letters of the word DISPLAY be arranged?

 (a) 720 
 (b) 1440
 (c) 2520 
 (d) 5040
 (e) None of these

Level: Easy

Answer: (d) 5040.

Solution: To find the number of unique arrangements of the letters in DISPLAY:

There are 7 letters in total.

There is no repetition of any letter, each letter appears only once.

As all letters are unique, we simply calculate 7!:

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040

So, the correct answer is (d) 5040.

Q2. In how many different ways can the letters of the word SMART be arranged?

 (a) 25 
 (b) 60
 (c) 180 
 (d) 200
 (e) None of these

Level: Moderate

Answer: (e) None of these.

Solution: Let's calculate the number of unique arrangements of the letters in SMART:

There are 5 letters in total.

All letters appear only once.

Since all letters are unique, we simply calculate 5!:

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

Therefore, the correct answer is (e) None of these.

Q 3. In how many different ways can the letters of the word FORMULATE be arranged?

 (a) 8100 
 (b) 40320
 (c) 153420 
 (d) 362880
 (e) None of these

Level: Easy

Answer: (e) None of these.

Solution: Let's find the number of unique arrangements of the letters in FORMULATE:

There are 9 letters in total.

The letter O appears twice, and the letter U appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 9! = 362880

Divide by the number of ways to arrange the Os (2! = 2) and the number of ways to arrange the Us (2! = 2) to account for the repeated letters.

9! / (2! * 2!) = 362880 / (2 * 2) = 362880 / 4 = 90720

So, the correct answer is (e) None of these.

Q 4. In how many different ways can the letters of the word GAMBLE be arranged?

 (a) 15 
 (b) 25
 (c) 60 
 (d) 125
 (e) None of these

Level: Easy

Answer: E

What are the must-do Permutation questions for the CAT exam?

Q5. In how many different ways can the letters of the word RIDDLED be arranged?

 (a) 840 
 (b) 1680
 (c) 2520 
 (d) 5040
 (e) None of these

Level: Difficult

Answer: (a) 840.

Solution: Let's find the number of unique arrangements of the letters in RIDDLED:

There are 7 letters in total.

The letter D appears three times, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 7! = 5040

Divide by the number of ways to arrange the Ds (3! = 6) to account for the repeated Ds.

7! / 3! = 5040 / 6 = 840

So, the correct answer is (a) 840.

Q6. In how many different ways can the letters of the word CREATE be arranged?

(a) 25
(b) 36
(c) 360
(d) 720
(e) None of these

Level: Moderate

Answer: (c) 360.

Solution: To calculate the number of unique arrangements of the letters in CREATE:

There are 6 letters in total.

The letter E appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 6! = 720

Divide by the number of ways to arrange the Es (2! = 2) to account for the repeated Es.

6! / 2! = 720 / 2 = 360

Therefore, the correct answer is (c) 360.

Q7. In how many different ways can the letters of the word TOTAL be arranged?

(a) 45
(b) 60
(c) 72
(d) 120
(e) None of these

Level: Moderate

Answer: (b) 60.

Solution: Let's find the number of unique arrangements of the letters in TOTAL:

There are 5 letters in total.

The letter T appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 5! = 120

Divide by the number of ways to arrange the Ts (2! = 2) to account for the repeated Ts.

5! / 2! = 120 / 2 = 60

So, the correct answer is (b) 60.

Q8. In how many different ways can the letters of the word OFFICES be arranged?

(a) 2520
(b) 5040
(c) 1850
(d) 1680
(e) None of these

Level: Moderate

Answer: (e) None of these.

Solution: Let's calculate the number of unique arrangements of the letters in OFFICES:

There are 7 letters in total.

The letter F appears twice, and the letter O appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 7! = 5040

Divide by the number of ways to arrange the Fs (2! = 2) and the number of ways to arrange the Os (2! = 2) to account for the repeated letters.

7! / (2! * 2!) = 5040 / (2 * 2) = 5040 / 4 = 1260

The correct answer is (e) None of these.

What were the previous year CAT Permutation  questions?

Q9. In how many different ways can the letters of the word BANANA be arranged?

 (a) 60 
 (b) 120
 (c) 360 
 (d) 720
 (e) None of these

Level: Difficult

Answer: (a) 60.

Solution: To find the number of unique arrangements of the letters in BANANA:

There are 6 letters in total.

The letter A appears three times and the letter N appears twice, while B appears once.

Calculate the total number of arrangements if all letters were different: 6! = 720

Divide by the number of ways to arrange the As (3! = 6) and the number of ways to arrange the Ns (2! = 2) to account for the repeated letters.

6! / (3! * 2!) = 720 / (6 * 2) = 720 / 12 = 60

Therefore, the correct answer is (a) 60.

Q10. In how many different ways can the letters of the word WEDDING be arranged?

 (a) 2500 
 (b) 2520
 (c) 5000 
 (d) 5040
 (e) None of these

Level: Moderate

Answer: (b) 2520.

Solution: Let's find the number of unique arrangements of the letters in WEDDING:

There are 7 letters in total.

The letter D appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 7! = 5040

Divide by the number of ways to arrange the Ds (2! = 2) to account for the repeated Ds.

7! / 2! = 5040 / 2 = 2520

Therefore, the correct answer is (b) 2520.

Q11. In how many different ways can the letters of the word INCREASE be arranged?

 (a) 40320 
 (b) 10080
 (c) 20160 
 (d) 64
 (e) None of these

Level: Easy

Answer: (c) 20160.

Solution: To calculate the number of unique arrangements of the letters in INCREASE:

There are 8 letters in total.

The letter E appears twice, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 8! = 40320

Divide by the number of ways to arrange the Es (2! = 2) to account for the repeated Es.

8! / 2! = 40320 / 2 = 20160

So, the correct answer is (c) 20160.

Q12. In how many different ways can the letters of the word ABSENTEE be arranged?

 (a) 512 
 (b) 6720
 (c) 9740 
 (d) 40320
 (e) None of these

Level: Moderate

Answer: (b) 6720.

Solution: To find the number of unique arrangements of the letters in ABSENTEE:

There are 8 letters in total.

The letter E appears three times, while the other letters appear once each.

Calculate the total number of arrangements if all letters were different: 8! = 40320

Divide by the number of ways to arrange the Es (3! = 6) to account for the repeated Es.

8! / 3! = 40320 / 6 = 6720

Therefore, the correct answer is (b) 6720.

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