Permutations

Treat the vowels as one block, giving (8!/(2!2!)) outside and (4!/2!) inside. Do not treat repeated letters as distinct.

Choose consonant positions in (^{11}C_6) ways and arrange the (5) vowels in (5!) ways. When relative order is fixed, only positions are chosen.

Divisibility by (8) is decided by the last (3) digits and the first digit cannot be (0). In such questions, count valid last (3)-digit endings first.

After fixing (A), the (2) seats adjacent to (A) are forbidden for (B) and (C). Remove forbidden positions first and then arrange the rest.

Subtract the together cases (10!/(4!4!)) from the total (11!/(4!4!2!)). For not-together cases, subtract the adjacent cases from the total.

Choose the correct letters in (^{7}C_2) ways and derange the remaining (5) letters in (44) ways. For exactly correct cases, all remaining objects must be wrong.

The number of necklaces is ((8-1)!/2) because both rotations and reflections are identical. Remember the difference between necklaces and circular seating.

Treat the three letters as one block, giving (7!) arrangements, and inside it (3!-1) valid orders. Count the internal arrangements of the block separately.

Take the number of letters as (2,3,4), then arrange positions, letters and digits separately. In mixed codes, first choose the position types.

Direct answer: Option C, 35,280. The number must have seven digits, be greater than 5,000,000, and be odd. Its first digit can be 5, 6, 7, or 8. Count by cases because the first digit also affects the available odd final digits. Case 1: first digit 5 or 7, two choices. The last digit must be odd; after choosing the first digit, three odd digits remain. The five middle positions can then be filled in 7P5=7×6×5×4×3=2,520 ways. This gives 2×3×2,520=15,120. Case 2: first digit 6 or 8, two choices. All four odd digits remain available for the last place, and the middle positions again have 2,520 arrangements. This gives 2×4×2,520=20,160. Total =15,120+20,160=35,280. Option A, 30,240, misses some cases. Option B, 33,600, uses an incorrect odd-digit count. Option C is correct. Option D, 40,320, overcounts. Memory cue: when the first and last conditions interact, split into cases.

First arrange the non-(E) letters in (6!/(2!2!)) ways and place (E)'s in (4) of the (7) gaps. With identical letters, gap selection is enough.

Fix (A), then (B) has (2) possible positions and the remaining (6) people arrange in (6!) ways. For circular gaps, fixing one person is easiest.

In the total (9!) arrangements, the (3!) orders of (1,2,3) and (2!) orders of (4,5) are equally likely. So use (9!/(3!2!)).

Add arrangements formed by smaller available letters at each position, then add (1). In rank problems, handle repeated letters carefully.

Arrange the other (7) people in (7!) ways and place (A,B,C) in the (8) gaps in (^{8}P_3) ways. The gap method is powerful for non-adjacent conditions.

Choose the (5) vowel positions in (^{13}C_5) ways and arrange consonants in (8!/2!) ways. When vowel order is fixed, do not count their separate orders.

The two specified men occupy the ends in (2!) ways, the remaining men in (4!) ways, and women in the (5) internal gaps in (5!) ways. Apply end restrictions first.

Arrange the (5) consonants in (5!) ways and place the vowels in (3) of the (6) gaps in (^{6}P_3) ways. If vowels are distinct, count both positions and order.

Arrangements with (A) before (B) are (9!/2), and adjacent cases with (A) before (B) are (8!). Handle order and adjacency separately.

Subtract the adjacent (A) cases (7!) from the total (8!/2!). For non-adjacent identical letters, subtraction is simple.

The number must be even and its digit sum must be divisible by (3). Divisibility by (6) needs both the (2) and (3) tests.

Arrange the boys around the circle in (6!) ways and place (5) girls in the (7) gaps in (^{7}P_5) ways. For circular gaps, arrange one group first.

Fix one book; for the other, (1) of the (5) positions is forbidden, and the remaining books arrange in (4!) ways. For opposite-position restrictions, fix one object.

The letter part has (^{7}P_3) ways, and in the digit part choose the repeated digit in (10) ways, the other digit in (9) ways and its position in (3) ways. Count code parts separately.

Arrange non-(T) letters in (6!/2!) ways and choose (3) of the (7) gaps for the (T)'s. Since the (T)'s are identical, only choose gaps.