Question 1

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Accepted Answer

Problem restatement Treat the vowels in 'OPTICAL' as a single block so that they always stay adjacent, and count the total distinct arrangements. Given data Word: OPTICAL (7 distinct letters). Vowels: O, I, A (3 vowels). Consonants: P, T, C, L (4 consonants). Concept/Approach Group the 3 vowels as one block [V]. Then arrange [V] with the 4 consonants (total 5 items), and finally permute the vowels inside [V]. Step-by-step calculation Arrange 5 items ([V], P, T, C, L): 5! = 120 Permute vowels within [V] (O, I, A): 3! = 6 Total arrangements = 120 × 6 = 720 Verification/Alternative No repeated letters, so no division by factorials for duplicates. The block method is exact. Common pitfalls Forgetting to multiply by the internal permutations of the vowels. Accidentally treating any letters as repeated (they are all distinct here). Final Answer 720

Question 2

A box contains 2 white, 3 black, and 4 red balls. In how many ways can 3 balls be drawn if at least one black ball must be included?

Accepted Answer

Problem restatement Count unordered draws of 3 balls from 9 distinct balls, with the condition that at least one is black. Given data Total balls = 2 (white) + 3 (black) + 4 (red) = 9. Draws are combinations (order does not matter). Constraint: include at least one black ball. Concept/Approach Use complement counting: total 3-ball combinations minus those with zero black balls. Step-by-step calculation Total ways = C(9, 3) = 84 Zero-black ways = choose from 2W + 4R = 6 balls ⇒ C(6, 3) = 20 Required ways = 84 − 20 = 64 Verification/Alternative Direct sum over cases with 1B, 2B, 3B yields the same count: C(3,1)C(6,2) + C(3,2)C(6,1) + C(3,3)C(6,0) = 3×15 + 3×6 + 1×1 = 45 + 18 + 1 = 64. Common pitfalls Treating balls as indistinguishable by color; here each physical ball is distinct, so use standard combinations. Final Answer 64

Question 3

Out of 7 consonants and 4 vowels, how many 5-letter words consisting of 3 consonants and 2 vowels can be formed?

Accepted Answer

Problem restatement Choose 3 consonants and 2 vowels and arrange the 5 chosen letters in all possible orders. Given data Consonants available = 7. Vowels available = 4. Word length = 5 (with 3 consonants + 2 vowels). Concept/Approach Independent selection followed by permutations: choose letters first, then arrange them (all distinct). Step-by-step calculation Choose consonants: C(7, 3) = 35 Choose vowels: C(4, 2) = 6 Arrange 5 letters: 5! = 120 Total = 35 × 6 × 120 = 25200 Verification/Alternative The multiplication principle applies: selections are independent and letters are distinct. Common pitfalls Multiplying by 5! before selecting the letters, or double-counting arrangements. Final Answer 25200

Question 4

In how many ways can the letters of the word 'LEADER' be arranged?

Accepted Answer

Problem restatement Count distinct permutations of 'LEADER' considering repeated letters. Given data Letters: L, E, A, D, E, R (6 letters total). Repetition: E occurs twice. Concept/Approach Permutations with repetition: total = 6! / 2!. Step-by-step calculation Total arrangements = 6! / 2! = 720 / 2 = 360 Verification/Alternative If all were distinct, we would have 720. Dividing by 2! corrects for swapping the two indistinguishable E's. Final Answer 360

Question 5

In how many different ways can the letters of the word 'DETAIL' be arranged so that the vowels occupy only the odd positions?

Accepted Answer

Problem restatement Place the vowels of 'DETAIL' in odd positions (1, 3, 5) and the consonants in the remaining even positions (2, 4, 6). Given data Word: D, E, T, A, I, L (6 distinct letters). Vowels: E, A, I (3 vowels). Odd positions available: 1, 3, 5 (exactly 3 slots). Concept/Approach Assign the 3 vowels to the 3 odd slots and the 3 consonants (D, T, L) to the 3 even slots independently. Step-by-step calculation Arrange vowels in odd slots: 3! = 6 Arrange consonants in even slots: 3! = 6 Total arrangements = 6 × 6 = 36 Verification/Alternative Because the counts match exactly (3 odd slots and 3 vowels), no combinations factor is needed; only permutations within positions. Common pitfalls Allowing vowels in even positions or forgetting to permute consonants separately. Final Answer 36

Question 6

In how many different ways can the letters of the word 'LEADING' be arranged so that the vowels always come together?

Accepted Answer

Problem restatement Count arrangements of 'LEADING' where the vowels form one adjacent block. Given data Word: L, E, A, D, I, N, G (7 distinct letters). Vowels: E, A, I (3 vowels). Consonants: L, D, N, G (4 consonants). Concept/Approach Block the vowels as [V]; arrange [V] with the 4 consonants (5 items), then permute vowels inside [V]. Step-by-step calculation Arrange 5 items: 5! = 120 Permute vowels (E, A, I): 3! = 6 Total arrangements = 120 × 6 = 720 Verification/Alternative All letters are distinct; no division by factorials for repeats is needed. Final Answer 720

Question 7

In how many ways can a committee of 5 men and 6 women be formed from 8 men and 10 women?

Accepted Answer

Problem restatement Select 5 men and 6 women independently from their respective pools. Given data Men available = 8; choose 5. Women available = 10; choose 6. Concept/Approach Use combinations and multiply independent choices. Step-by-step calculation Ways to choose men = C(8, 5) = 56 Ways to choose women = C(10, 6) = 210 Total committees = 56 × 210 = 11760 Verification/Alternative By symmetry, C(10, 6) = C(10, 4) = 210; the product remains 11760. Final Answer 11760

Question 8

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

Accepted Answer

Problem restatement Arrange 'CORPORATION' with all vowels in a single block. Account carefully for repeated letters. Given data Letters (11 total): C(1), O(3), R(2), P(1), A(1), T(1), I(1), N(1). Vowels: O, O, O, A, I (5 vowels, with O repeated 3 times). Consonants: C, R, R, P, T, N (6 letters, with R repeated 2 times). Concept/Approach Treat the 5 vowels as a single block [V]. First, arrange the 7 items: [V] plus the 6 consonants (with R repeated). Then multiply by the internal arrangements of the vowels (with 3 O's repeated). Step-by-step calculation Arrange 7 items: [V], C, R, R, P, T, N ⇒ 7! / 2! (for the two R's) = 5040 / 2 = 2520 Arrange vowels inside [V]: 5! / 3! (for three O's) = 120 / 6 = 20 Total arrangements = 2520 × 20 = 50400 Verification/Alternative Check counts: total letters 11; grouping vowels reduces to 7 items; duplicated R's and O's appropriately handled by dividing by factorials of repeats. Common pitfalls Forgetting to divide by 2! for the two R's in the outer arrangement. Forgetting to divide by 3! for the three O's inside the vowel block. Final Answer 50400

Question 9

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 9 that are divisible by 5 with no repeated digits?

Accepted Answer

Problem restatement Form 3-digit numbers from {2, 3, 5, 6, 7, 9} that are divisible by 5 and have no repeated digits. Given data Available digits: 2, 3, 5, 6, 7, 9. Divisibility by 5 ⇒ last digit must be 5 (0 is not available). No repetition allowed. Concept/Approach Fix the units digit as 5, then choose hundreds and tens from remaining distinct digits. Step-by-step calculation Units digit = 5 (1 way) Hundreds digit = choose from {2, 3, 6, 7, 9} ⇒ 5 ways Tens digit = choose from remaining 4 digits ⇒ 4 ways Total numbers = 1 × 5 × 4 = 20 Verification/Alternative Listing logic shows structure H–T–5; choices reduce sequentially without repetition. Final Answer 20

Question 10

In how many ways can a group of 5 men and 2 women be formed from 7 men and 3 women?

Accepted Answer

Problem restatement Select 5 men from 7 and 2 women from 3; multiply independent choices. Given data Men available = 7; choose 5. Women available = 3; choose 2. Concept/Approach Use combinations; choices are independent. Step-by-step calculation Ways to choose men = C(7, 5) = 21 Ways to choose women = C(3, 2) = 3 Total groups = 21 × 3 = 63 Verification/Alternative Equivalently, choose the two excluded men (C(7,2)=21) and the one excluded woman (C(3,1)=3) → same result. Final Answer 63

Question 11

From a group of 6 boys and 4 girls, four children are to be selected. In how many ways can the selection be made such that at least one boy is included?

Accepted Answer

Problem restatement Select 4 children from 6 boys and 4 girls with the restriction that the selection contains at least one boy. Given data Boys = 6 Girls = 4 Total to choose = 4 Concept/Approach Count total 4-child selections and subtract the disallowed case (all girls). Use combinations nCk. Step-by-Step calculation Total selections = ⁡10C4 = 210Disallowed (0 boy) = ⁡4C4 = 1Valid selections = 210 − 1 = 209 Verification/Alternative Casewise sum: (1B,3G) + (2B,2G) + (3B,1G) + (4B,0G) = ⁡6C1⁡4C3 + ⁡6C2⁡4C2 + ⁡6C3⁡4C1 + ⁡6C4⁡4C0 = 24 + 90 + 80 + 15 = 209. Common pitfalls Forgetting to subtract the all-girls selection, or miscounting with permutations instead of combinations. Final Answer 209

Question 12

In how many distinct arrangements of the letters of 'MATHEMATICS' do the vowels always occur together?

Accepted Answer

Problem restatement Arrange the 11 letters of 'MATHEMATICS' so that all vowels remain in one block. Given data Letters: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1) Vowels: A, A, E, I (4 vowels; A repeats) Concept/Approach Treat the 4 vowels as one super-block. Then arrange this block with the 7 consonants around it, accounting for repeated consonants. Multiply by the internal arrangements of the vowels (accounting for repeated A). Step-by-Step calculation Consonants: M, M, T, T, H, C, S (7 items with repeats M(2), T(2))Entities to arrange = 7 consonants + 1 vowel-block = 8Ways to arrange these 8 = 8! ÷ (2! 2!) = 40320 ÷ 4 = 10080Vowel-block arrangements (A, A, E, I) = 4! ÷ 2! = 12Total arrangements = 10080 × 12 = 120,960 Verification/Alternative Check counts of repeats carefully: two M's and two T's in the outer arrangement; two A's inside the block. Common pitfalls Ignoring duplicate letters (overcounting), or forgetting to multiply by internal arrangements of the vowel-block. Final Answer 120,960

Question 13

How many 4-letter words (with or without meaning) can be formed from the letters of 'LOGARITHMS' if no letter is repeated?

Accepted Answer

Problem restatement Count the 4-letter permutations from distinct letters of 'LOGARITHMS' with no repetitions. Given data Word has 10 distinct letters. Concept/Approach Choose and order 4 distinct letters from 10: permutations ⁡10P4. Step-by-Step calculation ⁡10P4 = 10 × 9 × 8 × 7 = 5,040 Common pitfalls Using combinations ⁡10C4 (ignores order) instead of permutations. Final Answer 5,040

Question 14

From a group of 7 men and 6 women, a 5-person committee is to be formed such that at least 3 men are included. In how many ways can this be done?

Accepted Answer

Problem restatement Select 5 people from 7 men and 6 women with at least 3 men on the committee. Given data Men = 7, Women = 6 Committee size = 5 At least 3 men Concept/Approach Casework on the number of men: 3 men + 2 women; 4 men + 1 woman; 5 men + 0 women. Step-by-Step calculation Case (3M,2W): ⁡7C3⁡6C2 = 35 × 15 = 525Case (4M,1W): ⁡7C4⁡6C1 = 35 × 6 = 210Case (5M,0W): ⁡7C5 = 21Total ways = 525 + 210 + 21 = 756 Common pitfalls Including the (2M,3W) case (violates the constraint) or using permutations instead of combinations. Final Answer 756

Question 15

Given (56)P(r+6) : (54)P(r+3) = 30800 : 1, find the integer value of r. (Use nPr = n! / (n − r)! and simplify the ratio carefully to isolate r.)

Accepted Answer

Introduction / Context: This question tests algebraic manipulation with permutations. Ratios of nPr expressions often telescope when written in factorial form, letting us solve for the unknown r quickly. Given Data / Assumptions: (56)P(r+6) : (54)P(r+3) = 30800 : 1. nPr = n! / (n − r)!, for valid integer ranges. We assume r is an integer making all factorial terms defined. Concept / Approach: Rewrite each permutation in factorial form and simplify the ratio. Many terms cancel, leaving a small product in r. Then equate to the given ratio to solve for r. Step-by-Step Solution: (56)P(r+6) = 56! / (56 − (r+6))! = 56! / (50 − r)!(54)P(r+3) = 54! / (54 − (r+3))! = 54! / (51 − r)!Ratio = [56!/(50 − r)!] / [54!/(51 − r)!] = (56!/54!) * [(51 − r)!/(50 − r)!]= (56 * 55) * (51 − r) = 3080 * (51 − r)Given 3080 * (51 − r) = 30800 ⇒ 51 − r = 10 ⇒ r = 41. Verification / Alternative check: Plug r = 41 back into the simplified expression: 3080 * (51 − 41) = 3080 * 10 = 30800, which matches the given ratio exactly. Why Other Options Are Wrong: 40 and 42 would give 3080 * 11 or 3080 * 9, not 30800. “None of these” is unnecessary because a listed option is correct. Common Pitfalls: Misplacing (n − r)! when expanding nPr, or forgetting that only adjacent factorial factors survive after cancellation. Final Answer: 41

Question 16

From 8 men and 7 women (15 persons total), how many distinct groups of 6 persons can be formed? (Order within the group does not matter.)

Accepted Answer

Introduction / Context: We are forming groups (unordered selections). This is a direct application of combinations: the number of ways to pick k people from n distinct people is C(n, k). Given Data / Assumptions: Total people = 8 + 7 = 15. Group size = 6. Order inside a group does not matter (pure selection). Concept / Approach: Use C(n, k) = n! / (k! (n − k)!). Here n = 15 and k = 6. Step-by-Step Solution: C(15, 6) = 15! / (6! * 9!)Compute via product form: (151413121110) / (654321) = 5005. Verification / Alternative check: Symmetry C(15, 6) = C(15, 9) holds; calculators or Pascal-style reasoning confirm 5005. Why Other Options Are Wrong: 5000 and 5050 are close but not exact; “None of these” is not needed because 5005 is precise. Common Pitfalls: Accidentally treating it as permutations (ordering groups), which would overcount. Final Answer: 5005

Question 17

There are 10 distinct oranges in a basket. In how many ways can 3 oranges be chosen? (Treat oranges as distinct items; order of selection does not matter.)

Accepted Answer

Introduction / Context: This is an elementary combinations problem: selecting k items from n distinct items without regard to order is counted by C(n, k). Given Data / Assumptions: n = 10 distinct oranges. k = 3 oranges chosen. Order is irrelevant. Concept / Approach: Use C(10, 3) = 10! / (3! * 7!). Step-by-Step Solution: C(10, 3) = (1098) / (321) = 120. Verification / Alternative check: Check with symmetry: C(10, 3) = C(10, 7) also equals 120. Why Other Options Are Wrong: 125, 140, and 110 are common miscomputations; 120 is the unique correct combination count. Common Pitfalls: Using permutations (10P3) instead of combinations, thereby overcounting by 3!. Final Answer: 120

Question 18

In a class of 25 distinct students, how many different 3-student committees can be formed? (Order of students within the committee does not matter.)

Accepted Answer

Introduction / Context: Forming committees is choosing subsets, not arranging sequences. The correct tool is combinations C(n, k). Given Data / Assumptions: n = 25 students. k = 3 students per committee. Concept / Approach: Compute C(25, 3) = 25! / (3! * 22!). Step-by-Step Solution: C(25, 3) = (252423) / (321) = (60023)/6 = 2300. Verification / Alternative check: Multiplying 2524*23 = 13,800 and dividing by 6 gives 2300 exactly. Why Other Options Are Wrong: Nearby values (2200, 2400, 3200) reflect arithmetic slips or using permutations. Common Pitfalls: Confusing ordered selections with unordered committee formation. Final Answer: 2300

Question 19

Eight people enter a lounge and each pair of people shake hands exactly once. What is the total number of handshakes?

Accepted Answer

Introduction / Context: Handshakes between pairs correspond to choosing 2 people from the group. This is the classic handshake (graph edges) problem. Given Data / Assumptions: n = 8 distinct people. Each handshake involves a unique pair; no repeats, no self-handshakes. Concept / Approach: The number of unique pairs from n is C(n, 2) = n(n − 1)/2. Step-by-Step Solution: C(8, 2) = 8*7/2 = 28. Verification / Alternative check: Interpret as edges in the complete graph K8; edges = C(8, 2) = 28. Why Other Options Are Wrong: 16, 36, and 56 are typical distractors; only 28 equals C(8, 2). Common Pitfalls: Double-counting ordered pairs (8P2 / 2! is the correct unordered count). Final Answer: 28

Question 20

If C(n+2, 8) : P(n−2, 4) = 57 : 16, find the integer value of n. (Use C(a, b) = a! / (b! (a − b)!) and P(a, b) = a! / (a − b)!)

Accepted Answer

Introduction / Context: We must simplify a ratio mixing a combination and a permutation. Converting each to factorials and canceling common factors typically yields a manageable polynomial equation in n. Given Data / Assumptions: C(n+2, 8) / P(n−2, 4) = 57/16. n is an integer large enough that all factorials are defined (n ≥ 19 from the solution). Concept / Approach: Write C(n+2, 8) = (n+2)! / (8! (n−6)!) and P(n−2, 4) = (n−2)! / (n−6)!. Then the ratio collapses to ((n+2)! / 8!) / (n−2)! = (n+2)(n+1)n(n−1) / 8!. Step-by-Step Solution: Let R = C(n+2, 8) / P(n−2, 4) = (n+2)(n+1)n(n−1) / 8!.8! = 40320. Set (n+2)(n+1)n(n−1) / 40320 = 57/16.Thus (n+2)(n+1)n(n−1) = 40320 * 57 / 16 = 2520 * 57 = 143640.Test n = 19: (21201918) = 420 * 342 = 143640 ⇒ works. Verification / Alternative check: Nearby n = 18 gives 116280; n = 20 gives 20212219 = larger than target, confirming n = 19 uniquely fits. Why Other Options Are Wrong: 17, 18, 20 do not satisfy the exact product 143640. Common Pitfalls: Forgetting that the (n−6)! terms cancel; mis-evaluating 8! or arithmetic with large products. Final Answer: 19

Permutation and Combination Questions