Permutations

Two groups are repeated (2) times each, so the number is (\frac{9!}{2!2!}=90720). In exams, put each repeated group factorial in the denominator.

Order matters in a signal, so (^{7}P_5=2520). In exams, words like signal indicate permutation.

Total circular arrangements are (5!), and together arrangements are (4!\cdot2!). So the answer is (120-48=72).

The pictures are selected and also ordered, so (^{10}P_3=720). In exams, use (nP_r) for selected and ordered cases.

The thousands place can be (6,8,9), and the remaining places have (5\cdot4\cdot3) ways. The total is (3\cdot5\cdot4\cdot3=180).

The (4) students receive one prize each from (6) prizes in an ordered way, so (^{6}P_4=360). In exams, distinct recipients make order important.

The direct answer is option A, 6720. The words “selected and arranged on a shelf” mean that both choosing and ordering matter. First choose 5 books from 8 and then arrange those 5 books: \(\binom{8}{5}\times5! = \frac{8!}{3!}=8P5=6720\). Equivalently, fill the five shelf positions successively: 8 choices for the first, 7 for the second, 6 for the third, 5 for the fourth, and 4 for the fifth, giving \(8\times7\times6\times5\times4=6720\). Option A is correct. Option B, 336, is \(8P?\) not the required complete count and does not represent selecting and ordering five books correctly; in fact \(\binom{8}{3}=56\), while \(8P5=6720\). Option C, 56, is \(\binom{8}{5}\), so it counts selection only and ignores shelf order. Option D, 40320, is \(8!\), which arranges all 8 books, whereas only 5 books are to be placed. Exam cue: if selected objects are also placed in positions, use permutation, not combination.

The speaking order gives different outcomes, so (^{7}P_3=210). In exams, treat performance order as permutation.

A number divisible by 5 must end in 0 or 5. If the units digit is 0, there are 5 choices (1 to 5) for the thousands digit, and the two middle places can be filled in \(4P_2=12\) ways from the remaining digits. This gives \(5\times12=60\). If the units digit is 5, the thousands digit cannot be 0, so it has 4 choices (1, 2, 3, 4); the middle places again have \(4P_2=12\) arrangements. This gives \(4\times12=48\). Hence, the total is \(60+48=108\). Exam tip: In cases involving 0, always ensure that the first digit is not 0.

The first position is already reserved for the specified object, so that object has no choice left to make. The remaining seven distinct objects must be placed in the remaining seven positions. Since every object is different, changing the order creates a new arrangement.

The number of orders of seven distinct objects is the factorial of seven: \(7!=7\times6\times5\times4\times3\times2\times1=5040\). Equivalently, the remaining positions can be filled in 7 choices, then 6, and so on. Therefore option A, 5040, follows. The other values do not represent the required number of arrangements after fixing the first object.

Subtract the cases where both particular objects occupy the two ends, which are (2!\cdot7!), from total (9!). The answer is (362880-10080=352800).

The number of seating (6) students in (6) distinct positions is (6!=720). In exams, use factorial when positions are distinct.

(5) of (7) children receive distinct toys, so (^{7}P_5=2520). In exams, distinct receivers make order count.

The particular coin has (2) end choices and the remaining coins are arranged in (9!) ways. The total is (2\cdot9!=725760).

Only rotations are the same, so the circular arrangement is ((7-1)!=720). In exams, check whether reflection is also considered the same.

The left-to-right order is important, so (^{6}P_4=360). In exams, standing order means permutation.

When one chapter is required to come first, its position is already decided. The remaining four distinct chapters may be placed in any order in the four remaining positions. Since the order matters, this is a permutation problem, not a selection problem. The fixed first chapter does not contribute any additional choice, because there is only one permitted chapter in that position.

The number of orders of four distinct objects is calculated by factorial: \\(4!=4\times3\times2\times1=24\\). Equivalently, there are four choices for the second position, then three, two, and one for the remaining positions. Thus the number of possible orders is 24, so option A follows. A result of \\(5!=120\\) would incorrectly treat the first chapter as freely choosable.

Changing top-to-bottom order changes the design, so (^{8}P_3=336). In exams, use permutation for ordered layers.

(5) questions are selected and written in order, so (^{7}P_5=2520). In exams, if writing order differs, use permutation.

The condition says that the four distinct mobiles must appear consecutively. Treat those four mobiles as one large block. Along with the three separate books, this gives four objects to arrange. The mobiles are still distinct, so their internal order must also be counted.

The four objects can be arranged in \(4!\) ways. Inside the mobile block, the four mobiles can be arranged in \(4!\) ways. Hence the total is \(4!\times4!=24\times24=576\). Therefore, option A is correct. Counting only the block arrangements would miss the different orders of the mobiles.

The calling order to the stage is important, so (^{9}P_4=3024). In exams, use permutation when order is specified.

The row arrangement of (10) distinct objects is (10!). In exams, use factorial directly when all objects are distinct.

The direct answer is option C: 1320 ways. The three posts are different, so assigning the same three students to different posts gives different results. Choose the president in 12 ways, the secretary from the remaining students in 11 ways, and the treasurer in 10 ways. Hence the total is \(12\times11\times10=1320\), or \({}^{12}P_3\). Option C is correct. Option A, 220, is \(\binom{12}{3}\), which chooses three students but does not assign posts. Option B, 1728, does not follow the decreasing-choice rule and effectively overcounts. Option D, 36, is far too small and does not represent the three distinct assignments. If A, B, and C are chosen, then changing who becomes president changes the result even when the same people are involved. Memory cue: different named posts mean order matters; use permutation, not combination.

The first digit cannot be (0), so (7\cdot7\cdot6\cdot5=1470). In exams, apply the leading-zero condition first.

There are (4) even choices for the last place and the remaining (4) places are filled in (^{7}P_4) ways. The total is (4\cdot840=3360).