Fundamental principle of counting

There are 5 choices for the hundreds place, then 4 remaining choices for the tens place and 3 for the units place. By the fundamental principle of counting, the total is 5 × 4 × 3 = 60 numbers.

The first digit has 4 choices (1, 2, 3, or 4). After that, 4 digits remain for the second place and 3 for the third. Thus, the total is 4 × 4 × 3 = 48.

The units digit must be even, so it has 3 choices: 2, 4, or 6. After choosing it, there are 5 choices for the hundreds digit and 4 for the tens digit. The total is 3 × 5 × 4 = 60.

All available digits are even. If the units digit is 0, there are 4 choices for the hundreds digit and 3 for the tens digit, giving 12 numbers. If the units digit is one of 2, 4, 6, or 8, there are 4 choices for it, 3 nonzero choices for the hundreds digit, and 3 choices for the tens digit. This gives 36 more numbers, for a total of 12 + 36 = 48.

The units digit must be odd, so it has 3 choices: 1, 3, or 5. The remaining three places can then be filled in 5, 4, and 3 ways. Thus, the total is 3 × 5 × 4 × 3 = 180.

The two letter positions can be filled in 4 × 3 ways, and the two digit positions in 3 × 2 ways. By the multiplication principle, there are 4 × 3 × 3 × 2 = 72 codes.

Each of the two letter positions has 5 choices because repetition is allowed, and the digit position has 4 choices. Therefore, there are 5 × 5 × 4 = 100 passwords.

The 3 distinct letters can be arranged in 6 × 5 × 4 ways, and the 2 distinct digits in 5 × 4 ways. Multiplying gives 6 × 5 × 4 × 5 × 4 = 2400 plates.

For the outward journey, there are 3 × 4 choices. On the return, there are 3 choices for the school-to-market segment (excluding the one used going out) and 2 choices for the market-to-home segment. Thus, the total is 3 × 4 × 3 × 2 = 72 ways.

There are 4 × 5 choices for a starter and a main course. Dessert has 4 possibilities: choose one of the 3 desserts or choose none. Thus, the total is 4 × 5 × (3 + 1) = 80 meals.

There are 5 × 6 ways to choose the two questions. Each pair can be presented in either of 2 languages, so the total is 5 × 6 × 2 = 60 choices.

The first digit cannot be 0, so there are 5 choices for it. After that, there are 5, 4, and 3 choices for the remaining places. Thus, the total is 5 × 5 × 4 × 3 = 300.

The units digit must be odd, giving 4 choices. After choosing it, 3 nonzero digits remain for the hundreds place. The tens place then has 3 choices, including 0 if it remains. Thus, 4 × 3 × 3 = 36.

There are 6 choices for the first place because 0 is not allowed there. Repetition is allowed, so each of the other three places has 7 choices. Therefore, 6 × 7 × 7 × 7 = 6 × 7³ = 2,058 codes.

The direct answer is option C: 48 codes. For the first position, any of the 3 colors may be chosen. After that, the next position cannot repeat the previous color, so it has 2 choices. The same reasoning applies to each of the remaining three positions: each has 2 choices based on the immediately preceding position. Therefore, the total is \(3\times2\times2\times2\times2=3\times2^4=48\). Option A, 81, would correspond to a different counting situation and does not enforce the no-equal-neighbours rule. Option B, 96, is too large and introduces no valid extra choice. Option C is correct. Option D, 24, is too small; it loses one factor of 2 or otherwise undercounts the four positions after the first. Notice that a color may reappear after one or more positions; only consecutive equality is forbidden. For example, red-blue-red is allowed. The exam cue is: first position has all choices, and every later position has total colors minus one choice.

Each badge is formed by independently choosing a shape, a color, and a symbol. By the fundamental principle of counting, the number of badges is 4 × 3 × 2 = 24.

The first place has 4 choices because P is excluded. Without repetition, the second place then has 4 choices and the third has 3. Thus, 4 × 4 × 3 = 48 codes can be formed.

Choosing a pen and a notebook gives 3 × 4 = 12 options. Buying a pencil gives 5 other options. Since these are alternative purchases, add them: 12 + 5 = 17.

A card is formed by independently choosing a membership type, a validity period, and a colour. By the fundamental principle of counting, the total number of options is \(2 \times 4 \times 3=24\). Option 18 may result from incorrectly adding choices; independent stages are multiplied. Exam tip: Multiply the numbers of choices when every choice at one stage can be paired with every choice at the next stage.

For each choice, select a class, then a row in that class, then a seat in that row. The number of choices is 3 × 6 × 4 = 72.

The direct answer is A: 18 numbers. To be greater than 50, the tens digit must be 5, 6, or 7. Thus there are 3 choices for the tens place. After choosing it, one digit has been used, so 6 of the remaining digits can be placed in the units position. Repetition is forbidden, so the total is \(3\times6=18\). Option A is correct. Option B, 21, would incorrectly allow 7 choices for the units digit for every tens digit, including reuse of the tens digit. Option C, 14, does not count all valid units choices. Option D, 28, is the total number of ordered two-digit selections without applying the greater-than-50 restriction. Notice that 50 itself cannot be formed because digit 0 is unavailable, and all numbers beginning with 5, 6, or 7 are greater than 50. Memory cue: choose the restricted first digit first, then count the remaining digits.

The hundreds digit must be 1 or 2 for the number to be 3-digit and less than 300, giving 2 choices. Each of the other two places has 6 choices because repetition is allowed. Thus, 2 × 6 × 6 = 72.

Each of the two letter places has 5 choices, and each of the two digit places has 6 choices. Since repetition is allowed, the total is 5 × 6 × 6 × 5 = 900 codes.

The first place has 4 choices. Each following place has 3 choices because it cannot match the symbol immediately before it. Thus, the total is 4 × 3 × 3 = 36 passwords.

Each department code can be paired with any of the 6 employee codes. Therefore, by the fundamental principle of counting, the total number of identities is \(5 \times 6=30\). The value \(5+6=11\) would result from addition, which is not applicable because one code from each group is selected together. Exam tip: Multiply the numbers of choices when a task is completed through successive steps.