Permutation and Combinations Problems with Solutions for SSC and Bank Exams

Permutation and combinations problems are not difficult to solve in aptitude. You just need to learn the correct way to solve problems. Here in this blog, you can resolve your permutation and combinations problems with solutions for ssc or bank exams. Learn and practice these permutation and combinations problems with solutions via different examples. So, improve your performance level with practice permutation and combination questions with answers for better rank in your competitive exams.

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Practice Permutation & Combination Problems with Solutions

$$ Q.1. The \ value \ of \ = \ ^{75}P_2 \ is  $$

(A) 2775                (B) 150                  (C) 5550               (D) None of these

Solution:

$$ ^{75}P_2 = {75!\over(75-2)!}={75!\over73!}={75× 74× 73!\over (73!)} = (75× 74)=5550.$$

Q.2. How many 4 letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed ?

(A) 40                    (B) 400                  (C) 5040                (D) 2520

Solution:

‘LOGARITHM’ contains 10 different letters.

Required number of word

 = Number of arrangements of 10 letters, taking 4 at a time

=  ¹⁰P₄ = (10×9×8×7)=5040

Q.3. How many words with or without meaning, can be formed by using all the letters of the word, ‘DELHI’, using each letter exactly once?

(A) 10                    (B) 25                     (C) 60                     (D) 120

Solution:

The word ‘DELHI’, contains 5 different letters.

Required number of words

= Number of arrangements of 5 letters, taken all at time

= ⁵P₅ = 5 ! = (5×4×3×2×1)=120.

Q.4.In how many ways can the letters of the word ‘APPLE’ be arranged 

(A) 720                  (B) 120                  (C) 60                     (D) 180

Solution:

The word ‘APPLE’ contains 5 letters, 1A, 2P, 1L and 1E.

∴ Required number of ways $$ ={5!\over(1!)(2!)(1!)(1!)}=60. $$

Q.5. In how many ways can the letters of the word ‘LEADER’ be arranged?

(A) 72                    (B) 144                  (C) 360                  (D) 720

Solution:

The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D, and 1R.

∴ Required number of ways $$ ={6!\over(1!)(2!)(1!)(1!)(2!)}=360. $$

Q.6. In how many different ways can the letters of the word ‘RUMOUR’ be arranged?

(A) 180                  (B) 90                     (C) 30                     (D) 720

Solution:

The word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1U.

∴ Required number of ways $$ ={6!\over(2!)(2!)(1!)(1!)}=180. $$

Q.7. How many words can be formed by using all the letters of the word, ‘ALLAHABAD’?

(A) 3780                (B) 1890                 (C) 7560                   (D) 720

Solution:

The word ‘ALLAHABAD’ contains 9 letters, namely 4A, 2L, 1H, 1B, and 1D

∴ Required number of words $$ ={9!\over(4!)(2!)(1!)(1!)(1!)}=7560. $$ 

Q.8. How many arrangements can be made out of the letters of the word ‘ENGINEERING’?

(A) 277200           (B) 92400              (C) 69300              (D) 23100

Solution:

The word ‘ENGINEERING’ contains 11 letters, namely 3E, 3N, 2G, 2I and 1R

∴ Required number of arrangements$$ ={11!\over(3!)(3!)(2!)(2!)(1!)}=277200. $$

Q.9. How many words can be formed from the letters of the word ‘SIGNATURE’ so that the vowels always come together?

(A) 720                  (B) 1440                (C) 69300              (D)17280

Solution:

The word ‘SIGNATURE’ contains 9 different letters.

When the vowels IAUE are taken together, they can be supposed to form an entity, treated as one letter.

Then, the letter to be arranged are SGNTR (IAUE).

These 6 letters can be arranged in ⁶P₆ = 6! = 720 ways.

The vowels in the group (IAUE) can be arranged amongst themselves in ⁴P₄ = 4! = 24 ways.

∴ Required number of words = (720×24) = 17280.

Q.10. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

(A) 120                  (B) 720                  (C) 4320                (D) 2160

Solution:

The word ‘OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) Can be arranged among themselves in 3! = 6 ways.

∴ Required number of words = (120×6) = 720.

I hope these solutions are beneficial for your competitive exams. If you have any doubt or want to ask anything regarding permutation and combinations problems with solutions, you can ask me in the comment section.