Permutations

The arrangement of (9) distinct objects is (9!). In exams, use factorial directly when all objects are distinct.

Order matters for different posts, so (^{8}P_4=1680). In exams, do not use combination when posts are different.

The hundreds place cannot be (0), so (6\cdot6\cdot5=180). In exams, always check the zero restriction on the first digit.

The direct answer is option A, 360. The units digit is fixed as 5, so the number automatically ends in 5. We must fill the remaining four positions using four different digits from the other six digits: 1, 2, 3, 4, 6, and 7. The first remaining position has 6 choices, then 5, then 4, then 3 choices. Therefore the count is ^6P_4=6×5×4×3=360. Option A is correct. Option B, 720, is twice the correct count and would overcount the arrangements. Option C, 120, is 5!, not the required selection and arrangement from six available digits. Option D, 840, has no valid count for these restrictions. The fixed final digit is not chosen again, and no repetition is allowed. Memory cue: fix the required last digit first, then permute the remaining available digits into the empty positions.

One letter is repeated (2) times, so the number is (\frac{8!}{2!}=20160). In exams, divide by factorials of repeated letters.

Apply the block method, a standard restricted-permutation technique. Since all three boys must sit together, regard them temporarily as one single block. The 7 girls plus this boys' block give 8 units, which can be arranged in 8! ways. Within the block, the 3 distinct boys can be ordered in 3! ways. By the multiplication principle, the total is 8! × 3! = 40320 × 6 = 241920. Hence option C is correct. Option A is half the correct result, option B omits the boys' internal arrangements, and option D is too small because it does not arrange all eight units properly.

Rotations are considered the same around a circular table, so the number is ((8-1)!=5040). In exams, use ((n-1)!) for circular table seating.

In a necklace, both rotation and reflection are the same, so (\frac{(6-1)!}{2}=60). In exams, distinguish necklace from circular seating.

Positions are ordered, so (^{10}P_2=10\cdot9=90). In exams, rank or position indicates permutation.

The direct answer is option A, 180. An even number must have an even units digit. From the given digits, the possible final digits are 2, 6, and 8, so there are 3 choices for the units place. After fixing that digit, 5 digits remain. The thousands, hundreds, and tens places must be filled without repetition: 5 choices, then 4, then 3. Thus the total is 3×5×4×3=180. Option A is correct. Option B, 360, forgets the restriction on the units digit and effectively overcounts. Option C, 240, does not follow the correct product of choices. Option D, 120, undercounts the possible choices. The first digit cannot be zero, but zero is not among the given digits, so there is no extra restriction. Exam cue: for an even number, choose the units digit first; then arrange the remaining digits.

There are (3) odd choices for the units place, then (4) valid choices for the thousands place after excluding (0). The total is (3\cdot4\cdot4\cdot3=144).

Treat the two subjects as two blocks, so the arrangement is (2!\cdot5!\cdot4!=5760). In exams, arrange both the blocks and the items inside blocks.

The direct answer is option A: 840 codes. A code has positions, so changing the order creates a new code. There are 7 choices for the first position, 6 for the second because repetition is forbidden, then 5 and 4. By multiplication, the total is \(7\times6\times5\times4=840\), or \({}^{7}P_4=\frac{7!}{3!}\). Option A is correct. Option B, 35, equals \(\binom{7}{4}\), which chooses four letters but ignores their order; that is unsuitable for codes. Option C, 2401, equals \(7^4\) and allows a letter to be repeated in every position, contrary to the condition. Option D, 210, is not the required ordered product and does not count all possible position sequences. The key distinction is that selection asks only who is chosen, while a code also records position. Memory cue: no repetition plus ordered code means \({}^{n}P_r\).

There are (7) letters and one letter is repeated (2) times, so (\frac{7!}{2!}=2520). In exams, count repeated letters carefully.

The chairs are distinct and (6) students are seated, so (^{8}P_6=20160). In exams, permutation may apply even when some places remain empty.

Treat the two particular people as one block, then (8!\cdot2!=80640). In exams, use the block method for together conditions.

The direct answer is option A, 282240. There are 9 distinct people, so without any restriction the number of row arrangements is \(9! = 362880\). Let the two particular people be P and Q. Count the unwanted arrangements in which they sit together: treat P and Q as one block. Then there are 8 objects to arrange, giving \(8!\) arrangements, and P and Q can exchange places inside their block in \(2!\) ways. Thus unwanted arrangements are \(8!\times2! = 40320\times2 = 80640\). Therefore wanted arrangements are \(9! - 8!\times2! = 362880-80640=282240\). Option A is correct because it gives this difference. Option B, 362880, counts every arrangement and therefore includes arrangements where the two people sit together. Option C, 80640, is the number of arrangements where they do sit together, the opposite condition. Option D, 40320, is only \(8!\), so it forgets the two possible orders inside the block. Memory cue: for “not together”, calculate total minus “together”; for “together”, use one block and multiply by \(2!\).

There are two possible patterns and each has (4!\cdot4!) arrangements. The total is (2\cdot4!\cdot4!=1152).

Arrange the red balls first in (5!) ways, then place (4) blue balls in (6) gaps in (^{6}P_4) ways. The total is (5!\cdot^{6}P_4=86400).

We choose ordered positions for (6) letters among (8) distinct boxes, so (^{8}P_6=20160). In exams, distinct boxes make it an arrangement.

The hundreds place can be (4,5,6), and the remaining places have (5\cdot4) ways. The total is (3\cdot5\cdot4=60).

The two particular books can occupy the ends in (2!) ways and the remaining books in (5!) ways. The total is (2!\cdot5!=240).

The particular person has (6) inner positions and the remaining people sit in (7!) ways. The total is (6\cdot7!=30240).

The three posts are different, so (^{11}P_3=990). In exams, different responsibilities make order important.

Use the multiplication principle while respecting both order and the restriction on the first position. Since the six distinct digits include 0, the first position cannot contain 0, so it has 5 choices. After selecting the first digit, one digit has been used; therefore the second position has 5 remaining choices, including 0 if it was not already selected. Repetition is not allowed, so the total is 5 × 5 = 25. Thus option B is correct. Option A incorrectly gives six choices for the second position, option C undercounts, and option D is 6 × 6, which permits repetition and also ignores the first-position restriction.