Question 1

Evaluate the permutation 5P2 (i.e., the number of ordered ways to choose 2 distinct items from 5 distinct items).

Accepted Answer

Introduction / Context:nP r counts ordered selections of r distinct items from n distinct items. Here, n = 5 and r = 2. Given Data / Assumptions: n = 5, r = 2. Formula: nP r = n! / (n − r)! = n * (n − 1) * … * (n − r + 1). Concept / Approach:Compute directly using the short product: 5 * 4. Step-by-Step Solution: 5P2 = 5 * 4 = 20. Verification / Alternative check:Factorial form: 5! / 3! = (120) / 6 = 20, consistent. Why Other Options Are Wrong:15 would be 5C2 (unordered); 18 and 122 are unrelated. Common Pitfalls:Confusing permutations (ordered) with combinations (unordered). Final Answer:20

Question 2

If C(50, r) = C(50, r+2), find r. (Use symmetry C(n, k) = C(n, n − k) and solve for k.)

Accepted Answer

Introduction / Context:Equal binomial coefficients at positions r and r+2 imply a reflection around the center via C(n, k) = C(n, n − k). We use this to relate r and r+2. Given Data / Assumptions: C(50, r) = C(50, r+2). 0 ≤ r, r+2 ≤ 50 (implicit domain). Concept / Approach:Nontrivial equality happens when r = 50 − (r+2). Solve for r. Step-by-Step Solution: Set r = 50 − (r + 2) ⇒ r = 48 − r ⇒ 2r = 48 ⇒ r = 24. Verification / Alternative check:Check: C(50, 24) = C(50, 26) by symmetry, consistent with the condition. Why Other Options Are Wrong:23, 22, 21 do not satisfy r = 50 − (r+2). Common Pitfalls:Assuming equality implies r = r+2 (impossible) instead of using symmetry across n/2. Final Answer:24

Question 3

Given nP3 = 9240, find n. (Recall nP3 = n * (n − 1) * (n − 2) for integers n ≥ 3.)

Accepted Answer

Introduction / Context:Here we invert a cubic-like product n(n − 1)(n − 2). Trying nearby integers quickly identifies the exact match. Given Data / Assumptions: nP3 = 9240. nP3 = n(n − 1)(n − 2). Concept / Approach:Test n = 22 because 222120 seems close to 9240; compute and confirm. Step-by-Step Solution: 22 * 21 = 462; 462 * 20 = 9240.Therefore n = 22. Verification / Alternative check:n = 21 gives 212019 = 7980; n = 23 gives 232221 = 10626, so 22 is correct. Why Other Options Are Wrong:20, 21, 23 do not produce 9240 exactly. Common Pitfalls:Using combinations instead of permutations; forgetting the product spans three consecutive integers. Final Answer:22

Question 4

In how many distinct arrangements can the letters of the word “INHALE” be written in a row? (All letters are distinct.)

Accepted Answer

Introduction / Context:Arranging n distinct letters in a row yields n! permutations. “INHALE” has 6 distinct letters. Given Data / Assumptions: Letters: I, N, H, A, L, E (6 distinct). Concept / Approach:Compute 6!. Step-by-Step Solution: 6! = 720. Verification / Alternative check:No repeated letters, so no division by factorials of multiplicities is needed. Why Other Options Are Wrong:360 or 120 would correspond to dividing by 2! or 3! which is inappropriate here; 650 is not a factorial value. Common Pitfalls:Misidentifying repeats; “INHALE” has none. Final Answer:720

Question 5

How many distinct arrangements of the letters in “ARMOUR” are possible? (R appears twice; other letters are distinct.)

Accepted Answer

Introduction / Context:When letters repeat, divide the total permutations by factorials of the multiplicities. “ARMOUR” has 6 letters with R repeated twice. Given Data / Assumptions: Total letters = 6; multiplicities: R × 2; A,M,O,U each × 1. Concept / Approach:Use 6! / 2! for the two indistinguishable R’s. Step-by-Step Solution: 6! / 2! = 720 / 2 = 360. Verification / Alternative check:Any attempt to treat both R’s as distinct would double-count each arrangement. Why Other Options Are Wrong:720 ignores repeated R; 540 and 300 do not correspond to the correct division. Common Pitfalls:Forgetting to divide by 2! for the duplicated letter. Final Answer:360

Question 6

How many distinct arrangements are possible for the letters of “STRESS”? (S appears 3 times; T, R, E appear once each.)

Accepted Answer

Introduction / Context:We handle repeated letters by dividing n! by the product of factorials of letter multiplicities. “STRESS” has 6 letters with S repeated thrice. Given Data / Assumptions: Total letters = 6; multiplicities: S × 3; T,R,E × 1 each. Concept / Approach:Compute 6! / 3! because only S repeats (three times). Step-by-Step Solution: 6! / 3! = 720 / 6 = 120. Verification / Alternative check:Any alternative count should match this canonical multiset-permutation formula. Why Other Options Are Wrong:720 ignores repeats; 360 and 240 reflect partial or incorrect divisors. Common Pitfalls:Miscounting repeated letters or dividing by extra factorials for letters that do not repeat. Final Answer:120

Question 7

How many distinct arrangements can be made from the letters of “BANKING”? (N appears twice; other letters are distinct.)

Accepted Answer

Introduction / Context:To count permutations of a word with repeated letters, divide n! by factorials of multiplicities. “BANKING” has 7 letters with N repeated twice. Given Data / Assumptions: Total letters = 7; multiplicity: N × 2. Concept / Approach:Compute 7! / 2!. Step-by-Step Solution: 7! / 2! = 5040 / 2 = 2520. Verification / Alternative check:Listing patterns confirms that exchanging the two N’s does not produce a new arrangement, hence the 2! divisor. Why Other Options Are Wrong:5040 ignores repetition; 2540 and 5080 are not valid factorial-based counts. Common Pitfalls:Overlooking exact multiplicities or dividing by the wrong factor. Final Answer:2520

Question 8

How many distinct arrangements of the letters in “FINANCE” are possible? (N appears twice; other letters are distinct.)

Accepted Answer

Introduction / Context:Again a multiset permutation: 7 letters with a duplication of N. The standard formula applies. Given Data / Assumptions: Letters: F, I, N, A, N, C, E (N × 2). Concept / Approach:7! / 2! counts distinct arrangements since interchanging the two N’s does not create a new word. Step-by-Step Solution: 7! / 2! = 5040 / 2 = 2520. Verification / Alternative check:Check multiplicities: only N repeats; no other repeated letters. Why Other Options Are Wrong:5040 ignores repetition; 1260 or 840 divide by too much. Common Pitfalls:Misidentifying repeat counts in the word. Final Answer:2520

Question 9

In how many distinct arrangements can the letters of “VENTURE” be written? (E appears twice; remaining letters are distinct.)

Accepted Answer

Introduction / Context:“VENTURE” contains 7 letters with E repeated twice. Use the multiset permutation formula. Given Data / Assumptions: Total letters = 7; multiplicity: E × 2; others distinct. Concept / Approach:Number of arrangements = 7! / 2!. Step-by-Step Solution: 7! / 2! = 5040 / 2 = 2520. Verification / Alternative check:Only E repeats; thus a single 2! divisor is needed. Why Other Options Are Wrong:5040 ignores repetition; 840 and 1260 are undercounts. Common Pitfalls:Over- or under-dividing by factorials when multiple letters repeat (not applicable here). Final Answer:2520

Question 10

Two countries meet with 12 delegates each. Every delegate of one country shakes hands with every delegate of the other country (no intra-country handshakes). How many handshakes occur?

Accepted Answer

Introduction / Context:This is a complete bipartite handshake count K(12,12): each of the 12 from Country A shakes with each of the 12 from Country B. Given Data / Assumptions: Country A: 12 delegates; Country B: 12 delegates. Only cross-country handshakes are counted. Concept / Approach:Each of the 12 delegates in A shakes with 12 in B, giving 12 * 12 handshakes. Step-by-Step Solution: Handshakes = 12 * 12 = 144. Verification / Alternative check:Graph-theoretic model: number of edges in K(12,12) is 12*12. Why Other Options Are Wrong:72 halves the count incorrectly; 288 doubles it by counting both directions as distinct handshakes. Common Pitfalls:Accidentally including intra-country pairs or double-counting the same handshake. Final Answer:144

Question 11

From 15 non-collinear points in a plane, how many distinct straight lines can be drawn using pairs of points?

Accepted Answer

Introduction / Context:Each pair of non-collinear points defines a unique line. With no three collinear, distinct pairs yield distinct lines. Given Data / Assumptions: 15 points; no three are collinear. Concept / Approach:Count pairs: C(15, 2). Step-by-Step Solution: C(15, 2) = 15*14/2 = 105. Verification / Alternative check:The non-collinearity ensures no pair duplicates an existing line. Why Other Options Are Wrong:120, 110, 115 are not equal to C(15, 2). Common Pitfalls:Forgetting the “no three collinear” condition; otherwise some pairs could lie on the same line and reduce the count. Final Answer:105

Question 12

A library has a copies of one title, b copies of each of two titles, c copies of each of three titles, and a single copy of d distinct titles (one each). In how many ways can all these books be arranged in a row? (Indistinguishable copies of the sam…

Accepted Answer

Introduction / Context:This is a multiset permutation problem. We have totals of identical copies by title. The denominator multiplies factorials of each title’s multiplicity. Given Data / Assumptions: Total books N = a + 2b + 3c + d. Indistinguishable copies for identical titles. Two titles have b copies each ⇒ (b!)^2; three titles have c copies each ⇒ (c!)^3. Concept / Approach:Number of distinct arrangements of N items with groups of indistinguishable items is N! divided by the product of factorials of each group size. Step-by-Step Solution: Arrangements = (a + 2b + 3c + d)! / [a! * (b!)^2 * (c!)^3 * (1!)^d].Since each of the d single-copy titles has multiplicity 1, their 1! factors are 1 and need not be written. Verification / Alternative check:Setting b = c = 0 reduces to (a + d)! / a!, consistent with a identical copies of one title and d singletons. Why Other Options Are Wrong:Options lacking (b!)^2 or (c!)^3 fail to account for indistinguishability across multiple titles of equal multiplicity. Common Pitfalls:Forgetting that “two books with b copies each” contributes (b!)^2 and “three books with c copies each” contributes (c!)^3. Final Answer:(a + 2b + 3c + d)! / {a! (b!)2 (c!)3}

Question 13

From 12 points in the plane, of which 7 are collinear and the rest are in general position, how many distinct triangles can be formed by choosing any 3 points?

Accepted Answer

Introduction / Context:Any 3 non-collinear points form a triangle. Triples that lie on the same straight line are degenerate and must be excluded. Given Data / Assumptions: Total points = 12. A set of 7 points are collinear; the remaining 5 are not collinear with them as a group. Concept / Approach:Total triples minus degenerate triples along the 7-point line gives the count of triangles. Step-by-Step Solution: Total triples = C(12, 3) = 220.Degenerate (collinear) triples = C(7, 3) = 35.Triangles = 220 − 35 = 185. Verification / Alternative check:Any triple including at least one of the 5 off-line points is non-collinear with the 7-point line, hence valid. Why Other Options Are Wrong:175, 115, 105 subtract too much or use the wrong base total. Common Pitfalls:Missing that only triples entirely within the 7-point collinear set fail to form a triangle. Final Answer:185

Question 14

How many 7-letter words consisting of exactly 4 consonants and 3 vowels can be formed from 12 distinct consonants and 4 distinct vowels if the first 4 positions are consonants and the last 3 positions are vowels (pattern C C C C V V V)?

Accepted Answer

Introduction / Context:The original options suggest a fixed-position pattern for consonants and vowels rather than free intermixing. We will explicitly adopt the pattern C C C C V V V to avoid ambiguity and align with the given numeric scale. Given Data / Assumptions: 12 distinct consonants available; choose and arrange 4 into the first four slots. 4 distinct vowels available; choose and arrange 3 into the last three slots. All letters are distinct; within each block, order matters. Concept / Approach:Number of choices for consonant block = 12P4. Number for vowel block = 4P3. Multiply blocks because they are independent under the fixed pattern. Step-by-Step Solution: Consonant block (C C C C): 12P4 = 1211109 = 11880.Vowel block (V V V): 4P3 = 43*2 = 24.Total words = 11880 * 24 = 285120. Verification / Alternative check:If consonants and vowels could appear in any positions (not fixed), the count would be 12C4 * 4C3 * 7!, which equals 9,979,200—far larger and inconsistent with the provided options. Hence the fixed-slot interpretation is appropriate. Why Other Options Are Wrong:251820, 258120, 281520 are near-misses derived from arithmetic slips or using combinations instead of permutations in one of the blocks. Common Pitfalls:Mixing up P and C; using 12C4 or 4C3 in place of 12P4 or 4P3 when the within-block order matters. Final Answer:285120

Question 15

Round-table seating with a restriction: Twelve persons are to be seated around a circular table. Two particular persons must not sit side by side. How many distinct circular arrangements are possible?

Accepted Answer

Introduction / Context:Circular permutations differ from linear ones because rotations are indistinguishable. This problem asks for the number of round-table arrangements of 12 people with a specific adjacency restriction on two identified persons. Given Data / Assumptions: Total persons = 12. Two particular persons must not be adjacent. All individuals are distinct; reflections are considered different unless stated otherwise (standard round-table convention counts rotations as the same, not reflections). Concept / Approach:Use “Total − Restricted.” First compute all circular arrangements without restrictions, then subtract the arrangements where the two specified people sit together as a block. For circular arrangements of n distinct people, the count is (n − 1)!. Step-by-Step Solution: Total circular arrangements = (12 − 1)! = 11!Treat the two as a single block + the other 10 ⇒ 11 items on a circleArrangements with the pair adjacent = (11 − 1)! * 2 = 10! * 2 (factor 2 for swapping within the pair)Required = 11! − 2 * 10! = (11 − 2) * 10! = 9 * 10! Verification / Alternative check:Anchor one person to eliminate rotation; counting linearly then adjusting leads to the same 9 * 10! figure. Why Other Options Are Wrong: 2 (10!) counts only adjacent cases. 10! omits most permissible configurations. 45 (8!) is unrelated to circular-block logic here. Common Pitfalls:Forgetting the factor 2 for the pair’s internal order, or using 12! instead of 11! for circular arrangements. Final Answer:9 (10!)

Question 16

Counting selections from bounded supplies: In how many ways can fruit be selected from 3 bananas, 4 apples, and 2 oranges, where each item can be chosen 0 up to its available count?

Accepted Answer

Introduction / Context:This is a standard “selections with upper bounds” problem where we count choices independently for each type and multiply. We are not forming permutations or combinations of identical items; we are counting how many different selection vectors (b, a, o) are possible given caps per type. Given Data / Assumptions: Bananas available: 0–3 (4 choices). Apples available: 0–4 (5 choices). Oranges available: 0–2 (3 choices). Choosing any combination (including choosing none) is allowed. Concept / Approach:For each fruit, the number of possible counts equals (max + 1). By the rule of product for independent decisions, multiply the per-type possibilities together. Step-by-Step Solution: Choices for bananas = 3 + 1 = 4Choices for apples = 4 + 1 = 5Choices for oranges = 2 + 1 = 3Total selections = 4 * 5 * 3 = 60 Verification / Alternative check:Enumerating edges (all-max, all-min, mixed) confirms counts fall within the independent ranges, reinforcing 60. Why Other Options Are Wrong:39, 315, or 512 do not match the product of simple ranges (4*5*3). 512 typically appears for unconstrained binary choices over 9 items, which is not this setting. Common Pitfalls:Confusing “ways to select” with permutations; forgetting to include the zero-choice option per fruit. Final Answer:None of these (the correct count is 60).

Question 17

Multiset permutations (expand a^3 b^2 c^4): When a^3 b^2 c^4 is written out fully as a string of 9 letters (aaa bb cccc), how many distinct arrangements are possible?

Accepted Answer

Introduction / Context:We seek the number of distinct permutations of a multiset where several letters repeat. The order matters, but identical letters are indistinguishable within each group. Given Data / Assumptions: Total letters = 9 (3 a’s, 2 b’s, 4 c’s). Letters within the same species are indistinct. Concept / Approach:Use the multinomial formula: total permutations of n objects with groups of identical items is n! divided by the product of factorials of each group size. Step-by-Step Solution: n = 9, counts: 3, 2, 4Distinct arrangements = 9! / (3! * 2! * 4!)Compute: 9! = 362880; divide by 3! = 6 ⇒ 60480Divide by 2! = 2 ⇒ 30240; divide by 4! = 24 ⇒ 1260 Verification / Alternative check:Sanity checks using smaller analogous sets (e.g., aabb) confirm the division by identical groups. Why Other Options Are Wrong:2520 doubles the correct value; 610 is not an integer factorization result from the required denominators. Common Pitfalls:Forgetting to divide by every identical-group factorial, or miscounting the total letters. Final Answer:1260

Question 18

Solving from ratios of consecutive combinations: If C(n, r−1) = 36, C(n, r) = 84, C(n, r+1) = 126 (repeated for clarity), confirm n and r and provide reasoning.

Accepted Answer

Introduction / Context:This reiterates and confirms the ratio-based method for adjacent binomial coefficients to ensure conceptual clarity for learners reviewing contiguous values in Pascal rows. Given Data / Assumptions:Same as earlier: 36, 84, 126 for C(n, r−1), C(n, r), C(n, r+1) respectively. Concept / Approach:Use identities of consecutive ratios to produce two linear equations in n and r, then solve. Step-by-Step Solution: (n − r + 1)/r = 84/36 = 7/3 ⇒ 3n + 3 = 10r(n − r)/(r + 1) = 126/84 = 3/2 ⇒ 2n = 5r + 3Solving gives n = 9, r = 3 Verification / Alternative check:Direct substitution matches all three numbers. Why Other Options Are Wrong:Other pairs fail one or more of the three equalities. Common Pitfalls:Mixing the ratio direction or losing the +1 term. Final Answer:n = 9, r = 3

Question 19

Equality of binomial coefficients (index transforms): Find all n such that C(35, n + 7) = C(35, 4n − 2).

Accepted Answer

Introduction / Context:Binomial symmetry says C(N, k) = C(N, N − k). Two binomial coefficients with the same top N are equal when their lower indices are equal or complementary with respect to N. Given Data / Assumptions:N = 35; indices k1 = n + 7, k2 = 4n − 2. Concept / Approach:Set up the two equality conditions: either k1 = k2 or k1 = 35 − k2. Solve both for integer n that keep indices within [0, 35]. Step-by-Step Solution: Case 1: n + 7 = 4n − 2 ⇒ 3n = 9 ⇒ n = 3Case 2: n + 7 = 35 − (4n − 2) = 37 − 4n ⇒ 5n = 30 ⇒ n = 6 Verification / Alternative check:Plugging back: C(35, 10) = C(35, 10) and C(35, 13) = C(35, 22) by symmetry — valid. Why Other Options Are Wrong:Singletons 3 or 6 omit the second valid value; 28 is irrelevant to the solutions. Common Pitfalls:Forgetting the complementary index possibility or neglecting domain limits. Final Answer:3, 6

Question 20

Permutation ratio identity: If (2n + 1)P(n − 1) : (2n − 1)P(n) = 3 : 5, find n.

Accepted Answer

Introduction / Context:We compare two permutations via factorial forms and simplify to a rational function of n. Solving the resulting polynomial yields n. Given Data / Assumptions:(2n + 1)P(n − 1) : (2n − 1)P(n) = 3 : 5. Concept / Approach:Use P(a, b) = a! / (a − b)!. Compute the ratio explicitly and reduce. Then equate to 3/5 and solve for n in integers. Step-by-Step Solution: R = [(2n + 1)! / (n + 2)!] / [(2n − 1)! / (n − 1)!] = [(2n)(2n + 1)] / [(n + 2)(n + 1)n]Set R = 3/5 ⇒ 5(2n)(2n + 1) = 3n(n + 1)(n + 2)Expand ⇒ 20n^2 + 10n = 3n^3 + 9n^2 + 6n ⇒ 3n^3 − 11n^2 − 4n = 0Factor ⇒ n(3n^2 − 11n − 4) = 0 ⇒ n = 4 (positive integer root) Verification / Alternative check:Substitute n = 4 into the original ratio and both sides simplify to 3/5. Why Other Options Are Wrong:6, 3, or 8 do not satisfy the polynomial; negative roots are not admissible. Common Pitfalls:Dropping factorial cancellations incorrectly, or mishandling the polynomial algebra. Final Answer:4

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