Fundamental principle of counting

Choose (2) or (3) from the first (3), then choose the remaining questions from the other (5). For an at least condition, split into cases.

Subtract all-math and all-physics selections from (\binom{11}{4}). The complement method is faster for such restrictions.

Choose (4)-digit sets whose sum is divisible by (3), then arrange them excluding leading zero. For divisibility, first check the digit sum.

Choose (3) men from (5) and (2) women from (4). An exactly condition usually gives one fixed case.

Subtract arrangements where (A,I) are together from total (6!). Treat together letters as one block for such conditions.

Going has (4 \times 5) ways and returning has (3 \times 4) ways. Multiply independent journey stages.

The top, middle, and bottom positions have (7,6,5) choices respectively. When order matters, multiply as an arrangement.

Choose (4) digits including (2) and (4), and in half of their arrangements (4) is left of (2). Symmetry saves time here.

The posts are distinct, so the number of ways is (7 \times 6 \times 5). Use arrangements, not combinations, when positions are different.

First select in (\binom{8}{4}\binom{5}{2}) ways and then arrange in (6!) ways. Handle selection and arrangement separately.

Check divisibility by (4) using the last two digits and fill the remaining (3) places for each valid ending pair. Fix the last two places first.

The main dish has (6-2=4) choices, so total meals are (5 \times 4 \times 4). Remove forbidden choices first.

Letters have (4 \times 3) ways and digits have (5 \times 5) ways. Check repetition rules separately for each part.

To reach the bottom-right corner, 3 right moves and 3 down moves are required, making 6 moves in all. The number of ways to choose the 3 positions for the right moves is \(\binom{6}{3}=20\); the remaining 3 positions are down moves. Therefore, the correct answer is 20. The option 24 can result from incorrectly counting move orders. Exam tip: for right-and-down grid paths, use combinations of the required right and down moves.

The two ends can be filled by (N,R) in (2!) ways and the middle (4) letters in (4!) ways. Fill fixed positions first.

Subtract arrangements where the two particular persons are together from (^{9}P_{4}). Treat the pair as a block when they are together.

The units digit must be (9), then the first digit has (4) choices excluding (0). Handle zero and odd-number conditions together.

The three subject blocks can be arranged in (3!) ways and books inside blocks in (4!3!2!) ways. Use the block method for together conditions.

The number of ways to choose (4) non-consecutive days is (\binom{10-4+1}{4}). Remember the gap method for non-consecutive selections.

Choose the (2) positions for (7) in (\binom{4}{2}) ways, and each remaining position has (9) choices. For exactly conditions, exclude the fixed digit from remaining places.

The valid cases are (3) girls (2) boys, (4) girls (1) boy, and (5) girls. Add all cases satisfying the greater-than condition.

There are (6) letters with (O) repeated twice, so the count is (\frac{6!}{2!}). Do not forget to divide by repeated letters.

First arrange the mathematics books in (4!) ways, then place the (3) English books in (5) gaps with order. The gap method is useful for separation restrictions.

Any (3) chosen digits give exactly one increasing order. When order is predetermined, combinations are enough.

Choose the remaining (3) players from (7), and choose the captain from those (3) players. Treat the compulsory member as fixed first.