Fundamental principle of counting

Each of the first two places has 3 choices, and each of the last two has 4 choices. Since repetition is allowed, the total is 3 × 3 × 4 × 4 = 3² × 4² = 144 codes.

The governing concept is the addition and multiplication principles used together for mutually exclusive cases. In the first case, the student buys a pen and a notebook, so there are 4 × 5 = 20 possible pairs. In the second case, the student buys one pencil, giving 3 possibilities. These cases are alternatives: a purchase in the first case is not the same type of choice as a pencil-only purchase, so their counts are added. Hence total options = 20 + 3 = 23, making option A correct. Option B counts only the first case, option C multiplies 4 by 3 and ignores notebooks, and option D adds all raw item counts without forming pairs.

To choose one seat, select 1 of the 3 classes, then 1 of the 7 rows, and then 1 of the 4 seats in that row. By the fundamental principle of counting, the total number of ways is \(3 \times 7 \times 4=84\). Option 42 counts only the classes and rows, \(3 \times 7\), and ignores the 4 seats in each row. Exam tip: multiply the number of choices at every successive stage.

The hundreds digit can be 1, 2, 3, or 4: it cannot be 0 in a 3-digit number, and it must be less than 5. Each of the other two positions has 8 choices, with repetition allowed. Therefore, 4 × 8 × 8 = 256 numbers are possible.

There are 5 × 4 = 20 filter-and-frame choices. The sticker choice has 4 possibilities: no sticker or 1 of the 3 stickers. Thus, 20 × 4 = 80 looks.

The two vowel positions can be filled in 4 × 3 ways, and the two consonant positions in 6 × 5 ways. By the multiplication principle, there are 4 × 3 × 6 × 5 = 360 codes.

Each setting consists of one choice from each of the four categories. By the multiplication principle, the total is 5 × 4 × 3 × 2 = 120 settings.

The hundreds digit must be 5, 6, 7, or 8, giving 4 choices. After choosing it, there are 6 choices for the tens digit and 5 for the units digit. Thus, 4 × 6 × 5 = 120 numbers are possible.

Each coin has 2 possible outcomes, so tossing 2 coins gives \(2\times2=4\) outcomes. A die has 6 possible outcomes. Therefore, by the fundamental principle of counting, the total number of outcomes is \(4\times6=24\), so option C is correct. Option 12 counts only one coin and the die, ignoring the second coin. Exam tip: For successive independent actions, multiply the number of possible outcomes at each stage.

Without restrictions, there are 5 × 7 = 35 packet-label pairs. Excluding the 3 disallowed pairs leaves 35 − 3 = 32 valid combinations.

There are 4 choices for the color. The two ordered shape positions can be filled in 5 × 4 ways because the shape cannot be repeated. Therefore, 4 × 5 × 4 = 80 signals are possible.

The first two positions can be filled in 3 × 2 ways, and the last two positions also in 3 × 2 ways. The digit sets for these two pairs are disjoint, so the total is 3 × 2 × 3 × 2 = 36 numbers.

There are 7 choices for the boy monitor and 5 choices for the girl co-monitor. By the multiplication principle, there are 7 × 5 = 35 ways to choose the pair.

The first position is fixed as M. Of the 7 available letters, 6 remain for the second position and then 5 for the third. Thus, 6 × 5 = 30 codes are possible.

Each option is formed by choosing one class, one subject, one test type, and one medium. By the multiplication principle, the total is 4 × 6 × 5 × 2 = 240.

There are 3 choices for the hundreds digit, 10 for the tens digit, and 4 for the units digit. These choices are independent, so the total is 3 × 10 × 4 = 120.

There are 3 independent choices for the language, 4 for the project, and 2 for the presentation mode. By the fundamental principle of counting, the total number of ways is \(3 \times 4 \times 2=24\). Hence, 24 is correct. Option 12 counts only the language and project choices \(3 \times 4\) and omits the 2 presentation-mode choices. Exam tip: For successive independent choices, multiply the number of options at each stage.

The first place has 5 choices. The second has 4 choices because it must differ from the first. Each of the remaining two places has 5 choices, so the total is 5 × 4 × 5 × 5 = 500.

For the outward trip there are 5 choices on A–B and 4 on B–C. On the return, there are 3 choices for C–B because the previously used service is excluded, and 5 choices for B–A. Thus, the total is 5 × 4 × 3 × 5 = 300 round trips.

One choice is made independently from each of the four categories. By the multiplication principle, the number of avatars is 4 × 5 × 3 × 2 = 120.

If the units digit is 0, the other places can be filled in 6 × 5 × 4 = 120 ways. If the units digit is 2, 4, or 6, there are 3 choices for it, 5 choices for the nonzero thousands digit, then 5 and 4 choices for the remaining places: 3 × 5 × 5 × 4 = 300. The total is 120 + 300 = 420.

There are 4 choices for the first letter, which fixes the last letter. Each of the 3 middle positions has 4 choices. Thus, the number of codes is 4 × 4³ = 256.

Without restrictions there are 3 × 4 × 5 = 60 combinations. For the one restricted type, 2 colors are unavailable, each with 5 possible validity periods, so 2 × 5 = 10 combinations are invalid. Therefore, 60 − 10 = 50 valid cards remain.

The hundreds digit must be 2, 3, 4, 5, or 6, giving 5 choices. Each choice makes the number greater than 200 and less than 700. With no repetition, there are then 7 choices for the tens digit and 6 for the units digit. The total is 5 × 7 × 6 = 210.

The first place has 5 choices and the second has 4 choices because its digit must differ from the first. Each of the last two places has 5 choices, since repetition is allowed there. Thus, the total is 5 × 4 × 5 × 5 = 500 PINs.