Q: Three-digit numbers from S = {2, 3, 4, 5, 7, 9} less than 400: How many distinct-digit three-digit numbers < 400 can be formed?

Introduction / Context: We form three-digit numbers from a given set with distinct digits and a numeric bound (< 400). The hundreds place determines the bound handling; then we count permutations for the remaining places. Given Data / Assumptions: S = {2, 3, 4, 5, 7, 9}; digits cannot repeat; number < 400. Concept / Approach: Constrain the hundreds digit first. Only 2 or 3 keep the number below 400. For each choice of hundreds, count permutations of tens and units from the remaining digits. Step-by-Step Solution: Hundreds choices: 2 options (2 or 3)For a fixed hundreds digit: tens = 5 choices, units = 4 choices (distinctness)Per hundreds choice: 5 * 4 = 20 numbersTotal = 2 * 20 = 40 Verification / Alternative check: Enumerating quickly for hundreds=2 and 3 confirms symmetry and count. Why Other Options Are Wrong: 20 counts only one hundreds choice; 80 or 120 overcount beyond the < 400 constraint. Common Pitfalls: Accidentally including 4,5,7,9 as hundreds, which violates the bound. Final Answer: 40

Q: Counting triangles with collinear points present: Out of 12 points, 7 are collinear. How many distinct triangles can be formed?

Introduction / Context: Triangles are determined by choosing any 3 non-collinear points. When many points lie on a single line, triples entirely from that line must be excluded because they do not form a triangle. Given Data / Assumptions: 12 total points; 7 lie on the same straight line; assume other combinations outside the 7 are in general position (no unintended collinearity). Concept / Approach: Total 3-point choices minus the degenerate 3-point choices from the collinear set gives the triangle count. Step-by-Step Solution: Total triples = C(12, 3) = 220Collinear triples among the 7 = C(7, 3) = 35Triangles = 220 − 35 = 185 Verification / Alternative check: No other line contains 3 or more collinear points by the statement, so only one subtraction is needed. Why Other Options Are Wrong: 201 or 158 misapply the subtraction; 19 is far too small. Common Pitfalls: Forgetting to subtract only degenerate triples; subtracting pairs (which still form triangles) is incorrect. Final Answer: 185

Q: Arranging JUDGE with vowels together: How many permutations of J, U, D, G, E place the two vowels together?

Introduction / Context: We want permutations of the letters of JUDGE with the vowels U and E adjacent. This is a classic “keep-a-group-together” arrangement question. Given Data / Assumptions: Letters: J, U, D, G, E (all distinct). Vowels: U, E; consonants: J, D, G. Concept / Approach: Treat the two vowels as a single block, then permute that block with the three consonants. Inside the block, the two vowels can swap. Step-by-Step Solution: Entities to arrange: [UE], J, D, G ⇒ 4 distinct entitiesArrange entities: 4! = 24Internal order of vowels: 2! = 2Total = 24 * 2 = 48 Verification / Alternative check: Complementary counting (total 5! minus permutations with vowels separated) also yields 48 after careful inclusion–exclusion. Why Other Options Are Wrong: 24 misses the internal swap; 60 or 120 overcount beyond constraints. Common Pitfalls: Forgetting the 2! within the vowel block. Final Answer: 48

Q: Committee from officers and jawans (at least one officer): From 4 officers and 8 jawans, how many 6-person committees include at least one officer?

Introduction / Context: We must count 6-person committees that include at least one officer. The complement (no officer) is easier to compute and subtract from the total committees possible. Given Data / Assumptions: Total people = 12 (4 officers + 8 jawans); committee size = 6. Concept / Approach: Total − (no officer). No-officer committees must be formed entirely from jawans. Step-by-Step Solution: Total committees = C(12, 6) = 924No-officer committees = C(8, 6) = 28Required = 924 − 28 = 896 Verification / Alternative check: Summing by officer count (1 to 4) reproduces 896. Why Other Options Are Wrong: 986, 886, 996 are off by ignoring the complement or miscomputing combinations. Common Pitfalls: Accidentally allowing 0 officers; arithmetic slips on C(12,6) or C(8,6). Final Answer: 896

Q: Simple committee count: From a group of 6 men and 4 women, how many 4-person committees can be formed (no constraints)?

Introduction / Context: With no restrictions, the number of committees depends only on total group size and committee size. Gender labels do not matter for counting unless constraints are imposed. Given Data / Assumptions: Total people = 10; committee size = 4. Concept / Approach: Use combinations C(10, 4) since order of selection does not matter. Step-by-Step Solution: C(10, 4) = 10*9*8*7 / (4*3*2*1) = 210 Verification / Alternative check: Symmetry: C(10, 4) = C(10, 6) = 210. Why Other Options Are Wrong: 225, 195, 185 do not equal C(10, 4). Common Pitfalls: Using permutations instead of combinations. Final Answer: 210

Q: Number of straight lines from 10 points with 7 collinear: How many distinct straight lines can be formed?

Introduction / Context: Distinct lines arise from pairs of points, but when many points are collinear, all pairs among them correspond to the same line and must not be overcounted. Given Data / Assumptions: 10 points total; 7 on one line; assume the other 3 are in general position relative to the 7 and each other. Concept / Approach: Total potential lines from all pairs minus the overcount among the collinear 7, then add back that single line once. Step-by-Step Solution: Total pairs = C(10, 2) = 45Pairs along the 7-point line = C(7, 2) = 21Distinct lines = 45 − 21 + 1 = 25 Verification / Alternative check: List categories: the one big line from the 7, lines from an off-line point to each of the 7, and lines among off-line points — totals reconcile to 25. Why Other Options Are Wrong: 26 or 21 result from missing either the subtraction or the add-back of one line. Common Pitfalls: Forgetting to add the unique collinear line back after subtracting its pairs. Final Answer: 25

Q: Permutations of BANANA: How many distinct permutations of the letters in BANANA are possible?

Introduction / Context: BANANA contains repeated letters, so we count distinct permutations by dividing total factorial by repeated-letter factorials. Given Data / Assumptions: Letters: A×3, N×2, B×1; total letters = 6. Concept / Approach: Distinct permutations = 6! / (3! * 2!). Step-by-Step Solution: 6! = 720Divide by 3! = 6 ⇒ 120Divide by 2! = 2 ⇒ 60 Verification / Alternative check: Sanity check with smaller multisets (e.g., AAB) supports the approach. Why Other Options Are Wrong: 270, 120, 360 are not equal to 6!/(3!*2!). Common Pitfalls: Missing one of the repeated-letter divisors. Final Answer: 60

Q: Handshake problem: In a room, everyone shakes hands with everyone else exactly once. If the total number of handshakes is 66, how many persons are in the room?

Introduction / Context: When each pair among n people shakes hands once, the total handshakes equal the number of 2-element subsets: C(n, 2). We invert this to find n from a given total. Given Data / Assumptions: Total handshakes = 66. Concept / Approach: Solve C(n, 2) = 66 ⇒ n(n − 1)/2 = 66 ⇒ n(n − 1) = 132, find integer n. Step-by-Step Solution: Try n = 12 ⇒ 12*11 = 132 (matches)Hence n = 12 Verification / Alternative check: Quadratic: n^2 − n − 132 = 0 ⇒ (n − 12)(n + 11) = 0 ⇒ n = 12. Why Other Options Are Wrong: 11 gives 55; 13 gives 78; 14 gives 91. Common Pitfalls: Using permutations n(n − 1) instead of combinations. Final Answer: 12

Q: Arrangements of CIVILIZATION: How many distinct permutations are there for the letters of “civilization”?

Introduction / Context: The word “civilization” has repeating letters, notably the letter i. Distinct permutations require dividing by factorials of identical letter counts. Given Data / Assumptions: Total letters = 12; the letter i appears 4 times; assume other letters are distinct for counting purposes. Concept / Approach: Distinct permutations = 12! / 4! due to the four indistinguishable i’s. Step-by-Step Solution: Compute symbolically: 12! / 4!No further identical-letter divisors are needed beyond i^4 for this count. Verification / Alternative check: Letter-by-letter counting confirms four i’s and the remaining letters each occur once in typical textbook treatments. Why Other Options Are Wrong: Subtracting 1 (as in options b/c) is unjustified; 13!/5! mismatches length and multiplicity. Common Pitfalls: Miscounting the number of i’s or injecting spurious duplicate counts for unique letters. Final Answer: 12! / 4!

Question 1

Linking permutations and combinations: If P(n, r) = 720 · C(n, r), find r.

Accepted Answer

Introduction / Context: Permutation and combination are related by P(n, r) = C(n, r) * r!. The equation given directly identifies r! as a numeric constant times C(n, r), enabling a quick solution for r. Given Data / Assumptions: P(n, r) = 720 · C(n, r). Concept / Approach: Use the identity P(n, r) = C(n, r) · r!. If C(n, r) ≠ 0, divide both sides by C(n, r) to isolate r!. Step-by-Step Solution: C(n, r) · r! = 720 · C(n, r)⇒ r! = 720 = 6!⇒ r = 6 Verification / Alternative check: Plug r = 6: indeed P(n, 6) = C(n, 6) · 6! = 720 · C(n, 6). Why Other Options Are Wrong: 5! = 120, 4! = 24, 7! = 5040; none equal 720. Common Pitfalls: Trying to solve for n unnecessarily; r is determined solely by 720. Final Answer: 6

Question 2

Three-digit numbers from S = {2, 3, 4, 5, 7, 9} less than 400: How many distinct-digit three-digit numbers < 400 can be formed?

Accepted Answer

Introduction / Context: We form three-digit numbers from a given set with distinct digits and a numeric bound (< 400). The hundreds place determines the bound handling; then we count permutations for the remaining places. Given Data / Assumptions: S = {2, 3, 4, 5, 7, 9}; digits cannot repeat; number < 400. Concept / Approach: Constrain the hundreds digit first. Only 2 or 3 keep the number below 400. For each choice of hundreds, count permutations of tens and units from the remaining digits. Step-by-Step Solution: Hundreds choices: 2 options (2 or 3)For a fixed hundreds digit: tens = 5 choices, units = 4 choices (distinctness)Per hundreds choice: 5 * 4 = 20 numbersTotal = 2 * 20 = 40 Verification / Alternative check: Enumerating quickly for hundreds=2 and 3 confirms symmetry and count. Why Other Options Are Wrong: 20 counts only one hundreds choice; 80 or 120 overcount beyond the < 400 constraint. Common Pitfalls: Accidentally including 4,5,7,9 as hundreds, which violates the bound. Final Answer: 40

Question 3

Counting triangles with collinear points present: Out of 12 points, 7 are collinear. How many distinct triangles can be formed?

Accepted Answer

Introduction / Context: Triangles are determined by choosing any 3 non-collinear points. When many points lie on a single line, triples entirely from that line must be excluded because they do not form a triangle. Given Data / Assumptions: 12 total points; 7 lie on the same straight line; assume other combinations outside the 7 are in general position (no unintended collinearity). Concept / Approach: Total 3-point choices minus the degenerate 3-point choices from the collinear set gives the triangle count. Step-by-Step Solution: Total triples = C(12, 3) = 220Collinear triples among the 7 = C(7, 3) = 35Triangles = 220 − 35 = 185 Verification / Alternative check: No other line contains 3 or more collinear points by the statement, so only one subtraction is needed. Why Other Options Are Wrong: 201 or 158 misapply the subtraction; 19 is far too small. Common Pitfalls: Forgetting to subtract only degenerate triples; subtracting pairs (which still form triangles) is incorrect. Final Answer: 185

Question 4

Arranging JUDGE with vowels together: How many permutations of J, U, D, G, E place the two vowels together?

Accepted Answer

Introduction / Context: We want permutations of the letters of JUDGE with the vowels U and E adjacent. This is a classic “keep-a-group-together” arrangement question. Given Data / Assumptions: Letters: J, U, D, G, E (all distinct). Vowels: U, E; consonants: J, D, G. Concept / Approach: Treat the two vowels as a single block, then permute that block with the three consonants. Inside the block, the two vowels can swap. Step-by-Step Solution: Entities to arrange: [UE], J, D, G ⇒ 4 distinct entitiesArrange entities: 4! = 24Internal order of vowels: 2! = 2Total = 24 * 2 = 48 Verification / Alternative check: Complementary counting (total 5! minus permutations with vowels separated) also yields 48 after careful inclusion–exclusion. Why Other Options Are Wrong: 24 misses the internal swap; 60 or 120 overcount beyond constraints. Common Pitfalls: Forgetting the 2! within the vowel block. Final Answer: 48

Question 5

Committee from officers and jawans (at least one officer): From 4 officers and 8 jawans, how many 6-person committees include at least one officer?

Accepted Answer

Introduction / Context: We must count 6-person committees that include at least one officer. The complement (no officer) is easier to compute and subtract from the total committees possible. Given Data / Assumptions: Total people = 12 (4 officers + 8 jawans); committee size = 6. Concept / Approach: Total − (no officer). No-officer committees must be formed entirely from jawans. Step-by-Step Solution: Total committees = C(12, 6) = 924No-officer committees = C(8, 6) = 28Required = 924 − 28 = 896 Verification / Alternative check: Summing by officer count (1 to 4) reproduces 896. Why Other Options Are Wrong: 986, 886, 996 are off by ignoring the complement or miscomputing combinations. Common Pitfalls: Accidentally allowing 0 officers; arithmetic slips on C(12,6) or C(8,6). Final Answer: 896

Question 6

Simple committee count: From a group of 6 men and 4 women, how many 4-person committees can be formed (no constraints)?

Accepted Answer

Introduction / Context: With no restrictions, the number of committees depends only on total group size and committee size. Gender labels do not matter for counting unless constraints are imposed. Given Data / Assumptions: Total people = 10; committee size = 4. Concept / Approach: Use combinations C(10, 4) since order of selection does not matter. Step-by-Step Solution: C(10, 4) = 10*9*8*7 / (4*3*2*1) = 210 Verification / Alternative check: Symmetry: C(10, 4) = C(10, 6) = 210. Why Other Options Are Wrong: 225, 195, 185 do not equal C(10, 4). Common Pitfalls: Using permutations instead of combinations. Final Answer: 210

Question 7

Number of straight lines from 10 points with 7 collinear: How many distinct straight lines can be formed?

Accepted Answer

Introduction / Context: Distinct lines arise from pairs of points, but when many points are collinear, all pairs among them correspond to the same line and must not be overcounted. Given Data / Assumptions: 10 points total; 7 on one line; assume the other 3 are in general position relative to the 7 and each other. Concept / Approach: Total potential lines from all pairs minus the overcount among the collinear 7, then add back that single line once. Step-by-Step Solution: Total pairs = C(10, 2) = 45Pairs along the 7-point line = C(7, 2) = 21Distinct lines = 45 − 21 + 1 = 25 Verification / Alternative check: List categories: the one big line from the 7, lines from an off-line point to each of the 7, and lines among off-line points — totals reconcile to 25. Why Other Options Are Wrong: 26 or 21 result from missing either the subtraction or the add-back of one line. Common Pitfalls: Forgetting to add the unique collinear line back after subtracting its pairs. Final Answer: 25

Question 8

Permutations of BANANA: How many distinct permutations of the letters in BANANA are possible?

Accepted Answer

Introduction / Context: BANANA contains repeated letters, so we count distinct permutations by dividing total factorial by repeated-letter factorials. Given Data / Assumptions: Letters: A×3, N×2, B×1; total letters = 6. Concept / Approach: Distinct permutations = 6! / (3! * 2!). Step-by-Step Solution: 6! = 720Divide by 3! = 6 ⇒ 120Divide by 2! = 2 ⇒ 60 Verification / Alternative check: Sanity check with smaller multisets (e.g., AAB) supports the approach. Why Other Options Are Wrong: 270, 120, 360 are not equal to 6!/(3!*2!). Common Pitfalls: Missing one of the repeated-letter divisors. Final Answer: 60

Question 9

Handshake problem: In a room, everyone shakes hands with everyone else exactly once. If the total number of handshakes is 66, how many persons are in the room?

Accepted Answer

Introduction / Context: When each pair among n people shakes hands once, the total handshakes equal the number of 2-element subsets: C(n, 2). We invert this to find n from a given total. Given Data / Assumptions: Total handshakes = 66. Concept / Approach: Solve C(n, 2) = 66 ⇒ n(n − 1)/2 = 66 ⇒ n(n − 1) = 132, find integer n. Step-by-Step Solution: Try n = 12 ⇒ 12*11 = 132 (matches)Hence n = 12 Verification / Alternative check: Quadratic: n^2 − n − 132 = 0 ⇒ (n − 12)(n + 11) = 0 ⇒ n = 12. Why Other Options Are Wrong: 11 gives 55; 13 gives 78; 14 gives 91. Common Pitfalls: Using permutations n(n − 1) instead of combinations. Final Answer: 12

Question 10

Arrangements of CIVILIZATION: How many distinct permutations are there for the letters of “civilization”?

Accepted Answer

Introduction / Context: The word “civilization” has repeating letters, notably the letter i. Distinct permutations require dividing by factorials of identical letter counts. Given Data / Assumptions: Total letters = 12; the letter i appears 4 times; assume other letters are distinct for counting purposes. Concept / Approach: Distinct permutations = 12! / 4! due to the four indistinguishable i’s. Step-by-Step Solution: Compute symbolically: 12! / 4!No further identical-letter divisors are needed beyond i^4 for this count. Verification / Alternative check: Letter-by-letter counting confirms four i’s and the remaining letters each occur once in typical textbook treatments. Why Other Options Are Wrong: Subtracting 1 (as in options b/c) is unjustified; 13!/5! mismatches length and multiplicity. Common Pitfalls: Miscounting the number of i’s or injecting spurious duplicate counts for unique letters. Final Answer: 12! / 4!

Question 11

DIRECTOR with vowels together: Arrange the letters of DIRECTOR so that all the vowels occur consecutively. How many ways?

Accepted Answer

Introduction / Context: We keep the vowels together as a block. DIRECTOR has vowels I, E, O (3) and consonants D, R, C, T, R (5, with R repeated twice). Count arrangements of the block with consonants, then the internal permutations of the vowels. Given Data / Assumptions: Word: D I R E C T O R; vowels = {I, E, O}; consonants = {D, R, C, T, R} with R repeated twice. Concept / Approach: Treat the vowel block as one entity. Then there are 6 entities (vowel block + 5 consonants), with a duplication among consonants (R×2). Arrange entities and multiply by the permutations within the vowel block. Step-by-Step Solution: Entities arrangements = 6! / 2! = 720 / 2 = 360 (due to two R’s)Vowel permutations inside block = 3! = 6Total = 360 * 6 = 2160 Verification / Alternative check: Explicit small-case analogs validate the block method with duplicates. Why Other Options Are Wrong: 4320 doubles by ignoring the repeated R; 2720 and 1120 are inconsistent with the duplication and block logic. Common Pitfalls: Forgetting to divide by 2! for the two R’s, or missing the 3! for the internal vowel permutations. Final Answer: 2160

Question 12

Arrangements of RECOVER: How many distinct permutations of the letters in RECOVER are possible?

Accepted Answer

Introduction / Context: RECOVER has repeated letters (R and E). We count distinct permutations by dividing 7! by the factorials of the repeated counts. Given Data / Assumptions: Total letters = 7; R occurs 2 times; E occurs 2 times; others are distinct. Concept / Approach: Distinct permutations = 7! / (2! * 2!). Step-by-Step Solution: 7! = 5040Divide by 2! * 2! = 4 ⇒ 5040 / 4 = 1260 Verification / Alternative check: Check letter counts directly: indeed two Rs and two Es. Why Other Options Are Wrong: 5040 ignores repetition; 1210, 1200 are not equal to 5040/4. Common Pitfalls: Omitting one of the repeated-letter divisors. Final Answer: 1260

Question 13

Arranging multi-volume books as inseparable blocks: There are 20 books: 4 singles and three multi-volume works of sizes 8, 5, and 3. In how many ways can all books be arranged on a shelf so that volumes of the same work stay together?

Accepted Answer

Introduction / Context: When multi-volume works must remain contiguous, treat each work as an inseparable block. The singles remain individual items. Then multiply by the internal permutations of volumes within each block (volumes are distinct). Given Data / Assumptions: Singles: 4 books; multi-volume works: sizes 8, 5, 3 (volumes distinct but required to be together). Concept / Approach: First, arrange “items” = 4 singles + 3 blocks = 7 items in 7! ways. Next, for each block, order the volumes internally: 8!, 5!, and 3!. Multiply all factors. Step-by-Step Solution: Arrange 7 items on shelf: 7!Order within 8-volume work: 8!Order within 5-volume work: 5!Order within 3-volume work: 3!Total = 7! * 8! * 5! * 3! Verification / Alternative check: Counting principle (blocks first, internal orders second) is standard and consistent. Why Other Options Are Wrong: 7! 8! 4! 3! uses 4! instead of 5!; 7! 6! 5! 3! incorrectly replaces 8! with 6!. Common Pitfalls: Forgetting to include internal permutations, or misreading the sizes of the works. Final Answer: 7! 8! 5! 3!

Question 14

Blocks of identical copies kept together: A library has two books each with three identical copies, and three other books each with two identical copies. In how many ways can all these books be arranged on a shelf so that copies of the same book are…

Accepted Answer

Introduction / Context: We have five distinct titles with multiple identical copies per title: two titles have 3 copies each, three titles have 2 copies each. Since identical copies within a title are indistinguishable and must be kept together, each title forms a single block on the shelf. Given Data / Assumptions: Titles: A, B with 3 identical copies each; C, D, E with 2 identical copies each. “Not separated” means all copies of a title appear contiguously, acting as one block. Identical copies ⇒ internal permutations within a block do not create new arrangements. Concept / Approach: Treat each title as one block. Then we simply arrange 5 distinct blocks on the shelf. Step-by-Step Solution: Number of blocks = 5Arrangements of 5 distinct blocks = 5! = 120No internal multiplicative factor: copies per block are identical, not distinct. Verification / Alternative check: Alternative reasoning confirms that any reordering within a block of identical copies is not distinguishable and thus does not increase the count. Why Other Options Are Wrong: 180, 160, 140 imply extra (nonexistent) internal permutations or incorrect block counts. Common Pitfalls: Treating identical copies as distinct (which would erroneously multiply the count). Final Answer: 120

Question 15

Seating in a row with a block condition: Four boys and two girls are to be seated in a straight line such that the two girls are always together (treated as an adjacent pair). In how many distinct linear arrangements can they be seated?

Accepted Answer

Introduction / Context: This is a classic permutations-with-a-block problem. When two specified people must sit together in a line, we can temporarily treat them as a single unit while counting arrangements, then multiply by the internal permutations of that block. Given Data / Assumptions: Individuals: 4 distinct boys and 2 distinct girls. Constraint: the two girls must be adjacent. Order matters, and all persons are distinct. Concept / Approach: Treat the two girls as one “block” to honor adjacency. Count permutations of the resulting units in a row. Multiply by internal arrangements of the two girls within their block. Step-by-Step Solution: Make a block G = (G1,G2). Units to arrange: {G, B1, B2, B3, B4} → 5! ways.Internal arrangements of (G1,G2) → 2! ways.Total arrangements = 5! * 2! = 120 * 2 = 240. Verification / Alternative check: Direct adjacency counting using positions for the pair also leads to the same product 5! * 2! = 240, confirming the result. Why Other Options Are Wrong: 120 counts the five units but misses the internal 2! for the girls. 720 is 6! ignoring the adjacency constraint. 148 is not a standard permutation count from this setup. Common Pitfalls: Forgetting the internal permutations of the two girls. Accidentally counting 6! without enforcing adjacency. Final Answer: 240

Question 16

Words from distinct letters with a vowel requirement: From the letters {a, b, c, d, e, f}, how many 3-letter words (ordered arrangements, no repetition) can be formed that contain at least one vowel?

Accepted Answer

Introduction / Context: We are forming ordered arrangements (permutations) of length 3 from 6 distinct letters, under the condition that each word contains at least one vowel. Using the complement (no vowels) simplifies the count substantially. Given Data / Assumptions: Alphabet set: {a, b, c, d, e, f} with vowels a, e. Length = 3; no repetition; order matters. At least one vowel must appear in the arrangement. Concept / Approach: Total 3-permutations from 6 letters = 6P3. Subtract arrangements with no vowels: choose only from {b, c, d, f} = 4P3. Use inclusion–exclusion via complement. Step-by-Step Solution: Total = 6P3 = 6 * 5 * 4 = 120No-vowel = 4P3 = 4 * 3 * 2 = 24At least one vowel = 120 − 24 = 96 Verification / Alternative check: Direct casework (1 vowel, 2 vowels) is longer but produces the same sum as the complement method; complement is cleanest here. Why Other Options Are Wrong: 72 and 48 are partial counts from incomplete casework. “None of these” is false because 96 is achievable by complement counting. Common Pitfalls: Treating the problem as combinations instead of permutations. Forgetting to exclude the no-vowel words entirely. Final Answer: 96

Question 17

Inviting friends from one side only: A man has 5 friends and his wife has 4 friends. They will invite friends from exactly one side (either his or hers), choosing one or more from that side. In how many ways can invitations be made?

Accepted Answer

Introduction / Context: The phrase “either of their friends, one or more” is commonly interpreted to mean: choose a nonempty subset from exactly one side (his side or her side), but not a mixture. We count nonempty subsets from each side and add them. Given Data / Assumptions: His friends: 5 distinct → nonempty subsets: 2^5 − 1 = 31. Her friends: 4 distinct → nonempty subsets: 2^4 − 1 = 15. Exactly one side is chosen. Concept / Approach: Add the counts from the two disjoint choices (his side OR her side). Step-by-Step Solution: Ways (his side) = 31Ways (her side) = 15Total = 31 + 15 = 46 Verification / Alternative check: If we had allowed mixing sides, the count would be 2^9 − 1 = 511, or (2^5 − 1)*(2^4 − 1) = 465 if at least one from each side. The given interpretation uniquely fits the provided options. Why Other Options Are Wrong: 31 counts only his side; 18 or 9 arise from misinterpreting the constraint. “None of these” is false since 46 is correct under the stated interpretation. Common Pitfalls: Including both sides simultaneously, which contradicts “either…one or more.” Final Answer: 46

Question 18

Counting strictly increasing triplets from 10 ordered positives: Given 10 positive real numbers n1 < n2 < n3 < … < n10, how many distinct triplets (ni, nj, nk) with i < j < k can be formed (i.e., strictly increasing triples by inde…

Accepted Answer

Introduction / Context: This is a combinations question: each strictly increasing triple corresponds to a set of three distinct indices i < j < k chosen from 10 positions. The actual values do not affect the count because the ordering is already strict by index. Given Data / Assumptions: Sequence of 10 numbers with n1 < n2 < … < n10. Triplet must respect i < j < k (strictly increasing by index). Concept / Approach: Choose 3 indices out of 10: combinations without order. Each choice yields exactly one increasing triplet. Step-by-Step Solution: Number of triplets = C(10,3) = 10 * 9 * 8 / (3 * 2 * 1) = 120 Verification / Alternative check: Any attempt to count by “sliding windows” would undercount (e.g., 8). The correct method uses combinations of indices, not adjacency of terms. Why Other Options Are Wrong: 45 and 90 are incorrect binomial coefficients for other parameters. 180 counts some permutations or misapplies arrangements. Common Pitfalls: Confusing “consecutive triplets” with “any strictly increasing triplets.” Final Answer: 120

Question 19

Three-letter passwords (no repetition) with at least one symmetric letter: Assume symmetric letters about a vertical axis are A, H, I, M, O, T, U, V, W, X, Y (11 letters). How many 3-letter passwords (A–Z, no repetition) contain at least one symmetr…

Accepted Answer

Introduction / Context: We work with uppercase English letters. A standard convention for “symmetric letters” (mirror symmetry about a vertical axis) is {A, H, I, M, O, T, U, V, W, X, Y}. We count 3-letter permutations (no repetition) containing at least one such letter using the complement method. Given Data / Assumptions: Total letters = 26; symmetric set size = 11; nonsymmetric = 15. Length = 3; no letter repeats; order matters. At least one symmetric letter required. Concept / Approach: Total permutations = 26P3. No-symmetric permutations = 15P3. Desired = 26P3 − 15P3. Step-by-Step Solution: 26P3 = 26 * 25 * 24 = 1560015P3 = 15 * 14 * 13 = 2730Count = 15600 − 2730 = 12870 Verification / Alternative check: Direct case splits by number of symmetric letters (1, 2, 3) lead to the same total; complement is faster and less error-prone. Why Other Options Are Wrong: 2730 is the complement (no symmetric letters). 1560000 assumes 26^3 with repetition, not 26P3. 990 is unrelated to this setup. Common Pitfalls: Using combinations instead of permutations. Assuming repetition allowed when it is prohibited. Final Answer: 12870

Question 20

Binary-answer test sequences: A test has 10 questions, each answered True (T) or False (F). Every candidate answers all questions. How many distinct T/F answer sequences are possible?

Accepted Answer

Introduction / Context: Each question admits 2 choices (T or F), independently across questions. Counting sequences of independent binary choices leads to a simple power rule, 2^n for n questions. Given Data / Assumptions: n = 10 questions. Choices per question = 2 (T or F), independent. Every candidate answers all questions. Concept / Approach: Use the multiplication principle: 2 * 2 * … * 2 (10 times) = 2^10. Step-by-Step Solution: Number of sequences = 2^10 = 1024 Verification / Alternative check: Binary coding analogy: each sequence corresponds to a 10-bit string; total 10-bit strings = 2^10 = 1024. Why Other Options Are Wrong: 512 = 2^9; too small. 20 and 40 are unrelated to 2^10. Common Pitfalls: Treating “questions” as dependent when they are independent. Final Answer: 1024

Permutation and Combination Questions