Permutations

Correct answer: A. (240)

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

Correct answer: B. (2520)

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

Correct answer: A. (3600)

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

Correct answer: A. (120)

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

Correct answer: A. (34650)

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.

Correct answer: A. (17280)

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

Correct answer: B. (300)

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.

Correct answer: A. (4989600)

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.

Correct answer: C. (840)

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

Correct answer: A. (2520)

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

Correct answer: B. (20160)

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.

Correct answer: B. (90720)

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.

Correct answer: A. (241920)

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

Correct answer: D. (216)

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.

Correct answer: A. (120960)

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

Correct answer: A. (6720)

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

Correct answer: C. (20)

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.

Correct answer: B. (840)

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.

Correct answer: C. (4368)

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

Correct answer: B. (360)

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

Correct answer: A. (840)

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.

Correct answer: A. (1440)

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.

Correct answer: D. (780)

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.

Correct answer: A. (15120)

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

Correct answer: B. (4320)

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.