Fundamental principle of counting

Correct answer: A. (384)

Choose the other (3) digits along with (1,2), then subtract arrangements with leading zero when zero is included. Check the zero-position restriction separately.

Correct answer: B. (21)

Choose (2) or (3) of the special sweets and fill the rest from ordinary sweets. Add valid cases for an at least condition.

Correct answer: B. (265)

There are (6) choices for the correct letter and the remaining (5) letters must be deranged in (!5=44) ways. Exactly-correct conditions often use derangements.

Correct answer: B. (96)

Choose (3) letters from (C,D,E,F) along with (A), then arrange the (4) letters in (4!) ways. Apply inclusion and exclusion restrictions first.

Correct answer: C. (295)

For each pair of colours, take all (4)-selections and subtract single-colour selections, then add the three pairs. Exactly (2) colours means at least one ball from each chosen colour.

Correct answer: A. (7200)

First choose the (3) positions for letters, then fill letters and digits in order. Deciding position types is the first step in mixed-code questions.

Correct answer: A. (112)

Subtract triangles containing both particular points from total (\binom{10}{3}). With the forbidden pair, the third point has (8) choices.

Correct answer: D. (720)

The word has seven letters: A, A, L, G, E, B and R. The two A’s are identical, and the condition requires them to remain next to each other. Treat the pair AA as one single block. The objects to arrange are then the block AA together with L, G, E, B and R, making six distinct objects in all.

These six objects can be arranged in \(6!\) ways because every object is different from the others. The block itself has no internal order to multiply by, since both letters inside it are identical A’s; changing one A with the other makes no new arrangement. Thus the number is \(6!=720\). Therefore option D is correct. A factor of 2 would be needed only if the two letters inside the block were different.

Correct answer: C. (360)

The thousands digit has (4) choices from (3,4,5,6), and the remaining (3) places are filled in (5 \times 4 \times 3) ways. For inequalities, the first place is decisive.

Correct answer: A. (275)

The valid cases are (1) teacher (3) students and (2) teachers (2) students. Form cases satisfying both at least conditions.

Correct answer: A. (336)

The first position has (8) choices, the second (7), and the third (6). Order matters for distinct positions.

Correct answer: B. (408)

Total valid numbers are (5 \times 5!), and subtract the cases where (0,5) are together. Check block cases carefully because of leading zero.

Correct answer: B. (18)

Pens are chosen in (\binom{4}{2}) ways and notebooks in (\binom{3}{2}) ways. Multiply independent selections.

Correct answer: A. (16)

Subtract selections containing both question (1) and question (2) from total (\binom{6}{3}). With the forbidden pair, the third question has (4) choices.

Correct answer: B. (36)

The two particular letters can occupy positions ((1,4)) or ((2,5)), and their internal order has (2!) ways. The remaining (3) letters are arranged in (3!) ways.

Correct answer: C. (2240)

The first and last colours can be (R,B) or (B,R), then the two middle positions are filled from the remaining (7) cards in order. Fix colour-based end positions first.

Correct answer: C. (60)

Choose (\binom{3}{2}) friends and (\binom{6}{3}) others. Exactly (2) means the third friend is excluded.

Correct answer: C. (168)

For two distinct hundreds and units digits, the larger must be in the hundreds place and it cannot be (0). Then the tens place has (6) remaining choices.

Correct answer: B. (325)

The valid cases are ((2,4)) and ((3,3)). Form cases only after combining the at least conditions of both sections.

Correct answer: D. (432)

There are (2) ways to decide which particular speaker goes in the first half, (3 \times 3) ways for their positions, and (4!) ways for the rest. Fix the sections and particular persons first.