Practice Questions
Concept Explanation
Permutations count arrangements of objects where order matters. Selecting r objects from n distinct objects and arranging them gives nPr.
Key Formulas
- nPr = n! / (n − r)!
- Arrangements of n distinct objects in a row = n!
- nP0 = 1, nPn = n!
Example
6P2 = 6!/4! = 6 × 5 = 30.
Common Trap
Permutation applies when order matters; do not use nCr for ranking problems.
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Q1 EasyIn how many ways can 5 books be arranged on a shelf?ExplanationArrangements of n distinct objects in a row = n!. Here n = 5, so number of ways = 5! = 5 × 4 × 3 × 2 × 1 = 120. Hence the answer is 120.
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Q2 EasyHow many different 2-letter arrangements can be formed from the letters {A, B, C, D, E} if no letter is repeated?ExplanationThis is a permutation of 2 letters from 5 distinct letters: 5P2 = 5! / (5 − 2)! = 5 × 4 = 20. Hence the answer is 20.
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Q3 EasyIn how many ways can 4 people stand in a queue?ExplanationArranging 4 distinct people in a line gives 4! = 24 ways. Hence the answer is 24.
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Q4 EasyHow many ways can the letters of the word DOG be arranged?ExplanationDOG has 3 distinct letters, so arrangements = 3! = 6. Hence the answer is 6.
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Q5 EasyIn how many ways can the letters of the word 'ORANGE' be arranged?ExplanationORANGE has 6 distinct letters. Number of arrangements = 6! = 720. Hence the answer is 720.
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Q6 MediumHow 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?ExplanationLOGARITHMS has 10 distinct letters. Number of 4-letter words without repetition = 10P4 = 10! / (10 − 4)! = 10 × 9 × 8 × 7 = 5040. Hence the answer is 5040.
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Q7 MediumWhat is the value of 6P3?ExplanationnPr = n! / (n − r)!, so 6P3 = 6 × 5 × 4 = 120. Hence the answer is 120.
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Q8 MediumIn how many ways can the letters of the word 'LEADER' be arranged?ExplanationLEADER has 6 letters with E repeated twice. Number of distinct arrangements = 6! / 2! = 720 / 2 = 360. Hence the answer is 360.
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Q9 MediumHow many ways can 3 students be seated in 8 distinct chairs (one student per chair)?ExplanationChoosing and assigning chairs is 8P3 = 8 × 7 × 6 = 336. Hence the answer is 336.
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Q10 MediumHow many ways can 6 runners finish in first and second place (no ties)?ExplanationOrdered selection of 2 finishers from 6: 6P2 = 6 × 5 = 30. Hence the answer is 30.
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Q11 HardIn how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?ExplanationLEADING has letters L, E, A, D, I, N, G. Vowels are E, A, I. Treat the 3 vowels as one bundle, so we arrange 5 units (vowel-bundle + L, D, N, G): 5! = 120. Inside the bundle, vowels can be arranged in 3! = 6 ways. Total = 5! × 3! = 120 × 6 = 720. Hence the answer is 720.
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Q12 HardIn how many ways can first, second and third prizes be awarded to 10 contestants if no contestant gets more than one prize?ExplanationOrder matters for ranked prizes, so use 10P3 = 10 × 9 × 8 = 720. Hence the answer is 720.
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Q13 HardIn how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?ExplanationMATHEMATICS has 11 letters: M,A,T,H,E,M,A,T,I,C,S with M, A, T each twice. Vowels are A, E, A, I (4 letters, A twice). Treat vowels as one bundle: remaining units = 7 consonants + 1 bundle = 8 units with M twice and T twice, so 8! / (2! × 2!) = 10080. Vowels inside the bundle: 4! / 2! = 12. Total = 10080 × 12 = 120960. Hence the answer is 120960.
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Q14 HardWhat is the value of 9P4?Explanation9P4 = 9 × 8 × 7 × 6 = 3024. Hence the answer is 3024.
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DS1 Easy
In a campus library quiz, how many ways can 5 distinct books be arranged on a shelf?
Statement IAll 5 books are to be used in the arrangement.
Statement IIThe books are arranged in a single row where order matters.
ExplanationStatement I alone with the shelf-arrangement context gives 5! = 120. Statement II alone does not specify how many books. Answer: A. -
DS2 Easy
For the value of nPr. What is that value?
Statement IIt is given that n = 7 and r = 3.
Statement IIIt is given that nPr = 210.
ExplanationFrom Statement I, 7P3 = 7 × 6 × 5 = 210. Statement II gives the value directly as 210. Each alone is sufficient. Answer: D.
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DS3 Medium
How many 2-person ordered arrangements can be formed from a group of people?
Statement IThere are 8 people and no person is repeated in an arrangement.
Statement IIThe order of the two selected people matters.
ExplanationStatement I alone does not confirm that order matters. Statement II alone lacks the group size. Together, arrangements = 8P2 = 8 × 7 = 56. Both are needed. Answer: C. -
DS4 Hard
How many linear arrangements of distinct objects are possible. Can the number be found?
Statement IThere are 6 distinct objects and all of them are arranged in a row.
Statement IIExactly 4 of the 6 objects are arranged in a row.
ExplanationStatement I alone gives 6! = 720. Statement II alone gives 6P4 = 6 × 5 × 4 × 3 = 360. Each statement alone answers its own fully specified arrangement count, so each is sufficient for the corresponding interpretation; however the two statements describe different problems. For standard DS where either statement may define the count: Answer D.