Fundamental principle of counting

Correct answer: A. (7 \times 6 \times 5 \times 4^3)

Letter choices decrease, while symbols give (4^3) because repetition is allowed. Read the repetition rule for each block separately.

Correct answer: B. (4 \times 3+3 \times 4=24)

There are two separate cases and each has (12) options. Add the cases for or, and multiply within each case.

Correct answer: A. (5 \times 4 \times 3=60)

Choose positions for the two distinct red balls and then the blue ball, giving (5,4,3) choices. Blank marks occupy the remaining positions automatically.

Correct answer: A. (36 \times 10^3)

The first two digits form an increasing pair in (36) ways. The remaining (3) positions independently give (10^3) choices.

Correct answer: C. (5 \times 4 \times 6 \times 2=240)

Every item is compulsory and independent, so all choices are multiplied. For complete kits, do not add choices.

Correct answer: A. (8 \times 7 \times 1 \times 7=392)

The third position is forced by the first letter, so it has (1) choice, and the fourth has (7) choices excluding the second letter. Count forced positions separately.

Correct answer: A. (2 \times 6 \times 6=72)

The hundred place can only be (2) or (3), and each remaining place has (6) choices. Apply the bound condition to the highest place.

Correct answer: B. (1 \times 5 \times 2+1 \times 5 \times 4=30)

The number of symbols changes with the color, so there are two cases. Uniform multiplication is not used when choices depend on earlier choices.

Correct answer: C. (7 \times 8 \times 9=504)

Choose the wicketkeeper first with (7) choices, then captain and vice-captain have (8) and (7) choices. Fill the most restricted post first.

Correct answer: A. (3^5-3^3=216)

From all (3^5) codes, subtract the (3^3) codes having (AA) in the first two positions. Subtract only the forbidden part.

Correct answer: A. (9 \times 1 \times 10^4)

The first digit is chosen in (9) ways and the second is fixed by it. A forced position has (1) choice.

Correct answer: B. (1 \times 2 \times 4 \times 6+2 \times 5 \times 4 \times 6=288)

The second-stage choices depend on the first selection, so separate the (1) and (2) first-stage cases. When dependence appears, use cases.

Correct answer: A. (10 \times 9 \times 8=720)

The shelves are different, so the first, second, and third shelves have (10,9,8) choices. For distinct positions, order matters.

Correct answer: D. (4 \times 9^3+6 \times 8 \times 9^2)

Count separately whether the first digit is (7) or not. For exactly (2) occurrences, choose positions carefully.

Correct answer: A. (6^3-6=210)

From all (6^3) rows, subtract the (6) rows where all three are identical. With a single prohibition, the complement method is quick.

Correct answer: B. (1 \times 3 \times 10^2+3 \times 5 \times 10^2=1800)

Class choices depend on the zone, so count North and other zones separately. For digits with repetition, the factor remains (10^2).

Correct answer: A. (6 \times 5 \times 4 \times 3=360)

Choose chairs for the four distinct students in order, giving (6,5,4,3) choices. Distinct chairs create an ordered selection.

Correct answer: A. (2^3 \times 4^3=512)

Each subject has (2 \times 4) choices and there are (3) subjects. Use powers for repeated independent stages.

Correct answer: A. (4 \times 3 \times 5 \times 4 \times 3=720)

First choose the position of the digit from the special group in (4) ways, then fill (3) and (5,4,3) choices. For exactly-one conditions, position choice is essential.

Correct answer: A. (4^5-3^5=781)

For at least one (4), subtract the (3^5) arrangements with no (4) from all (4^5). The complement method is easy for at-least questions.

Correct answer: A. (3 \times 2 \times 4^4=1536)

The first block has (3) and (2) choices, and each position in the last block has (4) choices. Keep the conditions of different blocks separate.

Correct answer: A. (18000)

Vowels can be chosen in (5\times4) ways and digits in (10\times9\times8) ways. In exams, multiply the choices of independent stages.

Correct answer: A. (26^4\times90)

There are (26^4) ways for the letters and (9\times9) ways for the digits. Count the restricted first digit separately.

Correct answer: B. (34000)

Split cases according to whether the first position is even or odd: (4\binom{4}{1}5\cdot5^3+5\binom{4}{2}5^2\cdot5^2=34000). Case division is safest for such restrictions.

Correct answer: B. (64)

A full plate can be chosen in (3\times5\times4=60) ways and only dessert in (4) ways. When choices are connected by either-or, use the addition rule.