Question 1

Forming 3-boy groups with an exclusion: From boys {A, B, C, D, E}, how many groups of 3 can be formed if C and D must not be in the same group?

Accepted Answer

Introduction / Context: This is combinations with an exclusion constraint: count all 3-person groups, then subtract the forbidden ones that include both C and D together. Given Data / Assumptions: Five distinct boys: A, B, C, D, E. We need groups (order does not matter) of size 3. Constraint: C and D cannot both appear in a chosen group. Concept / Approach: Total groups = C(5,3). Forbidden groups contain {C, D} plus one of the remaining three. Allowed = total − forbidden. Step-by-Step Solution: Total = C(5,3) = 10Forbidden = choose 1 from {A, B, E} with {C, D} fixed → 3Allowed = 10 − 3 = 7 Verification / Alternative check: List groups and strike those containing both C and D; seven remain. This matches the calculation. Why Other Options Are Wrong: 8 double-counts or mis-subtracts forbidden groups. 5 and 6 undercount the allowed combinations. Common Pitfalls: Treating ordered arrangements instead of combinations. Final Answer: 7

Question 2

Multiple-choice test outcomes (not all correct): A paper has 4 MCQs, each with 5 options and exactly one correct choice. How many answer patterns result in not getting all four answers correct?

Accepted Answer

Introduction / Context: We count all possible answer patterns and subtract the single perfect pattern where all answers are correct. This is a straightforward application of the complement principle. Given Data / Assumptions: 4 independent questions. Each question has 5 possible chosen options. Exactly one pattern corresponds to “all correct.” Concept / Approach: Total patterns = 5^4. Subtract the one all-correct pattern. Step-by-Step Solution: Total = 5^4 = 625Not all correct = 625 − 1 = 624 Verification / Alternative check: Alternatively, sum by number wrong (≥ 1) is longer; complement is minimal and exact. Why Other Options Are Wrong: 120 is unrelated (e.g., 5P3 * something). 1024 is 2^10 and not applicable. 19 is far too small. Common Pitfalls: Forgetting that exactly one pattern achieves full correctness. Final Answer: 624

Question 3

Linear arrangements avoiding adjacency of two specified people: Six distinct players stand in a line. In how many ways can they be arranged if Abhinav and Manjesh are never together?

Accepted Answer

Introduction / Context: To count arrangements where two specified people are not adjacent, use total permutations minus the arrangements where they are together (treated as a block) times their internal swaps. Given Data / Assumptions: Six distinct players; two are special (A, M). We consider linear orderings. Prohibited: A and M adjacent. Concept / Approach: Total linear orders = 6!. Adjacent count: treat (A,M) as a block with 2 internal orders. Not-adjacent = total − adjacent. Step-by-Step Solution: Total = 6! = 720Adjacent: block + 4 others → 5! arrangements, and (A,M) can be (A,M) or (M,A) → 2!, so 5!*2 = 120*2 = 240Not-adjacent = 720 − 240 = 480 Verification / Alternative check: Gap method (place 4 others, count gaps for A and M) yields the same 480 result, confirming correctness. Why Other Options Are Wrong: 120 and 240 correspond to partial counts (e.g., adjacent cases only). 360 still undercounts the exclusions. Common Pitfalls: Forgetting the 2 internal arrangements of the adjacent pair. Final Answer: 480

Question 4

Selecting three roads among 4 towns without forming a triangle: Four towns A, B, C, D (no three collinear). Choose 3 undirected roads (each joins a pair of towns) so that the chosen roads do not form a triangle. How many such selections are possible?

Accepted Answer

Introduction / Context: With 4 towns, the complete undirected graph K4 has 6 edges (roads). We must choose 3 edges that do not form a 3-cycle (triangle). This is a graph-subset counting exercise with exclusion of triangular 3-edge sets. Given Data / Assumptions: Vertices: 4 towns; edges: all unordered pairs (6 total). Select exactly 3 edges. Disallow any 3-edge selection that is exactly a triangle on some triple of towns. Concept / Approach: Total 3-edge subsets = C(6,3) = 20. Triangles in K4: each choice of 3 towns determines exactly 1 triangle; there are C(4,3) = 4 such triangles. Valid sets = total − triangles. Step-by-Step Solution: Total 3-edge selections = 20Forbidden (triangles) = 4Valid selections = 20 − 4 = 16 Verification / Alternative check: Enumerating non-triangular 3-edge graphs on 4 vertices matches 16: these are exactly the 3-edge forests or graphs containing at least one vertex of degree 3 without closing a cycle of length 3. Why Other Options Are Wrong: 12, 14, 18 are off by miscounting triangles or edges. “None of these” is false because 16 is correct. Common Pitfalls: Forgetting there are 4 distinct triangles in K4 (one per 3-vertex subset). Final Answer: 16

Question 5

Three dice rolled: at least one shows 2: Three fair six-faced dice are rolled. How many outcomes in the 6^3 sample space have at least one die showing 2?

Accepted Answer

Introduction / Context: Counting “at least one” events on dice is best done via the complement rule: subtract the count with no 2s from the total sample space of 6^3 equiprobable outcomes. Given Data / Assumptions: Three independent fair dice, faces {1,…,6}. We count ordered triples (d1, d2, d3). Concept / Approach: Total outcomes = 6^3. No-2 outcomes: each die has 5 allowed faces (exclude 2) → 5^3. At least one 2 = 6^3 − 5^3. Step-by-Step Solution: Total = 6^3 = 216No-2 = 5^3 = 125Count = 216 − 125 = 91 Verification / Alternative check: Inclusion–exclusion by exact number of 2s (1, 2, 3) yields the same 91; complement is simpler. Why Other Options Are Wrong: 36 and 81 are too small; 116 exceeds total remaining after subtraction. Common Pitfalls: Mixing “count of outcomes” with “probability.” Final Answer: 91

Question 6

5-digit numbers from {3, 4, 5, 6, 7} with repetition allowed: How many distinct 5-digit numbers can be formed using the digits 3, 4, 5, 6, 7 when repetition is allowed?

Accepted Answer

Introduction / Context: Each of the 5 positions can independently take any of the five allowed digits when repetition is allowed. The multiplication principle directly gives the count as 5^5. Given Data / Assumptions: Digits available: 5 (3, 4, 5, 6, 7). Number length: 5 digits. Repetition allowed; leading digit can be any of the five. Concept / Approach: Choices per position = 5; positions = 5. Total = 5^5. Step-by-Step Solution: Total 5-digit numbers = 5 * 5 * 5 * 5 * 5 = 5^5 = 3125 Verification / Alternative check: Counting by cases is unnecessary because every position has identical freedom (no restrictions on leading digit among the given five). Why Other Options Are Wrong: 625 = 5^4 (only 4 positions). 125 = 5^3 (only 3 positions). 3905 is not a power of 5 and does not arise here. Common Pitfalls: Accidentally disallowing repeats or restricting the leading digit. Final Answer: 3125

Question 7

Circular seating with a fixed seat for the host: A host and 8 guests are seated around a round table, but the host must occupy one specific seat. In how many ways can everyone be seated?

Accepted Answer

Introduction / Context: In circular permutations, fixing one position removes rotational symmetry. If the host is fixed at a particular labeled seat, the remaining 8 guests can be arranged freely around the table as in a line relative to that fixed reference seat. Given Data / Assumptions: One host assigned to a specific seat. 8 other distinct guests to arrange. Seats are effectively labeled due to the fixed reference. Concept / Approach: With host fixed, count permutations of the 8 remaining guests around the table. Step-by-Step Solution: Arrangements = 8! (no further symmetry remains) Verification / Alternative check: If instead the circle were unlabelled with no fixed person, arrangements would be (9−1)! = 8!. Here, explicitly fixing the host to a particular seat leads to the same count for the remaining guests, confirming 8!. Why Other Options Are Wrong: 9! would overcount by ignoring circular equivalence. 6! or 4! are unrelated to 8 free placements. Common Pitfalls: Confusing “host fixed seat” with “host fixed position up to rotation.” Both give 8! for the others. Final Answer: 8!

Question 8

Seating with a ladies-only seat among five seats: Two men (distinct) and one woman board a bus with 5 vacant seats, one seat reserved for ladies only. The woman may or may not take the reserved seat (men cannot). In how many ways can these three pas…

Accepted Answer

Introduction / Context: We count placements of three distinct passengers (M1, M2, W) into five seats where one is ladies-only (L). Men cannot occupy L. The woman can sit on L or any normal seat, and if she avoids L, that seat remains empty. Use clean casework by whether W uses L. Given Data / Assumptions: Seats: L (ladies-only) + 4 normal seats. M1, M2 cannot sit in L; W may or may not sit in L. Each person takes exactly one seat; two seats remain empty. Concept / Approach: Case 1: W sits in L; men fill 2 of 4 normal seats in order. Case 2: W sits in a normal seat; men fill 2 of the remaining 3 normal seats; L must stay empty. Step-by-Step Solution: Case 1: W→L (1 way). M1,M2 on 4 normal seats: P(4,2) = 4*3 = 12Case 2: W on a normal seat: 4 choices; then M1,M2 on remaining 3 normal seats: P(3,2) = 6 → 4*6 = 24Total = 12 + 24 = 36 Verification / Alternative check: Counting by first selecting seats then permuting people yields the same result; constraints force L to be taken by W or left empty. Why Other Options Are Wrong: 48 and 60 overcount by allowing men on L or by double-counting orders. 15 is far too small. Common Pitfalls: Accidentally placing a man in the ladies-only seat. Final Answer: 36

Question 9

Minimum matches in a single-elimination tournament with byes: An elimination tournament has 139 players. Some rounds may include a bye if there is an odd number. What is the minimum number of matches required to determine a single champion?

Accepted Answer

Introduction / Context: In any single-elimination tournament (knockout), each match eliminates exactly one player. Regardless of how byes are assigned, to go from n players down to 1 champion, we must eliminate n − 1 players, which requires n − 1 matches. Given Data / Assumptions: n = 139 entrants. Single-elimination; a bye is not a match and eliminates no one. Each match eliminates exactly one player. Concept / Approach: Matches needed = total eliminations = n − 1. Step-by-Step Solution: Required matches = 139 − 1 = 138 Verification / Alternative check: Construct any bracket: every non-champion loses exactly one match; count of losses equals number of matches. The champion has no loss. Thus total matches equals losers = 138. Why Other Options Are Wrong: 136, 137 undercount; 139 overcounts by one. Common Pitfalls: Thinking byes change the total count; they alter scheduling, not the total eliminations. Final Answer: 138

Question 10

Choosing one question from each of three exercises: A test picks one question each from Exercise 7 (12 questions), Exercise 8 (18 questions), and Exercise 9 (9 questions). In how many ways can the three-question paper be formed?

Accepted Answer

Introduction / Context: Selecting one item from each of several independent pools uses the multiplication principle. Here we choose one question from each of three exercises with given counts. Given Data / Assumptions: Exercise 7 has 12 questions. Exercise 8 has 18 questions. Exercise 9 has 9 questions. Choices are independent. Concept / Approach: Total papers = 12 * 18 * 9. Step-by-Step Solution: 12 * 18 = 216216 * 9 = 1944 Verification / Alternative check: No overlaps or constraints exist, so simple multiplication applies directly. Why Other Options Are Wrong: 2094, 1894, 2194 are arithmetic distractors, not equal to 12*18*9. Common Pitfalls: Adding the counts instead of multiplying across independent choices. Final Answer: 1944

Question 11

Handshake count among 12 people: If every pair of distinct persons among 12 shakes hands exactly once, how many handshakes occur?

Accepted Answer

Introduction / Context: Each handshake corresponds to an unordered pair of people. Counting unique handshakes among n participants equals the number of 2-combinations C(n,2). Given Data / Assumptions: n = 12 distinct people. One handshake per unordered pair. Concept / Approach: Count unordered pairs: C(12,2). Step-by-Step Solution: C(12,2) = 12 * 11 / 2 = 66 Verification / Alternative check: Degree-sum in complete graph K12: sum of degrees = 12*11; divide by 2 equals number of edges = 66 (handshakes). Why Other Options Are Wrong: 55, 44, 77 are values of C(n,2) for other n or guesses. Common Pitfalls: Double-counting ordered pairs; order does not matter for a handshake. Final Answer: 66

Question 12

Necklace/chain arrangements with 5 distinct beads: How many distinct chains (flip considered identical) can be formed from 5 different colored beads?

Accepted Answer

Introduction / Context: Counting circular bead arrangements where reflections (flips) are considered identical uses necklace or “free” circular permutations. For n distinct beads on a loop with flips identical, the count is (n − 1)! / 2 for n ≥ 3 (when no additional symmetries collide distinct colorings in general position). Given Data / Assumptions: n = 5 distinct colored beads. Rotations considered the same; reflections (mirror images) also considered the same. Concept / Approach: Free necklace count for distinct beads: (n − 1)! / 2. Step-by-Step Solution: (5 − 1)! / 2 = 4! / 2 = 24 / 2 = 12 Verification / Alternative check: Polya’s enumeration for all-distinct colors also yields (n−1)!/2 when flips are identified, matching 12 for n=5. Why Other Options Are Wrong: 24 equals circular without identifying flips; 18 and 30 are not standard counts here. Common Pitfalls: Forgetting to divide by 2 to identify mirror images as the same. Final Answer: 12

Question 13

Forming a committee from men and women: In how many ways can a committee of 3 men and 3 women be formed from 6 men and 5 women?

Accepted Answer

Introduction / Context: We independently choose required counts from men and women, then multiply by the multiplication principle because selections are independent categories (men vs women). Order within the committee does not matter (combinations, not permutations). Given Data / Assumptions: Men: choose 3 from 6 → C(6,3). Women: choose 3 from 5 → C(5,3). Total committees = product of these combinations. Concept / Approach: Use combinations because only membership matters. Multiply independent category choices. Step-by-Step Solution: C(6,3) = 20C(5,3) = 10Total = 20 * 10 = 200 Verification / Alternative check: Enumerating by first choosing all 6 and then picking which 3 are men also reduces to the same product once symmetry is accounted for; 200 is exact. Why Other Options Are Wrong: 180, 150, 210 are common distractors from misapplying permutations or adding counts. Common Pitfalls: Using permutations (ordering), which would overcount identical committees. Final Answer: 200

Question 14

Two series (A/B) for 12 students in two rows of six: Place 12 students in two rows of six each. Two booklet series (A and B) are used. No identical series may be side-by-side horizontally, and students sitting one behind the other must have the same…

Accepted Answer

Introduction / Context: We must assign series labels A/B to seats so that horizontally adjacent seats alternate (no identical side-by-side) while vertically aligned seats share the same series. This forces an ABABAB pattern across each row, with columns matched between rows. Then we assign specific students to A-labeled positions and B-labeled positions, respectively. Given Data / Assumptions: Two rows × six seats each (12 seats total). Adjacent horizontally cannot share series. Each column’s pair (front/back) must have the same series. Concept / Approach: Series layout: ABABAB or BABABA across columns (2 choices). Exactly 6 seats labeled A and 6 labeled B overall (pair per column). Choose which 6 students get series A, then permute within A seats; similarly for B. Step-by-Step Solution: Series pattern choices = 2 (start with A or B)Choose A-students: C(12,6)Permute A-students among 6 A-seats: 6!Permute B-students among 6 B-seats: 6!Total ways = 2 * C(12,6) * 6! * 6! Verification / Alternative check: Any alternative series layout violating AB alternation would create a horizontal clash. With columns locked to a single series, exactly two alternating templates exist, confirming the 2 factor. Why Other Options Are Wrong: 6! * 6! omits the choice of which 6 students are assigned to A vs B, and the pattern factor 2. 7! * 7! is unrelated to the structure of seats and series. Common Pitfalls: Forgetting the initial AB vs BA pattern choice or the combination step for selecting the A-group. Final Answer: 2 x 12C6 x (6!)2

Question 15

A department has 16 employees (8 men and 8 women). You must form a project team with exactly 3 men and 3 women. How many distinct teams are possible (order does not matter)?

Accepted Answer

Introduction / Context: This is a direct combinations problem: select a fixed count from each of two disjoint groups (men and women) where order is irrelevant and no role distinctions exist within a chosen team. Given Data / Assumptions: Men = 8, Women = 8. Select exactly 3 men and exactly 3 women. Teams are sets; internal order does not matter. Concept / Approach: Use combinations independently on each group and multiply: total ways = C(8,3) * C(8,3). This is the fundamental rule of counting for independent choices across disjoint sets. Step-by-Step Solution: C(8,3) = 8*7*6 / (3*2*1) = 56.Total teams = 56 * 56 = 3136. Verification / Alternative check: Symmetry implies selecting either gender first yields the same multiplication. No ordering or role labels are present, so there is no extra factorial factor. Why Other Options Are Wrong: 112 and 720 come from misusing permutations or partial counts; 112896 multiplies by spurious orderings. Common Pitfalls: Accidentally using permutations (which would imply ordered roles) or adding instead of multiplying across independent gender selections. Final Answer: 3136

Question 16

A man has 9 friends (4 boys and 5 girls). In how many ways can he invite a group that contains exactly 3 girls (with any number of boys, including possibly none)?

Accepted Answer

Introduction / Context: The invitation must include exactly 3 girls, while the number of boys is unconstrained. We count subsets: first fix the girls, then freely choose any subset of boys. Given Data / Assumptions: Girls = 5, Boys = 4. Pick exactly 3 girls. Pick any subset of the 4 boys (0 through 4 allowed). Concept / Approach: Multiply ways to choose girls by ways to (independently) choose boys. The number of subsets of 4 boys is 2^4. The number of 3-girl choices is C(5,3). Step-by-Step Solution: C(5,3) = 10.Subsets of 4 boys = 2^4 = 16.Total invitation groups = 10 * 16 = 160. Verification / Alternative check: Enumerating boy counts (0..4) gives Σ C(5,3)*C(4,k) = C(5,3) * 2^4 by the binomial theorem. Why Other Options Are Wrong: 320 double counts by also forcing boys; 200 and 80 arise from partial sums or mixing constraints. Common Pitfalls: Assuming a fixed total size or forgetting to include the empty boy subset. Final Answer: 160

Question 17

Four books A, B, C, and D are stacked vertically (top to bottom). In how many arrangements are A and B not adjacent to each other?

Accepted Answer

Introduction / Context: This is a standard “not adjacent” permutation constraint. We count all arrangements and subtract those with the forbidden adjacency. Given Data / Assumptions: Books are distinct: A, B, C, D. Linear (vertical) order matters. Adjacency refers to consecutive positions. Concept / Approach: Total permutations of 4 distinct books are 4! = 24. Adjacency count is found by treating the adjacent pair (A,B) or (B,A) as a single block and permuting blocks. Step-by-Step Solution: All arrangements = 4! = 24.Adjacent arrangements: form a block for A & B. Block permutations: 3! ways to place (AB) with C and D, times 2 internal orders (AB or BA) → 3!*2 = 12.Non-adjacent = 24 − 12 = 12. Verification / Alternative check: Direct enumeration or inclusion–exclusion on positions of A and B gives the same result. Why Other Options Are Wrong: 18, 14, 9 reflect subtracting the wrong adjacency count or forgetting the two internal orders of the (A,B) block. Common Pitfalls: Counting AB and BA as one, or forgetting there are only 3! placements of the block with C and D. Final Answer: 12

Question 18

Boxes numbered 1 to 5 are in a row. Each box is filled with either a red (R) or a blue (B) ball such that no two adjacent boxes contain blue balls. Balls of the same colour are identical. How many valid colourings are possible?

Accepted Answer

Introduction / Context: We are counting binary strings of length 5 using symbols {R,B} with the constraint that there are no consecutive B’s. Boxes are distinct (positions matter), colours are identical items. Given Data / Assumptions: Length n = 5. Alphabet {R,B}; constraint: no BB. All that matters is the pattern across positions. Concept / Approach: Let a(n) be the number of valid strings of length n with no consecutive B. Standard recurrence: a(n) = a(n−1) + a(n−2), with bases a(1) = 2 (R,B) and a(2) = 3 (RR, RB, BR). This is the Fibonacci-type count for “no adjacent 1s” problems. Step-by-Step Solution: a(1)=2, a(2)=3.a(3)=a(2)+a(1)=3+2=5.a(4)=5+3=8.a(5)=8+5=13. Verification / Alternative check: Constructive count by splitting into strings ending with R (a(n−1) choices) or with B (requires previous R, giving a(n−2)). Why Other Options Are Wrong: 10 and 8 undercount by truncating the recurrence; 22 is overcounting unconstrained patterns; 13 is exact. Common Pitfalls: Assuming indistinguishable boxes (they are positioned) or forgetting that pattern RB R… is allowed but BB is not. Final Answer: 13

Question 19

A child has 3 distinct marbles and 4 distinct pockets. In how many ways can the marbles be distributed among the pockets (pockets may be empty)?

Accepted Answer

Introduction / Context: This is a classic “balls into boxes” scenario with distinct balls (marbles) and distinct boxes (pockets). Each marble independently chooses a pocket; pockets may be empty. Given Data / Assumptions: Marbles are distinct (e.g., different colours). Pockets are distinct (different locations). Multiple marbles can go into the same pocket; empty pockets are allowed. Concept / Approach: For each of the 3 marbles, choose one of 4 pockets. By the multiplication principle, total outcomes = 4 * 4 * 4 = 4^3. Step-by-Step Solution: Choices per marble = 4.Marbles act independently: total = 4^3 = 64. Verification / Alternative check: Equivalent to functions from a 3-element marble set to a 4-element pocket set; the number of functions is 4^3. Why Other Options Are Wrong: 12 and 60 correspond to permutations/arrangements; 256 is 4^4 (as if 4 marbles). Common Pitfalls: Misinterpreting marbles as identical (which would be a stars-and-bars count) or forbidding empty pockets. Final Answer: 64

Question 20

How many distinct arrangements of the letters in the word “ASSASSINATION” are possible if all four S’s must be together as a single block?

Accepted Answer

Introduction / Context: We count permutations with repeated letters under a block constraint. The four S’s must appear contiguously, treated as a single super-letter. Given Data / Assumptions: Word length = 13 letters. Letter multiplicities (excluding the S-block): A×3, I×2, N×2, T×1, O×1; S is forced as one block. Concept / Approach: Replace the four S’s by one block “S⁴”. Then arrange this block with the remaining letters accounting for multiplicities. Use multinomial: total items = 10 (block + other letters). Count = 10! / (3! 2! 2!). Step-by-Step Solution: Items to arrange: {S⁴-block, A, A, A, I, I, N, N, T, O} (10 items).Number of arrangements = 10! / (3! * 2! * 2!) = 3628800 / 24 = 151200. Verification / Alternative check: Without the block constraint, total would be 13! / (4! 3! 2! 2!). Enforcing the block correctly reduces the count to 10! / (3! 2! 2!). Why Other Options Are Wrong: 10! ignores repeated letters; 3628800 is 10! (fails to divide by repeats); “14!(4!)” is not meaningful here. Common Pitfalls: Treating the S’s as distinct or forgetting to divide by identical letters’ factorials. Final Answer: 151200

Permutation and Combination Questions