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

If 56P r + 6 : 54P r + 3 = 30800 : 1 then the value of r is?

Accepted Answer

∵ 56Pr + 6 : 54Pr + 3 = 30800 : 1 ⇒ 56! / (50 - r)! = (30800 x 54!) / (51-r!) ⇒ 56 x 55 = 30800/(51 - r) ⇒ 51 - r = 10 ∴ r = 41

Question 16

How many groups of 6 persons can be formed from 8 men and 7 women?

Accepted Answer

Total no . of person = 8 + 7 = 15 No. of groups = 15C6 = 15! / {6! (15 - 6)!} = 15! / (6! 9!) = (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2 x 1) = 5005

Question 17

There are 10 oranges in a basket. Find the no. of ways in which 3 oranges are chosen from the basket?

Accepted Answer

Required number of ways = 10C3 = 10!/ {3! (10 - 3)!} = 10! / (3! 7!) = (10 x 9 x 8) / (3 x 2) = 120

Question 18

There are 25 students in a class. Find the number of ways in which a committee of 3 students is to be formed.?

Accepted Answer

Required number of ways = 25C3 = (25 x 24 x 23) / (1 x 2 x 3) = 2300

Question 19

8 men entered a lounge simultaneously. If each person shook hands with the other, then find the total no. of hand shakes?

Accepted Answer

Required no. of hand shakes = {8 x (8 - 1)}/2 = 28

Question 20

If n+2C 8 : n- 2P 4 = 57 : 16 then the value of n is?

Accepted Answer

n+2C8 : n-2P4 = 57 : 16 ⇒ {(n + 2 )! (n - 6)!} / {(n - 6)! (n - 2)! 8!} = 57/16 ⇒ (n + 2) (n + 1) n(n - 1) = 143640 ⇒ (n2 + n - 2) (n2 + n ) = 143640 ⇒ (n2 + n )2 - 2(n2 + n) + 1 = 143641 ⇒ (n2 + n - 1)2 = (379)2 [∵ n2 + n - 1 > 0] ⇒ n2 + n - 1 = 379 ⇒ n2 + n - 380 = 0 ⇒ (n + 20) (n - 19) = 0 ∴n = 19 ( Since n is not negative)

Permutation and Combination Questions