Fundamental principle of counting

The number must be even and the digit sum divisible by (3). Count choices for hundreds and tens according to the final even digit.

Letters have (26 \times 25) choices. Red gives (9 \times 9) digit choices, and the other two colors give (10 \times 9) choices each.

Of the 9 points, the 4 corner points are not allowed in the first position, so there are only 5 choices for the first point. The next three distinct points can then be chosen in 8, 7, and 6 ways respectively. Hence, the total number of patterns is \(5\times8\times7\times6=1680\). The value 1920 would result from incorrectly taking 8 choices instead of 5 for the first point. Exam tip: In pattern questions, order matters, so multiply the successive choices.

Choose (2) letters from (6) and ensure both appear in the three positions. Each pair gives (2^3-2) words.

Use the complement method. There are 9 choices for the first digit (1 to 9) and 10 choices for each of the remaining four positions, so the total number of PINs is \(9 \times 10^4=90000\). For PINs with no 7 at all, the first digit has 8 choices (1 to 9 except 7), while each remaining position has 9 choices. Thus, there are \(8 \times 9^4=52488\) such PINs. Therefore, the number containing at least one 7 is \(90000-52488=37512\). Note that 52488 counts PINs containing no 7. Exam tip: for “at least one” counting questions, try total minus none first.

Total paths are (10). If the first two are right, the remaining (1) right and (2) up moves can be arranged in (3) ways.

There are 21 consonants in the English alphabet. For the first position, B is forbidden, so there are 20 choices. After choosing the first consonant, repetition is not allowed; therefore the second, third, and fourth positions have 20, 19, and 18 choices respectively. The multiplication principle gives 20×20×19×18 = 136,800 codes. Hence option A is correct. A common mistake is to use 21 choices in the first position, which would ignore the restriction on B; another is to reduce the later choices incorrectly or to treat the code as an unordered selection rather than an ordered arrangement.

Total arrangements are (7 \times 6 \times 5 \times 4). For the thousands and units positions, exactly half have the thousands digit larger.

There are (30) ordered color choices. For the (5) choices starting with red, shape choices reduce from (9) to (6).

If the special photo is not used, arrangements are (6 \times 5 \times 4 \times 3). If used, it has (2) end choices and the rest have (6 \times 5 \times 4) choices.

The digit part has (10^3-10) choices. For letters, the first has (5) vowel choices and the second has (25) choices.

The first digit has (5) odd choices and the last has (5) even choices. The middle four positions are filled from the remaining (8) digits in order.

Count type distributions ((1,1,1)), ((2,1,0)), and ((1,2,0)). Their sum is (5 \times 4 \times 3+\binom{5}{2} \times 4+5 \times \binom{4}{2}).

Without the adjacency restriction, the number of four-letter codes is \(^{5}P_{4}=5\times4\times3\times2=120\). Now subtract the codes in which P and Q are adjacent. Treat P and Q as one block. The other two letters can be selected from the remaining three letters in 3 ways. For each selection, the three units can be arranged in \(3!\) ways, and the block can be PQ or QP in 2 ways. Thus, the forbidden codes are \(3\times3!\times2=36\). Hence, the required number is \(120-36=84\). The value 72 can result from incorrectly counting arrangements of all five letters instead of four-letter codes. Exam tip: for a “not adjacent” condition, use total arrangements minus the forbidden adjacent arrangements.

Treat the first and last positions first, then arrange the remaining seven distinct books. If the first book is mathematics and the last is physics, there are 5×4 choices. If the first is physics and the last is mathematics, there are 4×5 choices. Thus the two endpoint cases contribute 5×4 + 4×5 = 40 choices. After fixing those endpoints, the remaining seven books can be arranged in 7! ways. Therefore the total is 40×7! = 40×5040 = 201,600. Option A is correct. The original value 460,800 overcounts the arrangements; simply arranging all nine books does not enforce the different-subject endpoint condition.

There are \(7^3=343\) total indices for the three positions. The number of indices with no repeated letter is \(7\times6\times5=210\). Therefore, the number with at least one repeated letter is \(343-210=133\). Option 210 counts only the indices in which all three letters are different. Exam tip: For “at least one repetition” questions, it is usually quickest to subtract the complementary case where all entries are distinct.

Notebook has (6) choices and cover has (4) choices including none. Add valid cases separately for no pen and for choosing a pen.

The answer to the first question is fixed as correct. Therefore, exactly 3 of the remaining 5 answers must be correct so that the total number of correct answers is 4. The number of ways to choose these 3 questions is \(\binom{5}{3}=10\). Hence, 10 response sheets are possible. The value 15 comes from \(\binom{6}{4}\), but it ignores the condition that the first answer must be correct. Exam tip: When a position is fixed, handle it first and then use combinations for the remaining positions.

The thousands digit is from (1,2,3), and the units digit is odd from (1,3,5). Separate cases according to whether the thousands digit is odd.

First count ordered color pairs. From 5 colors without repetition, there are 5×4 = 20 ordered pairs. Among these, the pairs containing green number 2×4 = 8: green may be first or second, and the other position has 4 choices. For each of these 8 pairs, the round shape is forbidden, leaving 3 shapes, so they contribute 8×3 = 24 codes. The remaining 20−8 = 12 color pairs do not contain green and allow all 4 shapes, contributing 12×4 = 48. Hence the total is 24+48 = 72, so option D is correct. Option C subtracts only one shape overall instead of once for every green-containing pair.

Total assignments are (8 \times 7 \times 6). If both special students appear, choose the third from (6) students and arrange the (3) winners.

Fix the third ball as green and count first-place validity separately. The middle choices differ when the first ball is green or blue.

If star is not chosen, there are (4 \times 3 \times 2) answers. If star is chosen, the second position is fixed and the other two positions have (4 \times 3) choices.

For letters there are (26\times25\times24) choices and for digits (9\times10) choices. In exams multiply the choices at each stage.

The direct answer is option A, 60. The journey must make three consecutive choices: one route from A to B, one from B to C, and one from C to D. The multiplication principle says that when choices are made in successive stages, multiply their numbers. Therefore the total is 4 × 3 × 5 = 12 × 5 = 60. Option A is correct. Option B, 12, multiplies only the first two stages and forgets the five choices from C to D. Option C, 45, is not the product of all three route counts. Option D, 75, also has no valid counting basis and overcounts the journeys. Each complete journey is identified by one choice at every stage, so multiplication—not addition—is required. Memory cue: for an ‘and then’ sequence of independent choices, multiply; addition is used for alternatives, not successive stages.