Permutation – Solved Problems
Permutation deals with the arrangement of objects where the order of selection matters. Permutation problems are widely used in algebra, probability, and competitive examinations.
Key Formulas:
Permutation of r objects from n objects:
nPr = n! / (n − r)!
Factorial:
n! = n × (n−1) × (n−2) × … × 1
Permutation of r objects from n objects:
nPr = n! / (n − r)!
Factorial:
n! = n × (n−1) × (n−2) × … × 1
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Problem 1: Basic permutation
Find the number of permutations of 3 objects selected from 5 objects.
Solution:
n = 5, r = 3
nPr = 5! / (5 − 3)!
= 5! / 2!
= (5 × 4 × 3 × 2!) / 2!
= 60
n = 5, r = 3
nPr = 5! / (5 − 3)!
= 5! / 2!
= (5 × 4 × 3 × 2!) / 2!
= 60
Problem 2: Arrangement of letters
How many different arrangements can be made using the letters A, B, C?
Solution:
Number of letters = 3
Total permutations = 3! = 6
Number of letters = 3
Total permutations = 3! = 6
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Problem 3: Arrangement of digits
In how many ways can 4 different digits be arranged?
Solution:
Number of digits = 4
Permutations = 4! = 24
Number of digits = 4
Permutations = 4! = 24
Problem 4: Selecting positions
How many ways can a President, Vice-President, and Secretary be selected from 6 people?
Solution:
Since positions are different, order matters.
n = 6, r = 3
6P3 = 6! / 3!
= (6 × 5 × 4) = 120
Since positions are different, order matters.
n = 6, r = 3
6P3 = 6! / 3!
= (6 × 5 × 4) = 120
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Problem 5: Permutation with condition
In how many ways can the letters of the word “CAT” be arranged so that C is always at the beginning?
Solution:
Fix C at the first position.
Remaining letters = A, T
Number of arrangements = 2! = 2
Fix C at the first position.
Remaining letters = A, T
Number of arrangements = 2! = 2
Problem 6: Circular permutation
In how many ways can 5 people sit around a round table?
Solution:
For circular permutation:
Number of arrangements = (n − 1)!
= (5 − 1)! = 4! = 24
For circular permutation:
Number of arrangements = (n − 1)!
= (5 − 1)! = 4! = 24
Problem 7: Repetition allowed
How many 3-digit numbers can be formed using digits 1, 2, and 3 if repetition is allowed?
Solution:
Each position can be filled in 3 ways.
Total permutations = 3 × 3 × 3 = 27
Each position can be filled in 3 ways.
Total permutations = 3 × 3 × 3 = 27
Problem 8: Word problem
How many ways can 3 prizes be distributed among 5 students if each student can receive at most one prize?
Solution:
Distribution depends on order (distinct prizes).
Number of ways = 5P3
= 5! / 2!
= 60
Distribution depends on order (distinct prizes).
Number of ways = 5P3
= 5! / 2!
= 60
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Key Takeaways
- Order matters in permutations
- Use factorial notation carefully
- Circular permutations reduce arrangements
- Conditions can simplify the problem