Permutations

Total arrangements are (\frac{7!}{2!}) and together cases are (6!), so the difference is (2520-720). For not-together cases, subtract together cases from total.

The row may start with a boy or a girl, giving (2) patterns, each with (6!\cdot6!) orders. Equal numbers create two alternating patterns.

The last two digits can only be (25,75), and the first two places are filled in (6P2) ways. For divisibility by (25), fix the last two digits.

There are (7) letters with (S,C) repeated (3,2) times, so the count is (\frac{7!}{3!2!}). Do not count repeated letters as distinct.

The particular flag is fixed at the first position, and the remaining (3) places are filled in (9P3) ways. Lock the fixed position first.

Subtract adjacent cases (2\cdot7!) from total (8!). For two special objects not adjacent, subtract the block case from total.

Arrange mathematics books in (5!) ways and place (4) physics books in (6) gaps in (6P4) ways. Use the gaps method for separation.

The two particular letters are selected, choose (3) more from (7), and arrange the block and others in (4!\cdot2!) ways. In partial selection, fix chosen objects first.

Choose the other (4) digits from (8); (0) can occupy (4) non-first positions, and the rest arrange in (4!) ways. Count zero positions separately.

Place (T,E) at the two ends in (2!) ways and arrange the remaining (6) letters in (6!) ways. For endpoint constraints, fill endpoints first.

Treat the adjacent pair as a block giving (9!\cdot2), then subtract cases where the third person is at an end, (2\cdot8!\cdot2). For combined conditions, count a simple total and subtract forbidden cases.

Fix one particular student; the other is fixed opposite, and the remaining (6) students can be seated in (6!) ways. For opposite-seat circular problems, fix one person.

Total arrangements are (\frac{8!}{2!}), and adjacent (C) cases are (7!), so separated cases are (20160-5040). For separated objects, subtract adjacent cases from total.

Any chosen set of (5) digits has only (1) increasing order, so the count is (8C5). If the order is fixed, count only selection.

There are (11) letters and (B,I) each repeat (2) times, so the count is (\frac{11!}{2!2!}). Count letter frequencies carefully in long words.

Choose the student receiving (4) prizes in (5) ways and distribute as (\frac{12!}{4!(2!)^4}). For distinct objects in fixed group sizes, divide by group factorials.

Make the first pair a block giving (2\cdot5!), then subtract cases where the second pair is adjacent, (2\cdot2\cdot4!). Use block and subtraction together in mixed circular constraints.

If the last digit is (0), count (9P5); if it is (2,4,6,8), count (4\cdot8\cdot8P4). Separate zero and nonzero last-digit cases.

There are (11) letters with (N,E,D) repeated (3,3,2) times, so the count is (\frac{11!}{3!3!2!}). Write all repeated-letter frequencies.

In total (9!) orders, the two relative orders of captain and goalkeeper are equally likely, so half are valid. If relative order is fixed, divide by (2).

The positions of the two particular cards differ by (3), giving (5) position pairs and (2) orders, then the remaining (6!) arrangements. Count position pairs for exact gaps.

The vowels (E,E,I,O) form one block, so (6) outside units arrange in (6!) ways and the vowels arrange internally in (\frac{4!}{2!}) ways. Do not forget repeated vowels inside the vowel block.

Among total (10!) arrangements, the (3!) relative orders of the three special objects are equally likely, so the count is (\frac{10!}{3!}). If relative order is fixed, divide by that group's factorial.

Arrange the consonants in (\frac{6!}{3!2!}) ways and place the vowels in (7) gaps in (\binom{7}{5}\frac{5!}{3!2!}) ways. Use the gap method when vowels must not be adjacent.

Treat (A) and (B) as one block, so circular arrangements are (6!) and internal arrangements are (2!). Exam tip: use block method for together conditions.