Permutation and Combination

The word MUMBAI has 6 letters with M repeated twice. Number of arrangements = 6!/2! = 720/2 = 360. Option A (720) is the error of not accounting for the repeated M. Options C and D are incorrect calculations.

The word WEDDING has 7 letters with D repeated twice. Number of arrangements = 7!/2! = 5040/2 = 2520. We divide by 2! because the two D's are identical and swapping them doesn't create a new arrangement. This is the standard formula for permutations with repeated elements.

DARYABAD has 8 letters with A repeated twice and D repeated twice. Arrangements = 8! / (2! × 2!) = 40320 / 4 = 10080. Wait, let me recalculate: 8! = 40320, divided by 4 = 10080. Hmm, this doesn't match option A (3360). Let me check the word again: D-A-R-Y-A-B-A-D - that's 8 letters with A appearing 3 times and D appearing 2 times. So 8! / (3! × 2!) = 40320 / 12 = 3360.

PLANNING has 8 letters with 2 N's repeated. Total arrangements = 8!/2! = 40320/2 = 20160. However, on closer inspection, there's only one N repeated, so 8!/2! = 20160. Wait, let me recalculate: PLANNING = P-L-A-N-N-I-N-G has 8 letters with 3 N's. So 8!/3! = 40320/6 = 6720. The answer is correct.

The word HAPPINESS has 9 letters with P appearing twice and S appearing twice. Using the formula for permutations of multiset: 9!/(2! × 2!) = 362880/4 = 90720 distinct arrangements.

The word MISTAKE has 7 letters with 3 vowels (I, A, E) and 4 consonants. Treating the 3 vowels as one block, we have 5 items to arrange (4 consonants + 1 vowel block). These can be arranged in 5! = 120 ways. Within the vowel block, the 3 vowels can be arranged in 3! = 6 ways. Total arrangements = 120 × 6 = 720. Option B is correct.

Case 1: All 6 Maths faculty + 1 English from 3 = C(6,6) × C(3,1) = 1 × 3 = 3 ways. Case 2: All 4 Reasoning faculty + all 3 English faculty = C(4,4) × C(3,3) = 1 × 1 = 1 way. Total = 3 + 1 = 4 ways.

Committee of 7 with 1 English + remaining 6 split equally between Reasoning and Maths: 1 English (choose 1 from 3), 3 Reasoning (choose 3 from 4), 3 Maths (choose 3 from 6). Ways = C(3,1) × C(4,3) × C(6,3) = 3 × 4 × 20 = 240.

The word VARANSI has 7 letters with A and I repeated (2 vowels, 5 consonants). Total distinct arrangements = 7!/(2!) = 2520. Words starting with vowel: Choose first letter (A or I) = 2 choices, arrange remaining 6 letters with one repeated vowel = 6!/2! = 360. Probability = (2 × 360)/2520 = 720/2520 = 3/7.

LUCKNOW has 7 letters with 2 vowels (U, O) and 5 consonants. Treat vowels as one unit: we have 6 units (5 consonants + 1 vowel group). These can be arranged in 6! = 720 ways. Within the vowel group, U and O can be arranged in 2! = 2 ways. Total = 720 × 2 = 1440.

Treat boys as one block and girls as another block. Boys internally: 4! ways. Girls internally: 3! ways. The two blocks can be arranged in 2! ways. Total = 4! × 3! × 2! = 24 × 6 × 2 = 288. Option C is correct.

The word 'SHOULD' has 6 distinct letters. The number of arrangements is 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. The claimed answer D (720) is correct.

The word 'FOLDER' has 6 distinct letters. The number of arrangements of n distinct items is n!. Here, 6! = 720.

DICTIONARY has 10 letters. Vowels: I, I, O, A, Y (5 vowels). Treat the 5 vowels as one unit. We have 6 units to arrange (5 consonants + 1 vowel group). 6! ways = 720. The 5 vowels within their group can be arranged in 5! ways = 120. Total arrangements = 720 × 120 = 86400. However, standard approach counts only I, I, O, A as vowels (4 vowels): 7 units × 4!/2! (since I repeats) = 5040 × 12 = 60480.

We need to select 8 men from 12, with 2 particular men always included and 3 particular men always excluded. Since 2 are included and 3 are excluded, we have 12 - 2 - 3 = 7 men available to choose from. We need to select 8 - 2 = 6 more men from these 7. The number of ways is C(7,6) = 7.