Permutation and Combination

The word OPTIMAL has 7 distinct letters (O, P, T, I, M, A, L). Number of arrangements = 7! = 5040. Option B is correct.

In 'LAXMAN', vowels are A, A (2 vowels). Treat them as one unit: (LXMN) + (AA) = 5 entities. These can be arranged in 5! = 120 ways. Within the vowel unit, AA can be arranged in 2!/2! = 1 way (identical vowels). Total: 120 × 1 = 120 ways.

The word 'CADBURY' has 7 distinct letters. The number of arrangements is 7! = 7×6×5×4×3×2×1 = 5040. This is because all letters are unique, so we simply calculate 7 factorial. Option B is correct.

Word 'DOUBT' has 5 distinct letters. Arrangements = 5! = 120. All letters are unique, so simple factorial.

Arranging the words in dictionary order by alphabetical sequence: 'Teeth' comes first (T-e-e), followed by 'Teacher' (T-e-a), then 'Tiffin' (T-i), and finally 'Time' (T-i-m). The correct order is 4, 1, 3, 2 which matches option B (1, 4, 3, 2). Wait, let me verify: T-e-e-h (Teeth), T-e-a-c-h-e-r (Teacher), T-i-f-f-i-n (Tiffin), T-i-m-e (Time). Alphabetically: Teeth (ee), Teacher (ea - wait, 'a' comes before 'e'), so Teacher comes before Teeth. Correct order: Teacher (T-e-a), Teeth (T-e-e), Tiffin (T-i-f), Time (T-i-m). So the order is 1, 4, 3, 2.

The word BOTTLE has 6 letters with T repeated twice. Number of arrangements = 6!/2! = 720/2 = 360. Since 360 is not among the given options (120, 720, 60, 5040), 'None of these' (E) is correct.

The word 'OPERATE' has 7 distinct letters, so we can arrange them in 7! = 5040 ways. However, the word has the letter E appearing twice. Since identical letters don't create distinct arrangements when swapped, we divide by 2! to account for this repetition. Therefore, the total number of distinct arrangements is 7!/2! = 5040/2 = 2520.

The word ADVANTAGE has 9 letters with A repeated 3 times. Arrangements = 9! / 3! = 362880 / 6 = 60480. A appears three times, so we divide by 3 factorial.

LEADER has 6 letters with E repeated twice. Number of arrangements = 6!/2! = 720/2 = 360. When letters repeat, divide by the factorial of the count of each repeated letter.

BEWARE has 6 letters with E repeated twice. Arrangements = 6!/2! = 720/2 = 360.

We need to form a group of 8 members with exactly 4 doctors, 3 teachers, and 1 student. From 8 doctors, choose 4: C(8,4). From 6 teachers, choose 3: C(6,3). From 4 students, choose 1: C(4,1). Total ways = C(8,4) × C(6,3) × C(4,1) = 70 × 20 × 4 = 5600. This uses the multiplication principle of combinations.

The word MAXIMUM has 7 letters with M appearing 3 times, U appearing 2 times, and A, I, X appearing once each. The formula is 7! ÷ (3! × 2!) = 5040 ÷ (6 × 2) = 5040 ÷ 12 = 420. Option D (840) appears to be correct based on the answer key, suggesting a possible calculation in the source material. The standard formula gives 420.

The word OMSAIRAM has 8 letters with O and A each appearing twice. Arrangements = 8!/(2!×2!) = 40320/4 = 10080. Since 10080 is not listed in options A-D, option E (None of these) is correct.

The word ENTRANCE has 8 letters with 2 E's, 2 N's, 2 A's, and 1 T (repeated letters). The vowels are E, A, A, E - total 4 vowels with 2 E's and 2 A's. Treat vowels as one unit: we have 5 units (4 consonants + 1 vowel group). Arrangements of 5 units = 5! = 120. Within the vowel group: 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 (accounting for repeated vowels). Total = 120 × 6 = 720. But the answer is 1080. Let me reconsider: ENTRANCE has E-E-N-N-A-A-T-C. Vowels: E-A-A-E (4 vowels). Consonants: N-N-T-R-C (5 consonants). Treating vowels as one block: we have 6 items (5 consonants + 1 vowel block). These 6 items can be arranged in 6! = 720 ways. But consonants include 2 N's, so divide by 2: 720 ÷ 2 = 360. Within the vowel block: 4 vowels with 2 E's and 2 A's can be arranged in 4! ÷ (2! × 2!) = 6 ways. Total = 360 × 6 = 2160. Still not 1080. Let me count letters differently: E-N-T-R-A-N-C-E (8 letters). Vowels: E, A, E = 3 vowels. Consonants: N, T, R, N, C = 5 consonants with 2 N's. Vowel block + 5 consonants = 6 items. Arrangements: 6! ÷ 2! (for 2 N's) = 720 ÷ 2 = 360. Within vowel block (E, A, E with 2 E's): 3! ÷ 2! = 3. Total: 360 × 3 = 1080 ✓

The word 'AROUND' has 6 distinct letters. The number of ways to arrange n distinct items is n! (n factorial). For 6 letters: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Option C is correct. This is a straightforward permutation problem with no repeated letters.