Question 1

Twelve distinct beads are used to make a necklace. Considering rotations and reflections as the same design, how many distinct necklaces can be formed?

Accepted Answer

Introduction / Context: For circular arrangements of distinct items, rotational symmetry reduces linear permutations by a factor of n. When reflections (flips) also yield the same necklace, we divide by an additional factor of 2 for n > 2. Given Data / Assumptions: n = 12 distinct beads. Necklaces equivalent up to rotation and reflection (dihedral symmetry). Concept / Approach: Without symmetry: 12! linear orders. Circular (rotations identical) gives (12−1)! = 11! arrangements. Accounting for mirror images as identical divides further by 2: total = 11!/2. Step-by-Step Solution: Circular arrangements (no reflection quotient) = 11!.Accounting for flips (dihedral action) = 11!/2. Verification / Alternative check: Polya’s enumeration agrees with the dihedral reduction when all beads are distinct and n≥3. Why Other Options Are Wrong: 10!/2 corresponds to 11 beads; 12!/2 ignores circular equivalence; “Couldn’t be determined” is incorrect because assumptions are explicit. Common Pitfalls: Forgetting to divide by 2 for reflections when asked for necklaces (not oriented bracelets). Final Answer: 11!/2

Question 2

Twenty guests plus the host (21 people total) are to be seated at a single round table. How many distinct circular seatings are possible (rotations considered identical)?

Accepted Answer

Introduction / Context: For n distinct people around one round table, rotations are considered identical, so the number of seatings is (n−1)!. Given Data / Assumptions: Total people n = 21 (20 guests + 1 host). One round table; reflections considered distinct unless stated otherwise (standard convention). Concept / Approach: Fix one person's seat to break rotational symmetry; arrange the remaining n−1 people linearly. Step-by-Step Solution: Seatings = (21−1)! = 20!. Verification / Alternative check: If reflections were also identified (rare for seating), dihedral reduction would apply; but typical seating problems use pure circular equivalence. Why Other Options Are Wrong: 19! and 18! correspond to incorrectly subtracting more symmetries; “Couldn’t be determined” is incorrect given the standard model. Common Pitfalls: Confusing bracelets (with flips) and round seating (orientation matters, flips distinct). Final Answer: 20!

Question 3

In how many ways can 5 boys and 5 girls sit around a circular table so that boys and girls alternate in the seating?

Accepted Answer

Introduction / Context: We require strict alternation of genders around a circle. Seat one gender first to create fixed slots for the other gender. Given Data / Assumptions: 5 distinct boys, 5 distinct girls. Single round table; rotations identified (circular permutations). Alternation constraint: B–G–B–G–B–G–B–G–B–G. Concept / Approach: Arrange boys in a circle first: (5−1)! = 4!. Then place the 5 girls into the 5 gaps between boys: 5! ways. Step-by-Step Solution: Arrange boys: 4! = 24.Place girls in gaps: 5! = 120.Total = 24 * 120 = 2880. Verification / Alternative check: Starting with girls gives 4! ways similarly and then 5! for boys—same product. Alternation enforces unique interleaving slots. Why Other Options Are Wrong: Nearby values (2800, 2680, 2280) are arithmetic slips or stem from dividing by an extra symmetry that does not apply. Common Pitfalls: Using 5! for the first gender (forgetting circular reduction) or multiplying by 2 for reversing direction (unnecessary for standard circular seating). Final Answer: 2880

Question 4

From 18 distinct coloured beads, necklaces of length 12 are to be made. Each necklace uses 12 distinct beads selected from the 18. Considering rotation and reflection as the same design, how many distinct necklaces are possible?

Accepted Answer

Introduction / Context: The original stem was ambiguous. Under the Recovery-First Policy, we minimally clarify: pick any 12 distinct beads from 18, then arrange them as a necklace up to rotation and reflection. This yields a clean, teachable count consistent with necklace problems. Given Data / Assumptions (clarified): 18 distinct beads available. Each necklace uses 12 distinct beads. Designs equivalent under rotation and mirror reflection. Concept / Approach: Two stages: (1) choose the multiset of beads; (2) form the necklace quotienting by dihedral symmetry. Choosing 12 distinct beads from 18 gives C(18,12). Arranging 12 distinct beads on a necklace (rotations and reflections identified) gives 11!/2. Step-by-Step Solution: Selection count = C(18,12).Necklace count for a 12-bead set = 11!/2.Total designs = C(18,12) * (11!/2). Verification / Alternative check: Equivalently: choose 12 and then count circular arrangements divided by 2 for flips; beads are all distinct, so no additional stabilizers arise. Why Other Options Are Wrong: Using 10!/2 or 12!/2 mismodels the circular quotient; “None of these” is incorrect under the clarified reading. Common Pitfalls: Forgetting the reflection factor 1/2, or treating the 12-bead selection as ordered. Final Answer: C(18,12) * 11!/2

Question 5

Out of 4 men and 4 women, how many different committees of exactly 3 men and 2 women can be formed (order does not matter)?

Accepted Answer

Introduction / Context: We select fixed counts from each subgroup. Order of members on a committee is irrelevant, so use combinations. Given Data / Assumptions: Men = 4; choose 3. Women = 4; choose 2. Committees are sets (no roles or ordering). Concept / Approach: Compute C(4,3) * C(4,2) by independence across disjoint groups. Step-by-Step Solution: C(4,3) = 4.C(4,2) = 6.Total = 4 * 6 = 24. Verification / Alternative check: Enumerating men first, then women, or vice versa, yields the same product. Why Other Options Are Wrong: 20 and 16 are near-miss counts (e.g., forgetting some combinations); 15 is C(6,2) and unrelated. Common Pitfalls: Accidentally using permutations or mixing order into a committee selection. Final Answer: 24

Question 6

A cricket team of 11 players is to be chosen from 15 players, but one particular player must be included. In how many ways can the team be chosen?

Accepted Answer

Introduction / Context: A mandatory selection reduces the degrees of freedom by fixing one seat. We then choose the remaining players from the rest. Given Data / Assumptions: Total players = 15. Team size = 11. One particular player is always selected. Concept / Approach: If one player is fixed, we need to choose 10 more from the remaining 14. The number of such teams is C(14,10). Step-by-Step Solution: C(14,10) = C(14,4) (by symmetry) = 1001. Verification / Alternative check: Compute directly: 14*13*12*11 / (4*3*2*1) = 24024 / 24 = 1001. Why Other Options Are Wrong: Other values correspond to choosing 11 from 15 without the mandatory constraint or arithmetic slips. Common Pitfalls: Subtracting the mandatory player from total team size incorrectly (e.g., recalculating as C(15,11)). Final Answer: 1001

Question 7

How many triangles can be formed by choosing any 3 vertices of a 12-sided polygon (no three chosen vertices are collinear except at the polygon vertices)?

Accepted Answer

Introduction / Context: Any three non-collinear points determine a triangle. In a convex (or general position) polygon, any 3 distinct vertices form a valid triangle. Given Data / Assumptions: Polygon has 12 vertices. No three vertices are collinear beyond polygon vertices. All 3-vertex choices produce distinct triangles. Concept / Approach: Choose any 3 of the 12 vertices using combinations. Step-by-Step Solution: Number of triangles = C(12,3) = 12*11*10 / (3*2*1) = 220. Verification / Alternative check: For an n-gon, the general formula is C(n,3). Substituting n=12 yields 220. Why Other Options Are Wrong: 200 and 240 are common arithmetic slips; 260 is too high. Common Pitfalls: Subtracting “diagonals” or internal structures unnecessarily; triangles only depend on 3 vertices. Final Answer: 220

Question 8

How many diagonals does a hexagon have?

Accepted Answer

Introduction / Context: The number of diagonals of an n-gon equals the number of line segments between pairs of distinct vertices that are not sides. Given Data / Assumptions: n = 6 (hexagon), vertices in general position. Concept / Approach: Total segments between vertex pairs = C(n,2). Subtract the sides, which are n in number. Thus diagonals = C(n,2) − n = n(n−1)/2 − n = n(n−3)/2. Step-by-Step Solution: Diagonals = 6*(6−3)/2 = 6*3/2 = 9. Verification / Alternative check: List vertices and count non-adjacent pairs; you will find 9 unique diagonals. Why Other Options Are Wrong: 12 counts all segments to non-adjacent and adjacent; 10 and 6 are misapplications of the formula. Common Pitfalls: Forgetting to subtract the 6 sides from C(6,2) = 15; 15 − 6 = 9. Final Answer: 9

Question 9

In a plane there are 10 points, of which 5 are collinear and the other 5 are in general position (no extra collinearities). How many distinct straight lines can be formed by joining pairs of these points?

Accepted Answer

Introduction / Context: Lines determined by a finite point set are counted by considering all pairs, but collinear points collapse many pairs into the same line. We correct the overcount caused by the 5 collinear points. Given Data / Assumptions: Total points = 10. A single set of 5 collinear points; remaining 5 are in general position and not on that line. Concept / Approach: Start with all pairwise lines C(10,2), then adjust for the collinear set. Among the 5 collinear points, C(5,2)=10 pairs lie on the same single line; these 10 pairs would naively count as 10 different lines but yield only 1. Therefore, subtract 9 from the naive total. Step-by-Step Solution: Naive count = C(10,2) = 45.Within the 5 collinear points: 10 pairs but only 1 line ⇒ overcount = 10 − 1 = 9.Corrected total = 45 − 9 = 36. Verification / Alternative check: Count lines through the 5 collinear points (1), lines from each of the other 5 points to the collinear set (all distinct), and lines among the other 5—this decomposition matches 36. Why Other Options Are Wrong: 45 ignores the collinearity; 35 and 30 over-subtract. Common Pitfalls: Subtracting 10 instead of 9 (remember one line remains). Final Answer: 36

Question 10

Twelve distinct balls are to be split between two distinct boys (Rahul and Rohan) so that one boy receives 5 balls and the other receives 7 balls. In how many distributions is this possible?

Accepted Answer

Introduction / Context: We distribute distinct items between two labeled recipients with a fixed split of counts. There are two symmetric cases: Rahul gets 5 (Rohan 7) or Rahul gets 7 (Rohan 5). Given Data / Assumptions (clarified): 12 distinct balls; two distinct boys (Rahul, Rohan). One boy must get exactly 5 balls; the other must get 7 (not specified which). Concept / Approach: Case-split by who gets 5: choose 5 for Rahul (C(12,5)), or choose 5 for Rohan (another C(12,5)). These cases are disjoint and cover all valid allocations. Step-by-Step Solution: Number of 5-ball subsets = C(12,5) = 792.Two choices for who receives 5 ⇒ total = 2 * 792 = 1584. Verification / Alternative check: Equivalently, choose which 7 go to the 7-ball recipient: 2*C(12,7) = 2*792 = 1584 (since C(12,5)=C(12,7)). Why Other Options Are Wrong: Values near 1600 come from arithmetic slips; 1784/1560/1854 are not equal to 2*C(12,5). Common Pitfalls: Answering C(12,5)=792 only (implicitly fixing which boy gets 5) despite the statement allowing either boy to be the 5-ball recipient. Final Answer: 1584

Question 11

From 5 oranges, 4 mangoes, and 3 bananas, one fruit of each type is to be chosen. In how many distinct selections can this be done?

Accepted Answer

Introduction / Context: We perform independent choices from disjoint categories (oranges, mangoes, bananas). The total number of selections is the product of independent counts. Given Data / Assumptions: Oranges = 5, Mangoes = 4, Bananas = 3. Choose exactly one from each category. Concept / Approach: Use the multiplication principle across independent categories: total ways = 5 * 4 * 3. Step-by-Step Solution: 5 * 4 * 3 = 60. Verification / Alternative check: Enumerating choices per category confirms the product. Why Other Options Are Wrong: 120 doubles the count (perhaps ordering); 110 and 75 are unrelated combinations. Common Pitfalls: Confusing selection (unordered) with arrangement (ordered) across categories. Final Answer: 60

Question 12

A student attempts a set of 10 questions and may solve any subset, provided at least one question is attempted. In how many ways can the student choose which questions to solve?

Accepted Answer

Introduction / Context: Each question can be solved or not solved, leading to a subset selection model. We exclude the empty subset because at least one question must be attempted. Given Data / Assumptions: Total questions = 10. Any subset allowed except the empty subset. Concept / Approach: The number of all subsets of a 10-element set is 2^10. Excluding the empty set leaves 2^10 − 1. Step-by-Step Solution: 2^10 = 1024.Nonempty subsets = 1024 − 1 = 1023. Verification / Alternative check: Sum over k from 1 to 10 of C(10,k) equals 2^10 − 1 by the binomial theorem. Why Other Options Are Wrong: 1024 includes the empty subset; 1025 is impossible; 1000 is unrelated. Common Pitfalls: Forgetting to remove the empty selection when “at least one” is required. Final Answer: 1023

Question 13

Fifteen people are to be seated around two round tables with capacities 7 and 8. Assuming the two tables are distinguishable by size, in how many ways can they be seated (rotations at each table considered identical)?

Accepted Answer

Introduction / Context: We split people between two round tables of different sizes and then account for circular symmetry at each table. Since capacities differ (7 vs 8), the tables are distinguishable by size. Given Data / Assumptions: Total people = 15. Table A seats 7, Table B seats 8. Circular seatings: rotations identical at each table; reflections distinct (standard seating convention). Concept / Approach: First choose which 7 people sit at the 7-seat table: C(15,7) = C(15,8). Then arrange those 7 around a circle: (7−1)! = 6!. Arrange the remaining 8 around the other circle: (8−1)! = 7!. Step-by-Step Solution: Distribute people: C(15,7) (equivalently C(15,8)).Seat at 7-table: 6! arrangements.Seat at 8-table: 7! arrangements.Total = C(15,8) * 6! * 7!. Verification / Alternative check: Because table sizes differ, swapping entire tables cannot overcount; thus no division by 2 is needed. Why Other Options Are Wrong: 15!/(8!) is dimensionally inconsistent; 15C8*8! ignores circular reduction; the “7!/88!” option is malformed. Common Pitfalls: Dividing by 2 as if tables were identical when capacities already distinguish them. Final Answer: 15C8 x 6! x 7!

Question 14

The word 'Director' has 8 letters with R repeated twice (letters: D, I, R, E, C, T, O, R). In how many distinct arrangements of all letters are the three vowels (I, E, O) not all together in one single block?

Accepted Answer

Introduction / Context: This is a permutations-with-repetition problem. The word “Director” contains 8 letters with a repetition of R (twice). We need the number of arrangements where the three vowels (I, E, O) are not all together as one contiguous block. Note this condition does not forbid two vowels from being adjacent; it only excludes the case where all three vowels appear consecutively as one block. Given Data / Assumptions: Letters: D, I, R, E, C, T, O, R. Vowels: I, E, O (distinct). Consonants: D, C, T, R, R (R repeats). All 8 letters are used in each arrangement. Concept / Approach: Total arrangements minus arrangements with all three vowels together (as one block). This classical “complement” strategy is simpler than directly counting all acceptable patterns. Step-by-Step Solution: Total arrangements (accounting for repeated R): 8! / 2! = 20160.Count arrangements with vowels together as one block: Treat (I,E,O) as a single super-letter. Then we have 6 entities: [Vowel-Block] + D + C + T + R + R.Ways to arrange these 6 with two R’s: 6! / 2! = 360.Internal arrangements of vowels within the block: 3! = 6.Arrangements with all three vowels together: 360 * 6 = 2160.Required arrangements = 20160 − 2160 = 18000. Verification / Alternative check: Total 20160 is standard for 8 letters with a double letter. The “vowels-together” count uses block method with repeated consonant R handled by division by 2!. Why Other Options Are Wrong: 20160 counts all permutations; 16000 and 1800/1600 are not supported by the block-minus-complement calculation. Common Pitfalls: Misreading “never together” to mean “no two vowels adjacent.” Here the stem explicitly says “the three vowels are never together,” i.e., exclude only the single-block case. Final Answer: 18000

Question 15

Four married couples (4 men and their 4 wives) are available for mixed doubles tennis. A mixed doubles game uses two teams, each team being one man paired with one woman. What is the maximum number of distinct games possible if no team may consist o…

Accepted Answer

Introduction / Context: We must count distinct mixed-doubles games (two disjoint man–woman pairs) from 4 married couples, with the constraint that a husband and his own wife can never be on the same team. A person may appear in many different games across the count; we are counting all possible pairings of two teams adhering to the restriction. Given Data / Assumptions: Men: M1–M4; Women: W1–W4 with couples (Mi, Wi). Game = two disjoint pairs {(man, woman), (man, woman)}. No pair may be a married couple (Mi, Wi). Games are considered unordered collections of two teams (i.e., swapping the two teams does not produce a new game). Concept / Approach: Fix the set of 2 men and the set of 2 women and then count valid pairings between them. Sum over all choices of 2 men and 2 women. Step-by-Step Solution: Choose men: C(4,2) = 6. Choose women: C(4,2) = 6.For fixed {Mi, Mj} and {Wk, Wℓ}, there are 2 possible pairings: (Mi–Wk & Mj–Wℓ) or (Mi–Wℓ & Mj–Wk).Count valid pairings by cases:• If {Wk, Wℓ} = {Wi, Wj}: one pairing is forbidden (both married), the cross pairing is valid → 1 valid.• If exactly one of {Wk, Wℓ} is a spouse of a selected man: exactly 1 valid pairing avoids that marriage → 1 valid.• If none of {Wk, Wℓ} is a spouse of the chosen men: both pairings valid → 2 valid.Counting across the 6 women-choices for fixed men yields 7 valid pairings; across 6 men-choices: 6 × 7 = 42 games. Verification / Alternative check: The bipartite graph K4,4 minus the 4 marriage edges has exactly 42 matchings of size 2 (unordered), consistent with the above casework. Why Other Options Are Wrong: 12 and 36 undercount; 48 overcounts by treating invalid pairings as valid; 30 is ad hoc. Common Pitfalls: Confusing games with single pairs or double-counting by ordering the two teams. Final Answer: 42

Question 16

Seven-digit telephone numbers are formed with all digits distinct. If the leftmost (first) digit is 6 and the rightmost (last) digit is 5, how many such telephone numbers are possible?

Accepted Answer

Introduction / Context: This is a direct permutations problem with positional constraints on the first and last digits and “all digits distinct.” Given Data / Assumptions: Digits available: 0–9. Positions: D1 D2 D3 D4 D5 D6 D7. D1 fixed as 6; D7 fixed as 5. All digits in the 7-number must be distinct. Concept / Approach: After fixing D1 and D7, fill the five middle positions with distinct digits chosen and arranged from the remaining 8 digits. Step-by-Step Solution: Remaining pool after using 6 and 5: 8 digits.Number of ways to choose and arrange D2–D6: P(8,5) = 8 × 7 × 6 × 5 × 4 = 6720. Verification / Alternative check: Equivalent to placing any permutation of 5 distinct digits into the middle 5 slots with order mattering. Why Other Options Are Wrong: 120 is 5! (ignores choice of which digits); 30240 and 5040 come from other permutation counts not matching constraints; 100000 ignores “all distinct.” Common Pitfalls: Forgetting that 0 is allowed in middle positions and that digits cannot repeat. Final Answer: 6720

Question 17

In a plane there are 14 points, of which 4 are collinear and the remaining are in general position. How many distinct triangles can be formed using these 14 points as vertices?

Accepted Answer

Introduction / Context: The number of triangles from a set of points is normally C(n, 3), but any collinear triple cannot form a triangle. We adjust by subtracting those degenerate triples. Given Data / Assumptions: Total points n = 14. Exactly 4 are collinear; every other triple is non-collinear. Concept / Approach: Valid triangles = total triples − collinear triples. Only the special set of 4 contributes collinear triples. Step-by-Step Solution: Total triples: C(14, 3) = 364.Collinear triples among the 4: C(4, 3) = 4.Triangles = 364 − 4 = 360. Verification / Alternative check: If no three were collinear, C(14, 3) would be correct. One collinear block reduces the count by exactly its C(4, 3) degenerate triples. Why Other Options Are Wrong: 364 ignores collinearity; 368 and 365 exceed the possible total; 356 subtracts too many. Common Pitfalls: Subtracting C(4,2) instead of C(4,3) or double-counting. Final Answer: 360

Question 18

An exam has two sections with 3 and 5 questions respectively. “It is not necessary to attempt all the questions.” However, at least one question from each section is compulsory. In how many ways can a candidate select questions to attempt?

Accepted Answer

Introduction / Context: We are selecting subsets of questions from two independent sections, with the constraint that each chosen subset is non-empty (at least one question per section). Given Data / Assumptions: Section A: 3 questions. Section B: 5 questions. Any subset may be chosen, including multiple questions, but each section must contribute at least one chosen question. Concept / Approach: For a set of k items, non-empty subsets count to 2^k − 1. Multiply for independent sections. Step-by-Step Solution: Non-empty subsets of Section A: 2^3 − 1 = 7.Non-empty subsets of Section B: 2^5 − 1 = 31.Total ways = 7 × 31 = 217. Verification / Alternative check: Counting “at least one” directly avoids inclusion–exclusion over the complement event “attempt nothing from a section.” Why Other Options Are Wrong: 256 counts all subsets across both sections including the invalid case of choosing none from both; 320 and 38 are not consistent with the product 7 × 31. Common Pitfalls: Forgetting to exclude the empty subset in each section. Final Answer: 217

Question 19

There are 10 stations along a railway line. How many different one-way journey tickets (from one station to a different station) are needed by the authorities?

Accepted Answer

Introduction / Context: Each one-way ticket is specified by an ordered pair (start, destination) with start ≠ destination. We count ordered choices. Given Data / Assumptions: 10 stations labeled distinctly. Tickets are one-way (direction matters). Concept / Approach: For each start station, any of the remaining 9 stations can be the destination. Sum (or multiply) across starts. Step-by-Step Solution: Choices = 10 × 9 = 90. Verification / Alternative check: Unordered pairs would be C(10, 2) = 45; direction doubles this to 90, consistent with 10 × 9. Why Other Options Are Wrong: 91 and 92/93 are off by +1/+2/+3; 100 counts invalid same-station tickets. Common Pitfalls: Confusing one-way with round-trip or unordered counting. Final Answer: 90

Question 20

From a pool of 8 ladies and 7 gentlemen we must form a committee of 3 ladies and 4 gentlemen. Mrs. X refuses to serve if Mr. Y is a member of the same committee. How many committees are possible under this restriction?

Accepted Answer

Introduction / Context: This is a combinations problem with an exclusion constraint disallowing the simultaneous presence of two specific members (Mrs. X and Mr. Y) on the same committee. Given Data / Assumptions: Total ladies = 8; total gentlemen = 7. Committee must include 3 ladies and 4 gentlemen. Forbidden: committees that include both Mrs. X and Mr. Y together. Concept / Approach: Use total-minus-forbidden counting. Step-by-Step Solution: Unrestricted committees: C(8, 3) × C(7, 4) = 56 × 35 = 1960.Forbidden committees (contain both X and Y): choose the remaining 2 ladies from the other 7 → C(7, 2) = 21; choose the remaining 3 gentlemen from the other 6 → C(6, 3) = 20.Forbidden = 21 × 20 = 420.Valid committees = 1960 − 420 = 1540. Verification / Alternative check: Partition by whether Y is included: when Y is included, exclude X; when Y is not included, X is free — this yields the same total. Why Other Options Are Wrong: 1960 ignores the restriction; 3240 exceeds the total; other values do not match the subtraction. Common Pitfalls: Subtracting committees that contain X but not Y (which are allowed) by mistake. Final Answer: 1540

Permutation and Combination Questions