Permutations

There are (7) letters with (A) repeated twice, so (\frac{7!}{2!}=2520). In exams, divide by repeated letters.

Use the block method because the four girls must occupy consecutive seats. Treat all four girls as one combined block. This block together with the six distinct boys gives 7 distinct units, which can be arranged in 7! ways. The four girls can then be arranged internally within their block in 4! ways. By the multiplication principle, the total is 7! × 4! = 5040 × 24 = 120,960. Therefore option B is correct. Counting 10! would ignore the together condition, while omitting 4! would treat the girls as indistinguishable. The other numerical options arise from such incorrect arrangements or arithmetic errors.

Rotations are considered the same around a circular table, so the number is ((9-1)!=40320). In exams, fix one position in circular seating.

In a necklace, both rotation and reflection are the same, so (\frac{(8-1)!}{2}=2520). In exams, reduce reflections for necklaces.

The three medals are different ordered positions, so (^{14}P_3=14\cdot13\cdot12=2184). In exams, always count order for medals.

There are (2) odd choices for the unit place and the thousands place cannot be (0). The total is (2\cdot4\cdot4\cdot3=96).

Treat the two subjects as two blocks, so (2!\cdot5!\cdot4!=5760). In exams, also multiply internal arrangements in the block method.

Order matters in a password, so (^{9}P_4=3024). In exams, treat passwords or codes as permutations.

There are (11) letters with (I) four times, (S) four times and (P) twice. Hence the number is (\frac{11!}{4!4!2!}=34650).

The chairs are distinct and (7) students are seated, so (^{10}P_7=604800). In exams, count ordered positions even when some chairs remain empty.

Treat the two particular people as one block, then (9!\cdot2!=725760). In exams, include the internal arrangement of the block.

Subtract the together cases (9!\cdot2!) from total (10!). The answer is (3628800-725760=2903040).

Arrange the men first in (5!) ways, then place (4) women in (6) gaps in (^{6}P_4) ways. The total is (86400).

The ordered placement of (7) files in (9) distinct places is (^{9}P_7=181440). In exams, distinct shelves make assignment a permutation.

The direct answer is C: 480. The thousands digit must make the number greater than 5000. From the available digits, it can be 5, 6, 7 or 8: 4 choices. After choosing it, there are 6 unused digits for the hundreds place, then 5 and 4 choices. Thus the total is \(4\times6\times5\times4=480\). Option C is correct. Option A, 360, misses some arrangements. Option B, 720, incorrectly gives too many choices, as if every first digit were allowed or repetition were handled differently. Option D, 840, is also too large. The important cue is to restrict the first digit first, then multiply the remaining decreasing choices.

The two particular books occupy the ends in (2!) ways and the remaining books in (6!) ways. The total is (2!\cdot6!=1440).

The particular person has (7) inner positions and the others sit in (8!) ways. The total is (7\cdot8!=282240).

The two posts are different, so (^{13}P_2=13\cdot12=156). In exams, changing order changes the result for different posts.

For an even number, the units digit must be 0, 4, 6, or 8. If the units digit is 0, there are 5 choices for the hundreds digit and 4 for the tens digit, giving \(5\times4=20\) numbers. If the units digit is 4, 6, or 8, then in each case there are 4 non-zero choices for the hundreds digit and 4 choices for the tens digit, giving \(4\times4=16\) per case. Hence, the total is \(20+3\times16=68\). The count 64 does not handle the special case of 0 in the units place correctly. Exam tip: the hundreds digit of a three-digit number can never be 0.

The governing concept is the permutation of objects with repeated elements. BALLOON has seven letters: B, A, L, L, O, O, N. If all seven letters were different, the number of arrangements would be 7!. However, the two Ls are identical, so each distinct arrangement is counted 2! times in that total. The two Os are also identical and require division by another 2!. Therefore the number of distinct arrangements is 7!/(2! × 2!) = 5040/(2 × 2) = 1260. Hence option A is correct. Option C is simply 7! and fails to correct for either repeated pair. Option B divides by only one 2!, so it corrects for one repeated letter but not the other. Option D divides by an unnecessary additional factor and undercounts the arrangements.

Order matters in a signal, so (^{8}P_6=20160). In exams, use permutation for signal and code questions.

Arrange the boys around the circle in ((4-1)!) ways, then place the girls in (4) gaps in (4!) ways. The total is (144).

Treat the two people as one block, then (6) units have ((6-1)!) circular arrangements and (2!) internal ways. The total is (240).

Selection is followed by ordering, so (^{11}P_4=7920). In exams, use (nP_r) for selected and ordered cases.

The direct answer is option A, 360. A four-digit number greater than 7000 must have a thousands digit greater than 7. From the given digits, the possible thousands digits are 8 and 9 only, so there are 2 choices, not 3. After choosing that first digit, 6 digits remain for the hundreds place, then 5 for the tens place, and 4 for the units place. Hence the count is \(2\times6\times5\times4=240\), which is option D, not option A. Therefore the supplied answer is inconsistent. Option A, 360, would result from incorrectly allowing 7, 8, and 9 as the first digit; numbers beginning with 7 are not all greater than 7000, and in fact a number beginning with 7 can equal or exceed 7000 depending on its remaining digits, but with the exact digit set here every 7xxx number is greater than 7000 except 7000 itself, which cannot be formed because 0 is absent. Thus 7 should actually be allowed, giving 3 choices and \(3\times6\times5\times4=360\). Consequently option A is correct after all. Option B, 180, misses half the valid arrangements. Option C, 120, uses too few choices. Option D, 240, incorrectly excludes all numbers beginning with 7. Memory cue: compare the first digit first, and check carefully whether equality is possible with the remaining digits.