Practice Questions
Concept Explanation
Mixed P and C problems combine permutations, combinations, circular arrangements, and restrictions in multi-step placement scenarios.
Key Formulas
- nPr = n!/(n − r)!, nCr = n!/(r!(n − r)!)
- Circular: (n − 1)!; identical items: n!/(p!q!)
- Break problem into stages and multiply independent counts
Example
Choose 3 from 8, then arrange chosen 3 in a row: 8C3 × 3! = 8P3 = 336.
Common Trap
Decide at each step whether order matters before choosing nPr or nCr.
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Q1 EasyIn how many ways can the letters of the word 'APPLE' be arranged?ExplanationAPPLE has 5 letters with P repeated twice. Distinct arrangements = 5! / 2! = 120 / 2 = 60. Hence the answer is 60.
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Q2 EasyWhat is 8C3 + 8P3?Explanation8C3 = 56 and 8P3 = 336. Sum = 56 + 336 = 392. Hence the answer is 392.
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Q3 EasyHow many 4-letter codes can be formed from the English alphabet if repetition of letters is allowed?ExplanationEach of 4 positions can be any of 26 letters when repetition is allowed. Total = 26 × 26 × 26 × 26 = 26^4. Hence the answer is 26^4.
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Q4 Medium3 people are chosen from 8 and then arranged in a row. In how many ways can this be done?ExplanationChoose then arrange: 8C3 × 3! = 56 × 6 = 336, which equals 8P3 = 8 × 7 × 6 = 336. Hence the answer is 336.
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Q5 MediumIn how many ways can a team of 4 men and 2 women be formed from 8 men and 5 women?ExplanationRequired = 8C4 × 5C2 = 70 × 10 = 700. Hence the answer is 700.
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Q6 MediumHow many ascending 3-digit numbers can be formed from {1,2,3,4,5} without repetition?ExplanationAny 3 distinct digits have exactly one ascending order. So count = 5C3 = 10. Hence the answer is 10.
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Q7 MediumA code is formed using 4 letters from the 10 distinct letters of the word 'LOGARITHMS' with no letter repeated. How many such codes are possible?ExplanationLOGARITHMS has 10 distinct letters. Codes of 4 letters without repetition = 10P4 = 10 × 9 × 8 × 7 = 5040. Hence the answer is 5040.
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Q8 MediumWays to distribute 3 distinct prizes to 10 students (at most one prize each)?ExplanationOrder of prizes matters: 10P3 = 10 × 9 × 8 = 720. Hence the answer is 720.
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Q9 MediumChoose 2 from 5 and arrange them, then choose 2 from the remaining 3 and arrange them. Total ways?ExplanationFirst stage: 5P2 = 20. Second stage: 3P2 = 6. Total = 20 × 6 = 120. Hence the answer is 120.
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Q10 HardHow many different words can be formed with the letters of the word 'MISSISSIPPI'?ExplanationMISSISSIPPI has 11 letters: M × 1, I × 4, S × 4, P × 2. Distinct arrangements = 11! / (4! × 4! × 2!) = 39916800 / (24 × 24 × 2) = 39916800 / 1152 = 34650. Hence the answer is 34650.
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Q11 HardA committee of 3 is formed from 6 men and 4 women with more men than women. In how many ways?ExplanationCases: 3 men (6C3 = 20) or 2 men and 1 woman (6C2 × 4C1 = 15 × 4 = 60). Total = 20 + 60 = 80. Hence the answer is 80.
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Q12 HardOut of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?ExplanationSelect consonants and vowels: 7C3 × 4C2 = 35 × 6 = 210 groups. Arrange each group of 5 letters: 5! = 120. Total = 210 × 120 = 25200. Hence the answer is 25200.
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Q13 HardSelect 4 from 10 distinct people and seat them in a circle. Ways?ExplanationChoose 4: 10C4 = 210. Arrange in a circle: (4 − 1)! = 6. Total = 210 × 6 = 1260. Hence the answer is 1260.
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Q14 HardIn how many different ways can the letters of the word 'CORPORATION' be arranged so that the consonants always come together?ExplanationCORPORATION consonants: C, R, P, R, T, N (R × 2); vowels: O, O, A, I, O (O × 3). Treat the 6 consonants as one block → 6 units: (consonants) + 5 vowels, with O × 3 among the vowel units. Arrange units: 6! / 3! = 120. Arrange consonants inside the block: 6! / 2! = 360. Total = 120 × 360 = 43200. Hence the answer is 43200.
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DS1 Easy
In a campus mixed P&C drill, how many ways can 3 people be chosen from 8 and arranged in a row?
Statement IThere are 8 distinct people.
Statement IIExactly 3 are selected and then arranged.
ExplanationTogether: 8P3 = 336. Answer: C. -
DS2 Easy
How many arrangements of APPLE exist?
Statement IAPPLE has 5 letters with P appearing twice.
Statement IIAll letters of the word are used in each arrangement.
ExplanationStatement I alone: 5!/2! = 60. Statement II alone lacks the multiset. Answer: A.
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DS3 Medium
An HR quiz asks how many 4-letter codes from LOGARITHMS are possible without repetition.
Statement ILOGARITHMS has 10 distinct letters.
Statement IIEach code uses 4 distinct letters in order.
ExplanationTogether: 10P4 = 5040. Answer: C. -
DS4 Medium
For the number of arrangements of MISSISSIPPI.
Statement IThe word has 11 letters.
Statement III appears 4 times, S appears 4 times, and P appears twice.
ExplanationNeed both: 11!/(4!×4!×2!) = 34650. Answer: C.