Practice Questions
Concept Explanation
Word arrangement problems count permutations of letters with or without restrictions (fixed letters, vowels together, etc.).
Key Formulas
- All distinct letters: n!
- With repeats: n! / (p! q! …)
- Vowels together: treat block as one unit, then arrange units
Example
Arrangements of TABLE (5 distinct letters) = 5! = 120.
Common Trap
Divide by factorials of identical letters; treat letter blocks as single units when together.
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Q1 EasyIn how many ways can the letters of the word TABLE be arranged?ExplanationTABLE has 5 distinct letters. Number of arrangements = 5! = 120. Hence the answer is 120.
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Q2 EasyHow many arrangements of the word BALL are possible?ExplanationBALL has 4 letters with L repeated twice. Arrangements = 4! / 2! = 24 / 2 = 12. Hence the answer is 12.
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Q3 EasyIn how many ways can the letters of the word 'LEADER' be arranged?ExplanationLEADER has 6 letters with E repeated twice (L, E, A, D, E, R). Distinct arrangements = 6! / 2! = 720 / 2 = 360. Hence the answer is 360.
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Q4 EasyHow many arrangements of RADAR are possible?ExplanationRADAR has 5 letters: R × 2, A × 2, D × 1. Arrangements = 5! / (2! × 2!) = 120 / 4 = 30. Hence the answer is 30.
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Q5 EasyHow many arrangements of TEAM are possible with T fixed at the start?ExplanationFix T at the first place. Arrange E, A, M in the remaining 3 places: 3! = 6. Hence the answer is 6.
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Q6 MediumHow many words can be formed by using all letters of the word 'ALLAHABAD'?ExplanationALLAHABAD has 9 letters: A appears 4 times, L appears 2 times, and H, B, D once each. Distinct arrangements = 9! / (4! × 2!) = 362880 / (24 × 2) = 7560. Hence the answer is 7560.
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Q7 MediumIn how many different ways can the letters of the word 'LEADING' be arranged so that the vowels always come together?ExplanationLEADING has 7 distinct letters; vowels E, A, I. Treat the three vowels as one block → units L, N, D, G, (EAI) = 5 units, arranged in 5! = 120 ways. Vowels inside the block arrange in 3! = 6 ways. Total = 120 × 6 = 720. Hence the answer is 720.
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Q8 MediumHow many arrangements of the word BANANA are possible?ExplanationBANANA has 6 letters: A × 3, N × 2, B × 1. Arrangements = 6! / (3! × 2!) = 720 / (6 × 2) = 60. Hence the answer is 60.
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Q9 MediumHow many distinct arrangements of the word LETTER are possible?ExplanationLETTER has 6 letters: L, E, T, T, E, R → E appears twice and T appears twice. Distinct arrangements = 6! / (2! × 2!) = 720 / 4 = 180. Hence the answer is 180.
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Q10 MediumArrangements of ABCDE with A and E at the two ends equal:ExplanationEnds can be A…E or E…A (2 ways). Middle three letters B, C, D arrange in 3! = 6 ways. Total = 2 × 6 = 12. Hence the answer is 12.
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Q11 HardIn how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?ExplanationMATHEMATICS has vowels A, E, A, I (A twice). Treat AEAI as one block → 8 units: M, T, H, M, T, C, S, (AEAI) with M × 2 and T × 2. Arrange units: 8! / (2! × 2!) = 10080. Arrange vowels in the block: 4! / 2! = 12. Total = 10080 × 12 = 120960. Hence the answer is 120960.
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Q12 HardHow many distinct arrangements of the word COMMITTEE are possible?ExplanationCOMMITTEE has 9 letters: C, O, M, M, I, T, T, E, E → M × 2, T × 2, E × 2. Distinct arrangements = 9! / (2! × 2! × 2!) = 362880 / 8 = 45360. Hence the answer is 45360.
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Q13 HardHow many arrangements can be made with the letters of the word 'ENGINEERING'?ExplanationENGINEERING has 11 letters: E × 3, N × 3, G × 2, I × 2, R × 1. Distinct arrangements = 11! / (3! × 3! × 2! × 2!) = 39916800 / (6 × 6 × 2 × 2) = 39916800 / 144 = 277200. Hence the answer is 277200.
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Q14 HardIn how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?ExplanationCORPORATION: vowels O, O, A, I, O (O × 3). Treat OOAIO as one block → units C, R, P, R, T, N, (OOAIO) = 7 units with R × 2. Arrange units: 7! / 2! = 2520. Arrange vowels in the block: 5! / 3! = 20. Total = 2520 × 20 = 50400. Hence the answer is 50400.
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DS1 Easy
In a campus drill on word arrangements, how many distinct arrangements of TABLE exist?
Statement IAll 5 letters of TABLE are distinct.
Statement IINo letter is required to occupy a fixed position.
ExplanationStatement I alone: 5! = 120. Statement II alone does not identify the word or letter counts. Answer: A. -
DS2 Easy
For arrangements of BANANA.
Statement IBANANA has 6 letters.
Statement IIA appears 3 times and N appears 2 times.
ExplanationNeed length and repeats: 6!/(3!×2!) = 60. Answer: C.
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DS3 Medium
How many arrangements of LEADER are possible?
Statement IThe word LEADER has 6 letters with E appearing twice.
Statement IIRepetition of identical letters must be divided out using factorials.
ExplanationStatement I alone gives 6!/2! = 360. Statement II alone lacks the letter multiset. Answer: A. -
DS4 Medium
An HR quiz asks how many arrangements keep the vowels of LEADING together.
Statement ILEADING has vowels E, A, I that must form one block.
Statement IIThe consonants L, N, D, G are all distinct.
ExplanationNeed both: 5 units arrange in 5! and vowels in 3!, total 720. Answer: C.