Permutations

Treat the two books as one block giving (5!\times2!) arrangements. In exams, make a block for objects that must stay together.

There are (7) letters and (R) repeats twice, so the count is (\frac{7!}{2!}). Divide by factorials of repeated letters.

Total circular arrangements are (7!) and adjacent cases are (2\cdot6!), so the answer is (7!-2\cdot6!). In circular order, fix one position.

The last digit must be (5), and the remaining (3) places have (6P3) arrangements. Apply divisibility conditions first.

There are (11) letters with (I,S,P) repeated (4,4,2) times, so the count is (\frac{11!}{4!4!2!}). Do not count identical letters separately.

Arrange boys in (5!) ways and place (4) girls in (6) gaps in (6P4) ways. The gap method is most useful here.

The last digit is one of (1,3,5), and the first digit cannot be (0), giving (3\cdot5\cdot5\cdot4). Handle the leading zero condition separately.

There are (11) letters and (M,A,T) each repeat (2) times, so the count is (\frac{11!}{2!2!2!}). Correctly identify repeated letters.

Choosing and ordering (4) chairs together gives (7P4). Use permutation when distinct people occupy distinct positions.

Choose the remaining (3) digits from (7) digits and arrange all (5) digits in (5!) ways. Include compulsory digits first.

The two particular people occupy positions (4) apart, giving (5) position pairs and (2) orders, then arrange the remaining (7!). Count position pairs for distance constraints.

There are (9) letters and (M,T,E) each repeat (2) times, so the count is (\frac{9!}{2!2!2!}). Identical letters reduce permutations.

Treat the women as one block giving (8!) orders and (3!) internal orders. Use the block method for consecutive objects.

The last digit can be (0,2,4); separating the (0) and nonzero last-digit cases gives (120+96). Handle zero carefully at the first and last positions.

Treat the (5) vowels as one block, giving (5!) outside arrangements and (5!) internal arrangements. When vowels are together, make one vowel block.

With the particular letter included, choose (4) more from (7) and arrange the (5) in (5!) ways. Separate the compulsory object first.

Treat the two (N)s as one block; among (5) units, (A) repeats (3) times, so the count is (\frac{5!}{3!}). Track both blocks and repeated letters.

The thousands digit is (3,4,5,6), and the remaining (3) places can be filled in (7P3) ways. For range questions, fix the first digit first.

If both are selected, count (8C2\cdot4!); if both are not selected, count (8C4\cdot4!). Split such problems into cases and add.

For a necklace, the count is (\frac{(7-1)!}{2}) because reflection is also identical. Remember the difference between circular arrangements and necklaces.

Total arrangements are (\frac{7!}{2!}), and the (3) distinct vowels have (3!) equally likely orders, so divide by (3!). For order restrictions, divide by possible orders.

The two particular people are selected, choose (3) more from (6), then arrange the block and others in (4!\cdot2!) ways. Choose first, then arrange.

There are (14) valid ordered endings; endings without (0) give (5\cdot5P2) and endings with (0) give (6P3). For divisibility by (4), check the last two digits.

With the compulsory student, choose (5) more from the remaining (7) and arrange (6!) ways. Apply inclusion and exclusion conditions first.

Treat the three people as one block; (6) units around a circle give (5!), and the block has (3!) internal orders. Circular blocks reduce the number of units.