Permutations & Combinations
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Solved Examples
Study MaterialSolved Examples ā Permutations & Combinations
Solved examples help students understand the practical application of permutation and combination concepts in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, CAT, NDA, CDS, Insurance, Defence, and various aptitude examinations.
Topics Covered in Solved Examples
- Basic Permutation Problems
- Basic Combination Problems
- Arrangement of Letters and Numbers
- Circular Arrangement
- Committee Selection Problems
- Repeated Object Arrangements
- Digit Arrangement Problems
- Advanced Counting Techniques
Example 1: Basic Permutation
Question: In how many ways can 3 letters A, B, and C be arranged?
Solution:
Number of arrangements:
= 3!
= 3 Ć 2 Ć 1
= 6
Possible arrangements:
ABC, ACB, BAC, BCA, CAB, CBA
Answer: 6 ways
Example 2: Permutation Taking r Objects
Question: In how many ways can 2 letters be arranged from A, B, C, D?
Solution:
Using permutation formula:
4P2
= 4! / (4ā2)!
= 4! / 2!
= (4 Ć 3 Ć 2!) / 2!
= 12
Answer: 12 ways
Example 3: Basic Combination
Question: In how many ways can 2 students be selected from 5 students?
Solution:
Using combination formula:
5C2
= 5! / [2!(5ā2)!]
= 5! / (2! Ć 3!)
= (5 Ć 4) / 2
= 10
Answer: 10 ways
Example 4: Circular Arrangement
Question: In how many ways can 5 persons sit around a circular table?
Solution:
Circular arrangement formula:
= (nā1)!
= (5ā1)!
= 4!
= 24
Answer: 24 ways
Example 5: Word Formation
Question: In how many ways can the word CAT be arranged?
Solution:
Total letters = 3
All letters are different.
Number of arrangements:
= 3!
= 6
Answer: 6 ways
Example 6: Repeated Letters Arrangement
Question: In how many ways can the word BALL be arranged?
Solution:
Total letters = 4
L repeats 2 times.
Required arrangements:
= 4! / 2!
= 24 / 2
= 12
Answer: 12 ways
Example 7: Committee Selection
Question: In how many ways can a committee of 3 members be selected from 8 persons?
Solution:
Committee selection uses combination.
= 8C3
= 8! / [3!(8ā3)!]
= 8! / (3! Ć 5!)
= (8 Ć 7 Ć 6) / (3 Ć 2 Ć 1)
= 56
Answer: 56 ways
Example 8: Arrangement of Digits
Question: How many 3-digit numbers can be formed using digits 1, 2, 3, 4 without repetition?
Solution:
Required arrangements:
= 4P3
= 4! / (4ā3)!
= 4!
= 24
Answer: 24 numbers
Example 9: Arrangement with Zero
Question: How many 3-digit numbers can be formed using digits 0, 1, 2, 3 without repetition?
Solution:
First digit cannot be 0.
Choices for first digit:
= 3
Remaining digits:
= 3 Ć 2
Total numbers:
= 3 Ć 3 Ć 2
= 18
Answer: 18 numbers
Example 10: Selecting Boys and Girls
Question: In how many ways can 2 boys and 3 girls be selected from 5 boys and 6 girls?
Solution:
Ways to select boys:
= 5C2
= 10
Ways to select girls:
= 6C3
= 20
Total ways:
= 10 Ć 20
= 200
Answer: 200 ways
Example 11: Seating Arrangement
Question: In how many ways can 4 persons sit in a row?
Solution:
Linear arrangement:
= 4!
= 24
Answer: 24 ways
Example 12: Number Formation with Repetition
Question: How many 4-digit numbers can be formed using digits 1, 2, 3 if repetition is allowed?
Solution:
Repetition allowed:
= nr
= 34
= 81
Answer: 81 numbers
Example 13: Selecting All Objects
Question: In how many ways can all letters of the word DOG be arranged?
Solution:
Total letters:
= 3
All letters are different.
Required arrangements:
= 3!
= 6
Answer: 6 ways
Example 14: Large Combination Simplification
Question: Find the value of 10C8.
Solution:
Using symmetry property:
10C8 = 10C2
= 10! / [2! Ć 8!]
= (10 Ć 9) / 2
= 45
Answer: 45
Example 15: Circular Seating Restriction
Question: In how many ways can 6 persons sit around a round table?
Solution:
Circular arrangement:
= (6ā1)!
= 5!
= 120
Answer: 120 ways
Example 16: Selection of Leaders
Question: A captain and vice-captain are to be selected from 7 players. In how many ways can this be done?
Solution:
Positions are different.
Therefore permutation is used.
= 7P2
= 7 Ć 6
= 42
Answer: 42 ways
Example 17: Team Selection
Question: In how many ways can a cricket team of 11 players be selected from 15 players?
Solution:
Team selection uses combination.
= 15C11
= 15C4
= 1365
Answer: 1365 ways
Important Exam Tips
- If order matters ā Use permutation.
- If order does not matter ā Use combination.
- Use symmetry property for easy calculations.
- Memorize factorial values thoroughly.
- Practice repeated object problems regularly.
- Use circular arrangement formula carefully.
- Read question conditions properly.
Common Mistakes to Avoid
- Using permutation instead of combination.
- Ignoring repeated objects.
- Allowing zero at first position.
- Using wrong factorial calculations.
- Ignoring circular arrangement rules.
Practicing solved examples regularly improves conceptual clarity, logical thinking, and calculation speed in solving Permutations & Combinations aptitude questions in competitive examinations.