Practice Questions
Concept Explanation
When repetition is allowed, each position can be filled independently. When identical items exist, divide by factorials of repeat counts.
Key Formulas
- With repetition: nr arrangements of r objects from n types
- Identical items: n! / (p! q! r! …) for n objects with repeats
- Multiset formula removes overcounting from identical letters
Example
3-digit numbers using {1,2,3,4,5} with repetition: 53 = 125.
Common Trap
Use nr only when repetition is allowed; use n!/(p!q!) when items are identical.
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Q1 EasyHow many 3-digit numbers can be formed using the digits 1, 2, 3, 4 and 5 if repetition of digits is allowed?ExplanationEach of the 3 places can be filled in 5 ways when repetition is allowed. Total = 5^3 = 125. Hence the answer is 125.
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Q2 EasyHow many distinct arrangements of the letters of the word BOOK are possible?ExplanationBOOK has 4 letters with O repeated twice. Arrangements = 4! / 2! = 24 / 2 = 12. Hence the answer is 12.
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Q3 EasyHow many 3-letter codes can be formed from {X, Y, Z} if repetition is allowed?ExplanationEach of the 3 positions has 3 choices. Total = 3^3 = 27. Hence the answer is 27.
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Q4 EasyHow many 4-digit PINs can be formed using digits 0–9 if repetition of digits is allowed?ExplanationEach of 4 positions has 10 choices when repetition is allowed. Total = 10^4 = 10000. Hence the answer is 10000.
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Q5 MediumHow many 4-letter strings can be formed from {A, B, C} if repetition of letters is allowed?ExplanationEach of 4 positions has 3 choices. Total = 3^4 = 81. Hence the answer is 81.
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Q6 MediumHow many ways can 4 boxes be painted using 3 available colours if each box gets one colour and colours may be repeated?ExplanationEach box has 3 colour choices independently. Total = 3^4 = 81. Hence the answer is 81.
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Q7 MediumHow many distinct arrangements of the letters of the word LEVEL are possible?ExplanationLEVEL has 5 letters with L twice and E twice. Arrangements = 5! / (2! × 2!) = 120 / 4 = 30. Hence the answer is 30.
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Q8 MediumHow many distinct arrangements of the letters of the word PEPPER are possible?ExplanationPEPPER has 6 letters: P three times, E twice, and R once. Arrangements = 6! / (3! × 2!) = 720 / (6 × 2) = 720 / 12 = 60. Hence the answer is 60.
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Q9 HardHow 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 appear once each. Distinct arrangements = 9! / (4! × 2!) = 362880 / (24 × 2) = 362880 / 48 = 7560. Hence the answer is 7560.
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Q10 HardHow many arrangements can be made out of the letters of the word 'ENGINEERING'?ExplanationENGINEERING has 11 letters: E appears 3 times, N appears 3 times, G appears 2 times, I appears 2 times, and R appears once. Distinct arrangements = 11! / (3! × 3! × 2! × 2!) = 39916800 / (6 × 6 × 2 × 2) = 39916800 / 144 = 277200. Hence the answer is 277200.
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Q11 HardIn how many different ways can the letters of the word 'MISSISSIPPI' be arranged?ExplanationMISSISSIPPI has 11 letters: M once, I four times, S four times, and P twice. Distinct arrangements = 11! / (4! × 4! × 2!) = 39916800 / (24 × 24 × 2) = 39916800 / 1152 = 34650. Hence the answer is 34650.
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Q12 HardHow many distinct arrangements of the letters of the word 'BANANA' are possible?ExplanationBANANA has 6 letters: A three times, N twice, and B once. Arrangements = 6! / (3! × 2!) = 720 / (6 × 2) = 720 / 12 = 60. Hence the answer is 60.
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Q13 HardHow many distinct arrangements of the letters of the word 'SUCCESS' are possible?ExplanationSUCCESS has 7 letters: S three times, C twice, and U, E once each. Arrangements = 7! / (3! × 2!) = 5040 / (6 × 2) = 5040 / 12 = 420. Hence the answer is 420.
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Q14 HardIn how many ways can the letters of the word 'ASSISTANT' be arranged?ExplanationASSISTANT has 9 letters: A twice, S three times, T twice, and I, N once each. Distinct arrangements = 9! / (2! × 3! × 2!) = 362880 / (2 × 6 × 2) = 362880 / 24 = 15120. Hence the answer is 15120.
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DS1 Easy
How many 3-digit numbers can be formed from the digits 1 through 5?
Statement IDigits may be repeated in the number.
Statement IIEach number must use exactly 3 digits chosen from 1–5.
ExplanationStatement I alone does not state the length. Statement II alone does not say whether repetition is allowed (without repetition the count is 5P3 = 60; with repetition it is 5^3 = 125). Together, with repetition allowed, total = 5^3 = 125. Both are needed. Answer: C. -
DS2 Easy
How many 4-letter strings can be formed from a given set of symbols. What is that number?
Statement IThere are exactly 3 distinct symbols available.
Statement IIRepetition of symbols is allowed and each string must have length 4.
ExplanationStatement I alone does not specify length or whether repetition is allowed. Statement II alone does not give the number of symbols. Together, strings = 3^4 = 81. Both are needed. Answer: C.
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DS3 Medium
How many distinct arrangements of the word BANANA exist?
Statement IThe word BANANA has 6 letters in total.
Statement IIThe word contains 3 A's and 2 N's (and one B).
ExplanationStatement I alone does not give the repetition counts. Statement II alone does not explicitly confirm total length, though it implies 6 letters; strictly, both together give arrangements = 6! / (3! × 2!) = 60. Answer: C. -
DS4 Medium
How many distinct linear arrangements of coloured balls are possible.
Statement IThere are 5 balls of which 2 are red and 3 are blue.
Statement IIAll 5 balls are arranged in a single row.
ExplanationStatement I alone identifies the multiset of 5 balls, so arrangements = 5! / (2! × 3!) = 10. Statement II alone lacks the colour multiplicities. Answer: A.