Permutations

There are (3) position pairs at distance (5) for (S,Q), with (2) orders, and the remaining (6) places fill in (6!/3!) ways. For gap-distance questions, count position pairs first.

Treat the four specified people as one block with fixed internal order, so (9) units arrange in (9!) ways. If internal order is prescribed, do not multiply by (4!).

First count arrangements containing both (0) and (9), then subtract cases where they are adjacent. When (0) is included, handle the leading position carefully.

There are (5) odd positions among (9); place the (3) vowels in (^{5}C_3\cdot 3!/2!) ways and consonants in (6!/2!) ways. Decide position types first.

Fix (A); then (B) and (C) occupy the two equally spaced positions in (2!) ways. The remaining (6) people sit in (6!) ways.

By inclusion-exclusion, the count is (7!-3\cdot 6!+3\cdot 5!-4!). For multiple forbidden positions, this method is safe.

Treat the two (C)'s as a block, giving (8!/(2!2!)) arrangements, then subtract the cases with adjacent (A)'s, (7!/2!). For two conditions, first form the required block.

Add arrangements beginning with smaller available letters at each position and then add (1). Write the alphabetical order first when finding rank.

The order is fixed within the chosen positions, but the first digit cannot be (0). In such restrictions, check both order and leading zero.

For (A,B), use the (6) middle positions in (^{6}P_2) ways, then subtract adjacent cases (5\cdot 2\cdot 6!). Count middle-position and non-adjacent restrictions together.

Only the (3) vowel positions must be chosen, because both vowel and consonant orders are fixed. When orders are fixed, position selection gives the answer.

Total circular arrangements are (10!), and all-three-together cases are (8!\cdot 3!). For not-all-together cases, subtract the bad cases from total.

Inside the vowel block there are (5!/(2!3!)) arrangements, and outside there are (6!/2!) arrangements. Handle the block and repetitions together.

Treat each couple as a block; the (5) blocks arrange in (5!) ways and each block has (2) internal orders. Couples contribute (2^5) internal arrangements.

Choose one of the two specified letters in (2) ways, choose (4) more from the remaining (7), and arrange in (5!) ways. For exactly-one conditions, count both choices separately.

First include exactly one digit from (2,4,6), then make cases for an odd unit digit. Handle oddness and compulsory choice together.

Here (T) appears three times and (O,N,I) appear twice each, so the count is (12!/(3!2!2!2!)). For long words, make a frequency table.

Among the relative orders of (A,B,C), (A) is in the middle with probability (1/3), and (D) before (E) has probability (1/2). Hence the count is (10!/6).

The first digit must be even but not (0), the last digit must be odd, and the middle (4) places are filled from remaining digits. Always check the leading zero restriction.

First arrange red and blue balls in (^{11}C_5) ways, then choose (4) of the (12) gaps for green balls. With identical objects, position selection is the key.

Treat (A,B,C) as a block, so (6) units have (5!) circular arrangements, and internally (4) orders do not have (B) in the middle. Count internal orders separately.

Choose the (2) correctly assigned people in (^{8}C_2) ways and derange the remaining (6) with (D_6=265). Exactly (2) correct means all others are wrong.

Here (S) appears three times and (I) appears three times, so the count is (11!/(3!3!)). Divide by the factorial of each repeated group.

First take total arrangements with (6) before (8), (8!/2), then subtract half of the block arrangements where (2,4) are adjacent. Separate order and non-adjacency conditions.

There are (4) even positions among (9), and the (4) consonants occupy them in (4!) ways, while vowels fill the rest in (5!) ways. For position restrictions, first count the positions.