Practice Questions
Concept Explanation
When selections have restrictions, use complementary counting or split into valid cases.
Key Strategies
- Must include / must exclude: adjust r or use nCr on reduced set
- At least one: total − none
- Both roles filled from different pools: multiply independent choices
Example
Choose 3 from 7 with A mandatory: choose remaining 2 from other 6 → 6C2 = 15.
Common Trap
Subtract invalid cases from total rather than listing many valid cases when possible.
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Q1 MediumFrom 7 distinct books, 3 are to be chosen with one particular book always included. In how many ways can this be done?ExplanationFix the mandatory book, then choose the remaining 2 from the other 6: 6C2 = (6 × 5) / 2 = 15. Hence the answer is 15.
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Q2 MediumThere are 5 apples and 4 oranges. In how many ways can 3 fruits be selected so that at least one apple is included?ExplanationTotal ways to choose 3 from 9 fruits = 9C3 = 84. Ways with no apple (all oranges) = 4C3 = 4. At least one apple = 84 − 4 = 80. Hence the answer is 80.
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Q3 MediumChoose 2 people from 8 so that two particular people are never both selected. How many selections are valid?ExplanationTotal pairs = 8C2 = 28. Forbidden pair (the two particular people together) = 1. Valid = 28 − 1 = 27. Hence the answer is 27.
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Q4 MediumA committee of 3 is to be formed from 6 men and 4 women. In how many ways can it be formed if at least one woman must be selected?ExplanationTotal committees = 10C3 = 120. All-men committees = 6C3 = 20. At least one woman = 120 − 20 = 100. Hence the answer is 100.
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Q5 MediumA team of 4 is chosen from 10 people with 2 fixed members always included. How many ways?ExplanationInclude the 2 fixed members; choose 2 more from the remaining 8: 8C2 = 28. Hence the answer is 28.
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Q6 MediumFrom {1, 2, 3, 4, 5}, how many unordered pairs have an even sum?ExplanationEven sum ⇒ both odd or both even. Odds {1,3,5}: 3C2 = 3. Evens {2,4}: 2C2 = 1. Total = 4. Hence the answer is 4.
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Q7 MediumSelect 4 people from 12 with one particular person excluded. How many ways?ExplanationExclude that person; choose 4 from the remaining 11: 11C4 = (11 × 10 × 9 × 8) / (4 × 3 × 2 × 1) = 7920 / 24 = 330. Hence the answer is 330.
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Q8 MediumSelect 4 members from 6 engineers and 4 designers with exactly 2 engineers. How many ways?ExplanationChoose 2 engineers from 6 and 2 designers from 4: 6C2 × 4C2 = 15 × 6 = 90. Hence the answer is 90.
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Q9 MediumChoose 4 from 11 people with person A included and person B excluded. How many ways?ExplanationInclude A, exclude B; choose 3 more from the remaining 9: 9C3 = 84. Hence the answer is 84.
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Q10 MediumFrom 5 boys and 4 girls, a team of 3 is selected with at least 1 girl. In how many ways?ExplanationTotal = 9C3 = 84. No girl (all boys) = 5C3 = 10. At least one girl = 84 − 10 = 74. Hence the answer is 74.
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Q11 HardFrom a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?ExplanationCases: (3 men, 2 women) + (4 men, 1 woman) + (5 men). = (7C3 × 6C2) + (7C4 × 6C1) + 7C5 = (35 × 15) + (35 × 6) + 21 = 525 + 210 + 21 = 756. Hence the answer is 756.
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Q12 HardThree cards are drawn from a pack of 52. In how many ways can the hand contain at least one ace (4 aces in the pack)?ExplanationTotal 3-card hands = 52C3 = 22100. Hands with no ace = 48C3 = 17296. At least one ace = 22100 − 17296 = 4804. Hence the answer is 4804.
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Q13 HardOut of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?ExplanationSelect letters: 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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Q14 HardIn how many ways can a committee of 5 be formed from 6 men and 4 women so that at least 2 women are included?ExplanationValid: (3 men, 2 women) + (2 men, 3 women) + (1 man, 4 women) = (6C3 × 4C2) + (6C2 × 4C3) + (6C1 × 4C4) = (20 × 6) + (15 × 4) + (6 × 1) = 120 + 60 + 6 = 186. Hence the answer is 186.
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DS1 Easy
How many ways can 3 items be chosen from 7 if one specific item is mandatory?
Statement IOne particular item must be included.
Statement IIOrder of selection does not matter.
ExplanationStatement I alone: fix that item and choose 2 from 6 → 6C2 = 15. Statement II alone is incomplete. Answer: A. -
DS2 Easy
How many pairs from 8 people exclude a particular forbidden pair?
Statement IThere are 8 people and selections are of size 2.
Statement IITwo particular people cannot both be selected.
ExplanationI alone gives total pairs 8C2 = 28 but not the restriction. II alone lacks n. Together: 28 − 1 = 27. Answer: C.
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DS3 Medium
How many fruit selections of 3 include at least one apple?
Statement IThere are 5 apples and 4 oranges.
Statement IIExactly 3 fruits are selected from these 9.
ExplanationNeed both stock and size: 9C3 − 4C3 = 84 − 4 = 80. Answer: C. -
DS4 Medium
How many committees of 3 from 6 men and 4 women include at least one woman?
Statement IThere are 6 men and 4 women available.
Statement IIThe committee size is 3.
ExplanationTogether: 10C3 − 6C3 = 120 − 20 = 100. Neither alone suffices. Answer: C.