Fundamental principle of counting

Correct answer: B. (110)

Valid leader-coder pairs are (6\times8-(2\times3+4\times5)=22). Then multiply by the (5) choices for designer.

Correct answer: D. (240)

Letters have (5\times4=20) choices and the middle digits can be chosen in (4^2-2^2=12) ways. The complement method is fast for such restrictions.

Correct answer: B. (100)

Science-Skill selections are (1\times2+3\times6=20), and Arts has (5) choices. Add dependent subjects together first.

Correct answer: A. (3176)

When the last digit is (0), choices are (9\times8\times21), and when it is (5), choices are (8\times8\times26). Start divisibility conditions from the last digit.

Correct answer: C. (160)

Subject-set-version choices are (1\times4\times2+6\times4\times3=80), then there are (2) media choices. Treat the exceptional subject separately.

Correct answer: B. (2160)

If the first digit is even, there are (960) extensions, and if it is odd, there are (1200). Cases are necessary because of the (0) restriction on the first digit.

Correct answer: B. (3340)

Without the condition, total choices are (4\times6\times5\times7=3360), and (4\times5=20) choices are overcounted for the special pair. Hence (3340) is correct.

Correct answer: C. (55)

The first two layers have (3\times4-1=11) valid pairs, and the third layer has (5) choices. Subtract the blocked pair and then multiply.

Correct answer: C. 168

Use cases according to the city selected on day two. If the capital is selected on day two, day one can be any of the other 4 cities, day-one hotel choice has 3 possibilities, and day-two hotel choice has only 2 possibilities. This gives 4 × 3 × 2 = 24 plans. If a non-capital city is selected on day two, there are 4 choices for it, day one can be any of the remaining 4 cities, and both days have 3 hotel choices. This gives 4 × 4 × 3 × 3 = 144 plans. Adding disjoint cases gives 24 + 144 = 168, so option C is correct.

Correct answer: A. (15120)

Letters are chosen in (7\times6) ways and digits in (45\times8) ways. Choose the comparison-based pair first.

Correct answer: B. (156)

There are three cases depending on which counter gives the premium item, and the sum is (56+84+16=156). For exactly one condition, count each possible source separately.

Correct answer: C. (384)

The first (3) answers have (2^3-2=6) choices, and the middle (6) answers have (2^6) choices. Multiply by (1) for the fixed last answer.

Correct answer: A. (146250)

Letters have (26\times25) choices, and the vowel-digit part is formed in (5\times5+5\times4\times10=225) ways. Handle the dependent condition by vowel cases.

Correct answer: D. (144)

The die has (3) prime and (3) non-prime outcomes, so ((3\times5+3\times7)\times4=144). Count the type of outcome first.

Correct answer: B. (5500)

The choices for different questions are combined by multiplication, but the special condition depends on whether the first answer is D. Therefore, split the counting into two disjoint cases. If the first answer is D, it has 1 choice, the second question has 4 choices, the next three questions have \(5^3\) choices, and the last question has only 2 choices. This gives \(1\times4\times5^3\times2=1000\).

If the first answer is not D, it has 3 choices. The second question still has 4 choices, each of the next three has 5 choices, and the last question then has all 3 choices. This gives \(3\times4\times5^3\times3=4500\). The cases cannot overlap, so add them: \(1000+4500=5500\). Hence option B is correct.

Correct answer: C. (324)

The second-third stages can be formed in (1\times3+3\times5=18) ways. Then multiply by (3\times6) choices for the first and fourth stages.

Correct answer: B. (5040)

Digits are chosen in (40) ways and letters in (3\times7\times6=126) ways. Hence the total is (40\times126=5040).

Correct answer: C. (195)

Direct answer: option C, 195 selections. Use the fundamental principle of counting: when one choice is made after another, multiply the numbers of choices. For any of the three normal tracks, there are 6 session choices, 5 workshop choices, and 2 certificate choices, so the number is 3 × 6 × 5 × 2 = 180. In the fourth track, there are 3 sessions and 5 workshops, but no certificate choice; therefore it gives 3 × 5 = 15 selections. The cases are separate, so add them: 180 + 15 = 195. Option A, 210, is wrong because it does not use the correct total. Option B, 180, counts only the normal tracks and ignores the fourth track. Option C is correct because it includes both cases. Option D, 165, also results from an incorrect calculation. Memory cue: multiply within one case and add separate cases.

Correct answer: B. (8400)

Vowels are chosen in (5\times4) ways and consonants in (21\times20) ways. The total is (5\times21\times4\times20=8400).

Correct answer: C. 96

Count compatible morning-evening pairs first, then choose an afternoon class. If one of the 2 live morning classes is chosen, none of the 3 recorded evening classes may be selected, so there are 2 × 3 = 6 compatible pairs using the other 3 evening classes. If one of the remaining 3 morning classes is chosen, all 6 evening classes are allowed, giving 3 × 6 = 18 pairs. Thus there are 6 + 18 = 24 compatible morning-evening choices. Each can be combined with any of 4 afternoon classes, so the total is 24 × 4 = 96. Therefore option C is correct; simply multiplying all choices would incorrectly ignore the restriction.

Correct answer: A. (59049)

If the repeated last digit is (0), choices are (9^4), and if it is non-zero, choices are (9\times8\times9^3). The sum of both cases is (59049).

Correct answer: C. (288)

Subcategory choices are (6\times5+2\times3=36), and each has (4\times2) choices. The total is (36\times8=288).

Correct answer: A. (48)

There are (2) choices for the units digit and (4 \times 3 \times 2) ways for the remaining places. In exams apply the last digit condition first.

Correct answer: A. (216)

If the units digit is (0), there are (5 \times 4 \times 3 \times 2) ways; if it is (5), the first digit has (4) choices. Add the two cases carefully.

Correct answer: C. (2400)

Letters can be chosen in (5 \times 4) ways and digits in (6 \times 5 \times 4) ways. Multiply independent stages by the fundamental principle.