Fundamental principle of counting

To form one outfit, choose 1 of the 4 shirts, 1 of the 3 trousers, and 1 of the 2 belts. Since there is no colour-matching restriction, these choices are independent. Hence, the total number of outfits is \(4 \times 3 \times 2 = 24\). Option 12 counts only the shirt-trouser choices and ignores the belt choices. Exam tip: use the multiplication principle when each stage has independent choices.

Choosing a club, a sport, and a shift are three independent choices. For each club, there are 4 sport choices, and for each sport, there are 2 shift choices. Therefore, by the fundamental principle of counting, the total number of ways is \(3 \times 4 \times 2=24\). The value \(12\) counts only the club and sport choices and misses the shift choice. Exam tip: Multiply the number of choices when all selections are made together.

This is a direct application of the fundamental principle of counting. The journey has two successive stages: first, a train must be selected from A to B, and then a train must be selected from B to C. There are 4 choices for the first stage. For each of those choices, all 5 trains on the second route remain available, so each first-stage choice can be paired with 5 second-stage choices. Thus the total is 4 × 5 = 20 possible train combinations. Option B adds the choices instead of multiplying them, while C and D do not represent the required product. Therefore option A is correct.

There are 3 choices for the hundreds digit, 10 for the tens digit, and 3 for the units digit. By the multiplication principle, the total is 3 × 10 × 3 = 90 numbers.

The two vowel places can be filled in 3 × 2 ways, since a vowel cannot be reused. The consonant places can be filled in 5 × 4 ways. Therefore, the total is 3 × 2 × 5 × 4 = 120 codes.

There are 5 choices for a drink by itself. A drink-and-snack choice can be made in 5 × 4 = 20 ways. These are separate kinds of choices, so the total is 5 + 20 = 25.

The first position is already fixed as 1, so it has no remaining choice. Each of the other 4 positions can contain either 0 or 1, giving 2 choices per position. By the fundamental principle of counting, the total number of strings is \(2\times2\times2\times2=2^4=16\). The answer would be 32 only if all 5 positions had 2 choices each. Exam tip: first identify fixed positions, then multiply the choices for the remaining independent positions.

This is a restricted arrangement counted by the multiplication principle. The first position is fixed as A, so it has exactly 1 choice and cannot be counted as 6 choices. Because repetition is forbidden, A cannot be used again. For the second position, 5 letters remain available; after selecting one of them, 4 letters remain for the third position. Thus the total number of codes is 1 × 5 × 4 = 20. Option C is correct. The value 36 would allow unrestricted repeated choices, and 30 fails to account for the decreasing number of available letters. The value 15 would incorrectly treat the last two positions as an unordered pair, although their positions in a code are different.

Each of the first two places has 5 choices, and each of the last two has 4 choices. Since repetition is allowed, these choices remain available at every place. Thus, the total is 5 × 5 × 4 × 4 = 400 numbers.

Choosing one theme, one font, and one icon set involves independent choices. By the fundamental principle of counting, the total number of interfaces is \(3 \times 4 \times 5=60\). Option 12 results from calculating only \(3 \times 4\) and ignores the 5 icon-set choices. Exam tip: multiply the numbers of choices when every choice at one stage can be paired with every choice at the next stage.

The direct answer is B: 12 appointments. The posts are different, so the order matters. First choose the president in 4 ways. Once that student is selected, the same student cannot be secretary, leaving 3 choices for secretary. Therefore the total is \(4\times3=12\). Option A, 8, merely adds the two groups and does not form ordered president-secretary assignments. Option B is correct. Option C, 16, multiplies \(4\times4\) but wrongly permits the same student to hold both posts. Option D, 20, has no valid counting basis. For example, president 1 can be paired with three different secretaries, and the same count applies to each of four possible presidents. Since president and secretary are distinct roles, swapping them creates a different appointment. Memory cue: different posts mean ordered choices; remove one choice after the first person is selected.

The hundreds digit must be 4, 5, or 6, giving 3 choices. After choosing it, there are 5 choices for the tens digit and 4 for the units digit, since digits cannot repeat. Thus, the total is 3 × 5 × 4 = 60 numbers.

There are 4 × 6 = 24 possible packet–label pairs before restrictions. Removing the 2 distinct pairs that are not allowed leaves 24 − 2 = 22 valid choices.

The choices of class, subject, test type, and language can be made independently. By the fundamental principle of counting, the total is 3 × 5 × 4 × 2 = 120 combinations.

The units digit must be 2, 4, or 6, giving 3 choices. After choosing it, there are 6 choices for the hundreds digit and 5 for the tens digit. Thus, the total is 3 × 6 × 5 = 90.

The units digit must be 1, 3, or 5, giving 3 choices. For each choice, 5 nonzero digits remain for the hundreds place, and then 5 digits (including 0, if unused) remain for the tens place. Thus, the total is 3 × 5 × 5 = 75.

There are 5 choices for the first digit because it cannot be 0. After that, 5, 4, and 3 choices remain for the next three places. Therefore, the total is 5 × 5 × 4 × 3 = 300.

The two letter places can be filled in 5 × 4 ways, and the two digit places in 6 × 5 ways. By the fundamental principle of counting, the total is (5 × 4) × (6 × 5) = 600 codes.

The first place has 4 choices. Each of the remaining 4 places has 3 choices, since its color must differ from the immediately preceding one. Thus, the total is 4 × 3⁴ = 324 codes.

The first letter has 4 choices: C, D, E, or F. After choosing it, 5 letters remain for the second place and 4 for the third. Therefore, 4 × 5 × 4 = 80 codes are possible.

For the outbound trip, there are 3 × 5 choices. On the return, there are 4 choices for the library-to-coaching leg and 2 for the coaching-to-home leg, because the outbound routes cannot be reused. Thus, 3 × 5 × 4 × 2 = 120 round trips are possible.

There are 4 × 6 choices for the soup and main dish. For dessert, there are 4 possibilities: choose one of the 3 desserts or choose none. Therefore, 4 × 6 × 4 = 96 meal selections are possible.

The first position has 5 choices because 0 is not allowed there. Each of the other 3 positions has 6 choices, since repetition is allowed. Thus, 5 × 6³ = 1080 PINs are possible.

There are 2 independent choices in each of the 5 sections. By the fundamental principle of counting, the total number of selections is \(2\times2\times2\times2\times2=2^5=32\). The value \(25\) would arise if there were 5 choices in each of 2 stages; here, each section has only 2 choices. Exam tip: multiply the number of choices when selections at successive stages are independent.

The tens digit must be 6, 7, or 8. For each choice, the units digit has 7 remaining choices; when the tens digit is 6, all these choices make the number greater than 60. Thus, 3 × 7 = 21 numbers are possible.