Permutation and Combination - Aptitude Questions and Answers

Permutation and Combination are fundamental concepts in mathematics that deal with arranging and selecting objects from a set. These concepts are crucial in various fields, including probability theory, and statistics.

There are different types of questions that you can use to test your understanding, so start revising!

Prerequisites:

• Permutation
• Combination
• Tricks to Solve Permutation & Combination Questions

Solved Aptitude Questions on Permutations and Combinations

Q1: How many words can be formed by using 3 letters from the word "DELHI"? 

Solution: 

Here we will use the Permutations for this question. Formula for permutation is nPr, for this we have,

n = 5: Total 5 Letters

r = 3: Letters word we required

ⁿP_r = n!/(n-r)!

⁵P₃ = 5!/2! = 120/2 = 60

So, Total we can form 60 different permutation of word from Letter Delhi.

Q2: How many words can be formed by using the letters from the word "DRIVER" such that all the vowels are always together? 

Solution: 

In these types of questions, we assume all the vowels to be a single character, i.e., "IE" is a single character.

So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times.

Number of possible arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 × 2 = 120 

Q3: In how many ways, can we select a team of 4 students from a given choice of 15?

Solution: 

Number of possible ways of selection = ¹⁵C₄ = 15 ! / ((4 !) × (11 !))

Number of possible ways of selection = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1365 

Question 4: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?

Solution: 

Number of ways 3 boys can be selected out of 6 = ⁶ C ₃ = 6 ! / [(3 !) × (3 !)] = (6 × 5 × 4) / (3 × 2 × 1) = 20

Number of ways 2 girls can be selected out of 5 = ⁵ C ₂ = 5 ! / [(2 !) × (3 !)] = (5 × 4) / (2 × 1) = 10

Therefore, total number of ways of forming the group = 20 x 10 = 200 

Question 5: How many words can be formed by using the letters from the word "DRIVER" such that all the vowels are never together?

Solution: 

We assume all the vowels to be a single character, i.e., "IE" is a single character.

So, now we have 5 characters in the word, namely, D, R, V, R, and IE.

But, R occurs 2 times.

Number of possible arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 x 2 = 120 ,

Total number of possible words = 6! / 2! = 720 / 2 = 360

Therefore, the total number of possible words such that the vowels are never together 240

Practice Questions on Permutation and Combination

1. In how many ways can 7 people be seated in a row if 2 particular people must sit together?

2. How many 4-letter words (with or without meaning) can be formed using the letters of the word "APPLE", if repetition is not allowed?

3. A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must include at least 2 women?

4. How many ways are there to arrange the letters of the word "MISSISSIPPI"?

5. From a standard deck of 52 cards, in how many ways can 5 cards be selected?

6. Ten points are marked on a straight line. How many line segments can be formed using these points as endpoints?

7. In how many ways can 8 identical balls be distributed among 3 distinct boxes?

8. How many 6-digit numbers can be formed using the digits 0 to 9 if repetition is allowed and the number must be even?

9. In how many ways can 5 men and 5 women be seated in a row if all men sit together and all women sit together?

10. How many different necklaces can be made using 6 distinct beads?

Answer Keys

• 720 ways
• 60 ways
• 240 ways
• 34,650 ways
• 2,598,960 ways
• 45 line segments
• 36 ways
• 45,000 numbers
• 14,400 ways
• 60 necklaces

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• Aptitude