Permutation and Combination

Total arrangements of COFFEE = 6!/(2! × 2!) = 180 (two F's, two E's). Treat vowels (O, E, E) as one unit: arrangements = 4! × 3!/(2!) = 72. Vowels never together = 180 - 72 = 144. Option C is correct. Options A, B, D are miscalculations using wrong factorial formulas.

Option A is correct. Dictionary order: Garba (2) comes first, then Garden (3), then Gargle (4), then Garland (1), then Garnish (5). This follows alphabetical ordering where we compare letter by letter: Garba < Garden < Gargle < Garland < Garnish.

The word 'HANDLE' has 6 distinct letters. The number of arrangements of n distinct items is n!. So 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Option A is correct. This tests fundamental permutation concepts - arranging distinct items where order matters.

The word BOOKLET has 7 letters. Treating B and T as a single unit (they must always come together), we have 6 units to arrange: (BT), O, O, K, L, E. The number of arrangements is 6!/2! = 720/2 = 360 (dividing by 2! because O is repeated twice). Within the (BT) unit, B and T can be arranged in 2! = 2 ways. So total arrangements = 360 × 2 = 720.

The word EQUATION has 8 distinct letters (E, Q, U, A, T, I, O, N). The number of arrangements is 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320. This is a standard permutation problem where all letters are different.

PRODUCER has 8 letters with vowels O, U, E (3 vowels) and consonants P, R, D, C, R (5 consonants). Treat the 3 vowels as one unit. We have 6 units total (5 consonants + 1 vowel group). These can be arranged in 6! ways = 720. Within the vowel group, 3 vowels can be arranged in 3! ways = 6. Total arrangements = 720 × 6 = 4320. But wait, PRODUCER has two Rs. We need to account for this. Consonants are P, R, D, C, R - with R repeated. Arrangements of consonants = 5!/2! = 60. Total = 60 × 6 × 6 = 2160. This matches option B.

The word DETAIL has 6 letters with 2 vowels (E, A, I) and 3 consonants (D, T, L). Odd positions are 1, 3, 5 - we choose 3 of these for vowels: 3P3 = 6 ways. Even positions 2, 4, 6 get consonants: 3P3 = 6 ways. Total = 6 × 6 = 36 arrangements. The vowels must occupy odd positions, not all odd positions.

MARVELS has 7 distinct letters. Number of arrangements = 7! = 5040. This is straightforward permutation of all letters without repetition. Option B matches 5040.

The word JUSTIFY has 7 distinct letters (all letters are unique). Number of arrangements = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. No letters repeat, so no division is needed.

Total arrangements of 6 letters = 6! = 720. Treat vowels (O, E) as one group: 5 items × 2! = 240 arrangements with vowels together. Subtract from total: 720 - 240 = 480 arrangements where vowels are separated.

The word DETERMINE has 9 letters with E appearing 3 times. Total arrangements = 9!/3! = 362880/6 = 60480. Divide by 3! because the three E's are identical and can be rearranged among themselves without creating a new arrangement.

The word EPENTHESIS has 10 letters with E appearing 3 times, T appearing 2 times, S appearing 2 times, and vowels E, E, I (3 vowels). Treat vowels as one unit: 8 total units (7 consonants + 1 vowel group) with E(3), T(2), S(2) repetitions. Internal arrangement of vowels = 3!/3! = 1. Total = 8!/(3! × 2! × 2!) × 1 = 40320/(6 × 2 × 2) = 40320/24 = 1680. This doesn't match options. Wait - rechecking: vowels are E, E, I = 3 letters, but E repeats. Vowels together as one unit means we have EPNTHSS + (EEI) = 8 units total. 8!/(2! × 2!) for consonants (E, T, S in consonants). Actually need to be more careful about which letters repeat where. Let me recount: EPENTHESIS - E(3), P(1), N(1), T(2), H(1), S(2), I(1). Vowels = E, E, I. Consonants = P, N, T, H, T, S, S (7 letters with T twice, S twice, E once removed). So consonants arrange as 7!/(2! × 2!) = 5040/4 = 1260. Vowels internally: 3!/3! = 1. Total = 1260 × 1 = 1260. Still doesn't match. Rereading the question - it asks specifically for arrangements where vowels come together. The calculation I did gives 1260, but option B is 10080. Let me verify: 10 letters total, E(3), T(2), S(2). If vowels must be together, treat them as one block. But actually 10080 = 10!/12 = 3628800/12... hmm. Actually 10080 × 12 = 120960, not matching. Let me reconsider: maybe the word has different letters? EPENTHESIS: E-P-E-N-T-H-E-S-I-S. Counting: E=3, P=1, N=1, T=2, H=1, S=2, I=1. Total 10. 10!/(3! × 2! × 2!) = 3628800/(6 × 2 × 2) = 3628800/24 = 151200 total arrangements. With vowels together: treat E,E,I as one unit. We have 8 units. But E appears both in vowels and... no wait. Let me recalculate. Total without restriction: 10!/(3! × 2! × 2!) = 151200. With vowels together: The 3 vowels (E, E, I) form one block. Within the block, they can be arranged in 3!/3! = 1 way (since E repeats). Total arrangements = 8! × 1 = 40320. But this doesn't account for the repeated E's outside the vowel block. Actually, I need to reconsider what the vowel block contains vs consonants. Consonants: P, N, T, H, T, S, S (7 letters, T twice, S twice). Vowels: E, E, I (3 letters, E thrice). The consonants arrange in 7!/(2! × 2!) = 1260 ways. Place the vowel block in one of 8 positions (before, between, or after consonants). Total = 1260 × 8 = 10080. Yes, that's option B.

Number of ways to select 2 men from 13 = 13C2 = 13!/(2! × 11!) = (13 × 12)/(2 × 1) = 78.

Treat the 5 vowels (E, U, A, I, O) as one block. Then we have 4 consonants + 1 vowel block = 5 units to arrange in 5! = 120 ways. Within the vowel block, 5 vowels can be arranged in 5! = 120 ways. Total arrangements = 120 × 120 = 14400. This is not listed in options A-D, so E (None of these) is correct.

Arranging alphabetically: Bandage (a-n-d-a), Bangle (a-n-g), Bangalore (a-n-g-a), Bantam (a-n-t), Banquet (a-n-q). After 'Ban', compare: d (Bandage) < g (Bangle/Bangalore) < q (Banquet) < t (Bantam). Among Bangle (g-l-e) and Bangalore (g-a-l-o): 'l-e' comes before 'a-l' since e < l. Order: Bandage, Bangle, Bangalore, Banquet, Bantam. This gives 3, 2, 5, 1, 4.