Fundamental principle of counting

Use the multiplication principle and split according to the final digit, because an even number must end in 0, 2, 4, or 6. If the final digit is 0, the first digit has 6 nonzero choices, and the remaining three positions can be filled in 5P3 = 5×4×3 = 60 ways, giving 360. If the final digit is 2, 4, or 6, there are 3 choices for it; the first digit then has 5 nonzero choices, and the middle three positions have 5P3 = 60 arrangements, giving 3×5×60 = 900. The total is 360+900=1260, so option A is correct. The smaller listed values overlook some valid cases.

The direct answer is option A, 26 × 25 × 9 × 9 × 8. Apply the multiplication principle: multiply the choices at each successive position. The first uppercase letter has 26 choices. The second must be different, so it has 25 choices. For the first digit position, zero is forbidden, leaving 1 through 9, or 9 choices. The next digit must be distinct from the first, so 9 choices remain, including zero if unused. The last digit then has 8 choices. Thus the total is 26 × 25 × 9 × 9 × 8. Option B allows repeated letters and allows zero first, so it is wrong. Option C wrongly reduces the letter choices and digit choices. Option D incorrectly allows 10 choices for the first digit. Remember: distinct means subtract already used choices; a position-specific restriction must be applied at that position.

For the onward journey, there are 4 choices from A to B and 5 choices from B to C, giving \(4\times5=20\) ways. On the return journey, 4 routes remain from C to B after excluding the route already used, and 3 routes remain from B to A. Hence, the total number of round trips is \(4\times5\times4\times3=240\). Option 120 does not correctly account for the available choices on both return segments. Exam tip: multiply the independent choices for each segment, subtracting a forbidden route only from its own segment.

The thousands digit is from (3) to (8) and the units digit is odd. Split cases according to whether the thousands digit is odd or even.

Total letter arrangements are (26^3) and no-vowel arrangements are (21^3). Multiply the difference by (10 \times 9) digit choices.

Add selections from exactly two categories and all three categories. This gives ((6 \times 5)+(6 \times 4)+(5 \times 4)+(6 \times 5 \times 4)).

For divisibility by 5, the units digit must be 0 or 5. If the units digit is 0, there are 5 choices (1 to 5) for the thousands digit, followed by \(4\times3\) arrangements for the two middle places: \(5\times4\times3=60\). If the units digit is 5, the thousands digit cannot be 0, so it has 4 choices (1 to 4), and the middle places can be filled in \(4\times3\) ways: \(4\times4\times3=48\). Hence, the total is \(60+48=108\). Option 96 is incorrect because the restriction on 0 in the thousands place is not the same in both ending-digit cases. Exam tip: When 0 is among the digits, handle the leading-digit restriction separately.

There are 4 ways to choose a mathematics book and 3 ways to choose a science book, giving \(4+3=7\) choices. For a language book, one of the 2 dictionary types must also be chosen; hence the valid language selections are \(5\times2=10\). Therefore, the total number of valid selections is \(7+10=17\). The option 15 can result from not applying the multiplication rule correctly to the language-book-and-dictionary case. Exam tip: add mutually exclusive cases connected by “or”, and multiply choices that must be selected together.

There are 26 choices for the first letter and 25 choices for the distinct second letter, giving \(26\times25=650\) letter pairs. The four-digit part has \(10^4=10000\) possible strings. Of these, exactly 10 strings—\(0000,1111,\ldots,9999\)—have all four digits identical, so valid digit strings are \(10000-10=9990\). Hence, the total is \(26\times25\times9990=6493500\). Option B incorrectly includes the 10 forbidden identical-digit strings. Exam tip: Count all cases first and subtract the restricted cases when the restriction is simple.

Subtract the forbidden (1 \times 1 \times 6) from total (7 \times 4 \times 6). Identifying the forbidden case is the quick exam method.

Separate cases according to whether the repeated digit is (0). Do not forget the first-place nonzero restriction.

Divisibility by (4) depends on the last two digits. Count valid ending pairs while separating cases where (0) remains for the first place.

Each of the five sections has two choices: filled or blank. Without any restriction, the multiplication principle gives two choices for each section, so there are \(2^5=32\) patterns. However, the completely blank form is not allowed, so one pattern must be removed. This leaves \(32-1=31\) nonempty patterns.

Now exclude the patterns in which both the first and second sections are filled. Once those two sections are fixed as filled, each of the remaining three sections can independently be filled or left blank, giving \(2^3=8\) forbidden patterns. Therefore the required number is \(31-8=23\). Thus option C is correct. Option 22 would result from an incorrect subtraction, while 24 does not account properly for the restrictions.

Subtract arrangements with blue on top, (5 \times 4), from total (6 \times 5 \times 4). Subtracting forbidden cases is simpler.

The independent choices from sections A and C contribute 2 and 3 possibilities respectively. In section B, the number of unrestricted pairs is C(4,2) = 6. Exactly one pair, namely the first two questions, is forbidden, so 6 − 1 = 5 valid pairs remain. By the multiplication principle, the total number of complete selections is 2 × 5 × 3 = 30. Therefore option B is correct. Counting all six pairs without removing the prohibited pair would give 36, which is the likely distractor. The restriction concerns only section B and does not affect the choices in sections A or C.

Choose (2) even digits from (4) and (3) odd digits from (5). Then arrange the selected (5) digits in (5!) ways.

The letters have (26 \times 25) choices. For the three digits, the last has (5) choices and the other two have (10,10) choices.

Since E is prohibited, each of the four positions can use only A, B, C, or D, giving 4⁴ = 256 unrestricted cards. From these, subtract cards in which A never appears. If A is absent as well, every position has only B, C, or D, so there are 3⁴ = 81 such cards. By complementary counting, the required number is 4⁴ − 3⁴ = 256 − 81 = 175. Thus option A is correct. Option B counts all cards after excluding E but fails to enforce the requirement that A must occur; option C is only the excluded complement, and option D is an arithmetic or counting distractor.

Subtract the cases containing three consecutive (1)'s from total (2^6) by organized counting. For short strings, position-based counting is reliable.

The units digit is fixed as (9). The thousands digit is from (2,4,6,8), and the middle two places use the remaining (4) digits.

There are (4) jacket choices because wearing none is also a choice. Subtract forbidden (1 \times 1 \times 4) from total (5 \times 4 \times 4).

The three letters have (26 \times 25 \times 24) choices. The two symbol positions have (4 \times 4) choices.

This is a (5)-position binary sequence with no consecutive stairs. A short recurrence count gives (13).

The first question has (3) choices, the second has (2), the third has (3), and the fourth matches the third. Thus (3 \times 2 \times 3).

The six ordinary colors give (6 \times 5) choices. Add (3) choices for gold and (2) choices for silver.