Practice Questions
Concept Explanation
Combinations count selections where order does not matter. Choosing r objects from n distinct objects gives nCr.
Key Formulas
- nCr = n! / (r! (n − r)!)
- nCr = nC(n − r)
- nC0 = nCn = 1
Example
8C3 = 8!/(3!×5!) = 56.
Common Trap
Use nCr when order does not matter; nPr when order matters.
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Q1 EasyIn a campus quiz, how many ways can 2 items be chosen from 6 distinct items if order does not matter?ExplanationSelections where order does not matter use combinations: nCr = n! / (r!(n − r)!). Here 6C2 = (6 × 5) / (2 × 1) = 15. Hence the answer is 15.
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Q2 EasyHow many combinations of 4 students can be selected from 10 students?ExplanationOrder does not matter, so use 10C4 = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 5040 / 24 = 210. Hence the answer is 210.
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Q3 EasyIn how many ways can 3 books be chosen from 8 books?Explanation8C3 = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56. Hence the answer is 56.
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Q4 EasyHow many ways can exactly 1 item be selected from 9 distinct items?Explanation9C1 = 9. Hence the answer is 9.
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Q5 EasyWhat is the value of 9C0?ExplanationBy definition, nC0 = 1 for every n ≥ 0. So 9C0 = 1. Hence the answer is 1.
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Q6 EasyThe number of ways to choose r items from n distinct items when order does not matter is:ExplanationWhen order does not matter, selections are combinations: nCr = n! / (r!(n − r)!). (nPr counts ordered selections.) Hence the answer is nCr.
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Q7 EasyHow many ways can 3 mangoes be chosen from 7 distinct mangoes?Explanation7C3 = (7 × 6 × 5) / (3 × 2 × 1) = 35. Hence the answer is 35.
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Q8 MediumA placement test asks: how many ways can a team of 3 players be selected from 10 players?Explanation10C3 = (10 × 9 × 8) / (3 × 2 × 1) = 120. Hence the answer is 120.
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Q9 MediumEvaluate 7C4 for a screening worksheet.Explanation7C4 = 7C3 (since nCr = nC(n − r)) = (7 × 6 × 5) / (3 × 2 × 1) = 35. Hence the answer is 35.
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Q10 MediumIn how many ways can a team of 11 be chosen from 15 players?Explanation15C11 = 15C4 (because 15C11 = 15C(15 − 11)) = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 32760 / 24 = 1365. Hence the answer is 1365.
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Q11 MediumIn an aptitude drill, compute 12C2.Explanation12C2 = (12 × 11) / 2 = 66. Hence the answer is 66.
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Q12 MediumThe number of ways to select 2 cards from a pack of 52 cards is:Explanation52C2 = (52 × 51) / 2 = 1326. Order of drawing does not matter for a combination. Hence the answer is 1326.
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Q13 MediumHow many subsets does a set of 5 distinct elements have (including empty and full set)?ExplanationEach element may be in or out of a subset: 2^5 = 32 subsets. Equivalently, sum of 5Ck from k = 0 to 5 equals 32. Hence the answer is 32.
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Q14 HardIn how many ways can 5 cards be selected from a pack of 52 cards?Explanation52C5 = 52! / (5! × 47!) = (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1) = 311875200 / 120 = 2598960. Hence the answer is 2598960.
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DS1 Easy
In a campus selection problem, how many ways can 3 items be chosen from 8?
Statement IOrder of selection does not matter.
Statement IIItems are distinct and none is repeated.
ExplanationStatement I (with the usual distinct-items assumption) gives 8C3 = 56. Statement II alone does not fix whether order matters. Statement I alone is sufficient. Answer: A. -
DS2 Easy
What is the value of nCr?
Statement In = 10 and r = 4.
Statement IIIt is given that nCr = 210.
ExplanationI: 10C4 = 210. II: directly states the value is 210. Either alone is sufficient. Answer: D.
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DS3 Easy
Is nCr equal to nC(n − r)?
Statement In = 8 and r = 3.
Statement IIn = 8 and n − r = 5.
ExplanationI: 8C3 = 56 and 8C5 = 56, so equal. II: choosing r = 3 is the same as leaving out 5, so 8C3 = 8C5. Either alone shows equality. Answer: D. -
DS4 Medium
How many unordered 2-person teams can be formed from a group?
Statement IThere are 9 people in the group.
Statement IIEach team has exactly 2 members and order within a team does not matter.
ExplanationI alone lacks confirmation that pairs are unordered combinations of size 2. II alone lacks n. Together: 9C2 = 36. Answer: C.