Fundamental principle of counting

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.

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

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.

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.

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.

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.

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

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.

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.

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

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

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

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

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

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.

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.

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

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.

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

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.