Fundamental principle of counting

Correct answer: B. (1740)

First count all even numbers and subtract those without (3). The complement method is efficient for inclusion conditions.

Correct answer: C. (23520)

First arrange the consonants and place vowels in the gaps formed. Do not forget division due to repeated letters.

Correct answer: A. (3850)

Subtract the cases with (4,5,6) women from total (\binom{15}{6}). For at most conditions, complement counting is often simpler.

Correct answer: B. (15)

After fixing the first position as (0), place (4) non-adjacent ones in the remaining (9) positions. Use the gap method for non-adjacent selections.

Correct answer: C. (1680)

First choose (5)-digit sets whose sum is divisible by (9), then arrange each set in (5!) ways. Start divisibility problems with the digit sum.

Correct answer: C. (32)

Use inclusion-exclusion to subtract paths passing through the forbidden points from total paths. For grid paths, use (\binom{m+n}{m}).

Correct answer: C. (57600)

First count all mixed passwords and subtract cases where both first and last positions are digits. Keep position selection and object arrangement separate.

Correct answer: B. (150)

Subtract distributions with at least one empty student from total (3^5). Inclusion-exclusion is useful when each receiver must get at least one item.

Correct answer: C. (432)

Treat the girls as one block and subtract arrangements where the two particular boys are together. Use the block method with complement counting.

Correct answer: C. (1000)

The valid colour distributions are only ((2,2,1)) and ((3,1,1)), where the first place is for red. For colour restrictions, list valid cases first.

Correct answer: C. (960)

Treat the two adjacent persons as a block and subtract cases where the third person is adjacent to that block. In circular arrangements, fix one rotation.

Correct answer: C. (103)

Convert the restrictions on (x_1) and (x_2) into non-negative variables and take cases for (x_3). For bounded variables, add case counts.

Correct answer: C. (420)

Subtract selections containing (3) or (4) points from the set of (5) collinear points from total (\binom{12}{4}). A quadrilateral cannot have any (3) collinear points.

Correct answer: B. (604800)

Place (0) and (5) in the middle (6) positions with order, then fill the remaining positions with other digits. Apply end-position restrictions first.

Correct answer: C. (119)

Choose (6) books from each pair of subjects and add the three results. Here, selecting all (6) from one subject is impossible.

Correct answer: C. (12)

First arrange (B,N,N), then place the (3) identical (A)'s in the (4) gaps formed. Handle repeated letters and gaps carefully.

Correct answer: B. (1260)

First choose the positions in (^{7}P_3) ways, then choose (2) players from the remaining (4). Identify ordered and unordered stages separately.

Correct answer: C. (129600)

The letters have (6 \times 5) ways. For digits, choose the positions of the equal pair, choose the repeated digit, and fill the other two positions with distinct digits in order.

Correct answer: C. (14400)

First arrange (4) girls in (5) odd positions, then arrange the boys in the remaining (5) positions. Apply position restrictions first.

Correct answer: C. (66)

First consider non-consecutive selections of (4) numbers and then separate those with even sum. Parity classification helps in such questions.

Correct answer: C. (10160640)

First arrange the (9) books in a row and place (2) dividers in the (8) gaps. The divider method is useful for labeled shelves.

Correct answer: C. (207360)

First choose (2) non-adjacent vowel positions among (6), then fill vowels and consonants. With repetition allowed, each chosen position keeps the same number of choices.

Correct answer: D. (1456)

First choose (3) tasks for (A), then distribute the remaining (5) tasks between (B,C) with (B) getting at least (2). Apply the fixed person's condition first.

Correct answer: C. (3060)

The valid last-two-digit endings are (25,50,75), and the presence of (7) is checked in each case. For divisibility by (25), fix the last two digits first.

Correct answer: B. (120)

Subtract the paths passing through both points twice because exactly one point is required. Combine grid-path counting with inclusion-exclusion.