Fundamental principle of counting

There are 8 choices for the first-prize winner. Since that participant cannot also receive the second prize, 7 choices remain for it. The prizes are distinct, so the total is 8 × 7 = 56 outcomes.

The last digit has 3 choices: 1, 3, or 5. After choosing it, the first digit has 5 nonzero choices among the remaining digits. The middle places then have 5 and 4 choices. Thus, the total is 3 × 5 × 5 × 4 = 300 numbers.

The two digit positions can be filled in 5 × 4 ways, since the digits must be distinct. The two letter positions can be filled in 4 × 3 ways. Therefore, the total is (5 × 4) × (4 × 3) = 240 codes.

The direct answer is option B, 30 options. There are two separate cases: taking only a drink, or taking a drink together with one snack. The first case has 6 choices. In the second case, choose 1 of 6 drinks and 1 of 4 snacks, giving \(6\times4=24\) choices. Since the cases cannot happen together, add them: \(6+24=30\). Option A, 24, counts only drink-and-snack combinations and misses the drinks taken alone. Option B is correct because it includes both cases. Option C, 10, incorrectly adds 6 and 4 without forming pairs. Option D, 36, incorrectly counts all choices as if they were additional independent choices. Memory cue: disjoint cases are added; successive choices are multiplied.

Choose the color for the first place in 3 ways; this fixes the color in the third place. The second and fourth places each have 3 choices. Thus, there are 3 × 3 × 3 = 27 signals.

The bus, meal-pack, and museum choices are independent stages. By the fundamental principle of counting, the number of plans is \(4 \times 5 \times 3 = 60\). Therefore, option C is correct. Adding them to get \(4+5+3=12\) is incorrect because a plan requires one choice from each of the three categories. Exam tip: Multiply the numbers of choices when one option is selected at every stage.

By the fundamental principle of counting, multiply the choices for all five places: 2 × 2 × 2 × 5 × 5 = 2³ × 5² = 200 codes.

The units digit must be odd, so it has 4 choices: 1, 3, 5, or 7. After choosing it, the thousands, hundreds, and tens places have 7, 6, and 5 choices, respectively. Thus, the total is 4 × 7 × 6 × 5 = 840 numbers.

For a 4-digit number to be less than 4000, its thousands digit must be 1, 2, or 3, giving 3 choices. The remaining places then have 6, 5, and 4 choices without repetition. Therefore, 3 × 6 × 5 × 4 = 360 numbers.

Count two cases. If the units digit is 0, the hundreds and tens places have 7 and 6 choices, giving 7 × 6 = 42 numbers. If the units digit is 2, 4, or 6, there are 3 choices for it; the hundreds place then has 6 nonzero choices and the tens place has 6 choices, giving 3 × 6 × 6 = 108. The total is 42 + 108 = 150 numbers.

The two letter positions can be filled in 7 × 6 ways, and the three digit positions in 5 × 4 × 3 ways because repetition is not allowed within either group. By the multiplication principle, the total is 7 × 6 × 5 × 4 × 3 = 2520 codes.

Since repetition is allowed, the letter positions have 4³ possibilities and the digit positions have 6² possibilities. Thus the total is 4³ × 6² = 64 × 36 = 2304 passwords.

The first position has 5 choices. Each of the remaining 5 positions has 4 choices, since it must differ from the color immediately before it. Therefore, the number of sequences is 5 × 4⁵ = 5120.

There are 6 choices for the first letter, since A and B are excluded. With no repetition, there are then 7 choices for the second letter and 6 for the third. The total is 6 × 7 × 6 = 252 codes.

For the outward trip there are 4 × 6 choices. On the return, there are 5 choices from the library to the center (excluding the route just used) and 3 from the center to home. Thus there are 4 × 6 × 5 × 3 = 360 round trips.

There are 5 × 4 choices for the starter and main dish. The optional dessert has 3 + 1 choices (including none), and the optional drink has 2 + 1 choices. Therefore, the total is 5 × 4 × 4 × 3 = 240 selections.

The first position has 8 choices (1 through 8), each of the 3 middle positions has 9 choices, and the last position has 5 even choices (0, 2, 4, 6, 8). Since repetition is allowed, the total is 8 × 9³ × 5 = 29160 PINs.

The governing concept is the fundamental principle of counting: when independent choices are made in succession, the total number of complete outcomes is found by multiplication. From each of the four sections, one of three questions can be chosen, giving 3 × 3 × 3 × 3 = 3^4 = 81 possible question selections. For every one of these selections, either one of two languages may be selected. Hence the total number of ways is 3^4 × 2 = 81 × 2 = 162. Therefore option C is correct. Addition would be inappropriate because a complete outcome includes a choice from every section and a language simultaneously. Option A counts only question selections and omits the language choice; the remaining options use an incorrect product.

Direct answer: option C, 24 numbers. A two-digit number has a tens digit and a units digit. To be greater than 70, its tens digit must be 7, 8, or 9. If the tens digit is 7, the units digit can be any of the other eight digits: 1,2,3,4,5,6,8,9. The same eight choices are available when the tens digit is 8 or 9, because repetition is forbidden. Therefore the count is 3×8=24. Option A, 18, undercounts the available units choices. Option B, 21, also does not use the full 3×8 count. Option C, 24, correctly counts all choices without repetition. Option D, 27, would incorrectly allow nine units choices for each tens digit, including repetition of the selected tens digit. Note that 70 itself cannot be formed because digit 0 is not supplied, and all numbers beginning with 7, 8, or 9 are greater than 70.

The 3 letter positions can be filled in 5³ ways because repetition is allowed. The 2 symbol positions can be filled in 4 × 3 ways because symbols cannot repeat. Thus, the total is 5³ × 4 × 3 = 1500 codes.

The three posts are distinct, so assigning the same three students to different posts can produce different appointments. This is an ordered selection without repetition. There are 7 choices for president. After that student is used, 6 choices remain for vice-president, and then 5 choices remain for secretary. By the multiplication principle, the number of appointments is 7 × 6 × 5 = 210. Equivalently, it is the permutation ⁷P₃. Option D, 343, incorrectly allows repetition by using 7³; options B and C do not match the required product. Therefore option A is correct.

The outfit choices are 5 × 3 × 2 = 30. Choosing a bag instead gives 7 further choices. These are separate alternatives, so the total is 30 + 7 = 37 choices.

To specify one seat-shift, choose a block, a row in that block, a seat in that row, and a shift. By the fundamental principle of counting, the total number of ways is \(5 \times 6 \times 8 \times 2=480\). Option 240 misses the 2 choices of shift. Exam tip: Multiply the numbers of choices at successive independent stages.

The thousands digit can be 2, 3, 4, or 5, giving 4 choices. Each of the remaining three places can be any of the 10 digits. Thus, the total is 4 × 10 × 10 × 10 = 4000.

There are 6 × 5 = 30 filter-and-frame combinations. For each combination, there are 5 sticker choices: use none or choose 1 of the 4 stickers. Therefore, 30 × 5 = 150 looks are possible.