Fundamental principle of counting

Correct answer: A. (5 \times 4 \times 5 \times 4 \times 3 \times 2=2400)

There are (5) and (4) choices for odd digits and (5,4,3,2) choices for even digits. Filling places one by one is the safest exam method.

Correct answer: B. (26 \times 25 \times 24 \times 23 \times 10^3)

Letters cannot repeat so the choices decrease and digits can repeat so the factor is (10^3). Read each restriction separately first.

Correct answer: A. (5 \times 4 \times 3 \times 2=120)

The (4) ordered positions have (5,4,3,2) choices. When order matters use the multiplication principle.

Correct answer: C. 108

A number divisible by 5 must end in 0 or 5. If it ends in 0, there are 5 choices for the thousands digit and then 4 and 3 choices for the two middle positions: 5 × 4 × 3 = 60. If it ends in 5, the thousands digit cannot be 0, so it has 4 choices from 1, 2, 3, 4; the middle positions then have 4 × 3 choices: 4 × 4 × 3 = 48. Hence, the total is 60 + 48 = 108. Option 60 counts only the case where the last digit is 0. Exam tip: whenever 0 is available, separately check that the leading digit is not 0.

Correct answer: B. (4 \times 8 \times 7 \times 6 \times 3)

The middle three digits cannot repeat so their choices are (8,7,6). Multiply the boundary choices as well.

Correct answer: A. (4 \times 5+5 \times 3+4 \times 5 \times 3=95)

There are three separate cases and their results are added. Within each case the choices are multiplied.

Correct answer: B. (9 \times 10 \times 10 \times 5=4500)

The last even digit has (5) choices from (0,2,4,6,8). With repetition allowed the middle places still have (10) choices each.

Correct answer: C. (7 \times 6 \times 5=210)

The posts are distinct so order matters and the choices are (7,6,5). For posts the multiplication principle applies directly.

Correct answer: A. (26 \times 25 \times 9 \times 10^3)

The letters have (26) and (25) choices while the first digit has (9) and the rest have (10) choices. Identify the zero restriction separately.

Correct answer: A. (3 \times 4 \times 5 \times 4=240)

Direct answer: option A is correct, giving 240 arrangements. First choose 1 mathematics book from 3: 3 choices. Then choose 1 science book from 4: 4 choices. For history, choose and arrange 2 different books from 5: 5 choices for the first history position and 4 for the second, giving \(5\times4=20\). By the fundamental counting principle, multiply the independent stages: \(3\times4\times5\times4=240\). Option A shows this exact calculation. Option B uses \(3\times4\times10=120\); 10 is the number of unordered pairs from 5, but the question says “in order,” so each pair has two possible orders and must be counted as 20. Option C, \(3\times4\times5=60\), forgets the second ordered history choice. Option D adds choices instead of multiplying them, although every stage must be completed. Memory cue: sequential choices multiply; ordered selection of two from five is \(5\times4\), not \(\binom52\).

Correct answer: B. (9 \times 8 \times 5=360)

First place the unit digit with (5) odd choices, then the hundred and ten places have (8) and (7) choices. Handle the last-digit restriction first.

Correct answer: A. (26 \times 25 \times 24 \times 10 \times 9)

The three letters have (26,25,24) choices and the two digits have (10,9) choices. Count each type at its fixed positions.

Correct answer: B. (4^2 \times 5^3=2000)

Each question is an independent stage, so the count is (4 \times 4 \times 5 \times 5 \times 5). For mandatory answers multiply, do not add.

Correct answer: B. (6 \times 4 \times 3 \times 5 \times 4=1440)

First choose the (2) positions for even digits in (6) ways, then fill them with (4,3,5,4) choices. Keep position selection and digit filling separate.

Correct answer: B. (1 \times 5 \times 6+3 \times 2 \times 6=66)

The second-stage choices depend on the first choice, so cases must be separated. Do not use one uniform product for dependent stages.

Correct answer: B. (8 \times 7 \times 6=336)

The three roles are different, so the choices are (8,7,6). When roles have different names, order matters.

Correct answer: A. (6 \times 5 \times 6^3=6480)

The first letter has (6) choices and the last has (5), while each middle place has (6) choices. Reduce choices only where the condition applies.

Correct answer: B. (5 \times 21 \times 10 \times 9 \times 8)

The letter choices are (5) and (21), and the digit choices are (10,9,8). Keep letter restrictions separate from digit restrictions.

Correct answer: A. (2^4 \times 3^2=144)

The four binary questions give (2^4) and the two three-option questions give (3^2). Use multiplication for all independent responses.

Correct answer: A. (5 \times 6 \times 5 \times 4=600)

The thousands digit is one of (3,4,5,6,7), then (6,5,4) choices remain. Apply the inequality condition at the first place.

Correct answer: B. (3 \times 2 \times 3^4=486)

The first color has (3) choices and the last has (2), while the (4) middle boxes are independent. The inequality applies only to the ends.

Correct answer: A. (6 \times 26 \times 26 \times 10 \times 9)

First choose the (2) letter positions in (6) ways and arrange uppercase-lowercase in (2) ways; thus the correct count should be (6 \times 2 \times 26^2 \times 10 \times 9). Among the options (6 \times 26 \times 26 \times 10 \times 9) is incomplete.

Correct answer: A. (10 \times 5 \times 4 \times 21 \times 20 \times 19)

First choose the (2) vowel positions in (10) ways, then fill vowels and consonants in order. Separate position selection from letter filling.

Correct answer: B. (360)

Count separately the cases when the last digit is (0) and when it is (2,4,6). For zero restrictions, casework is safer.

Correct answer: A. (5 \times 5 \times 4=100)

There are (5) choices for first place, then (5) and (4) choices remain. Remove the forbidden choice and fill ordered positions.