Fundamental principle of counting

Correct answer: B. (31)

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.

Correct answer: C. (325)

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

Correct answer: D. (252)

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

Correct answer: A. (60)

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

Correct answer: A. (480)

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

Correct answer: A. (240)

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

Correct answer: A. (210)

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

Correct answer: B. (300)

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

Correct answer: A. (210)

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

Correct answer: B. (504000)

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

Correct answer: D. (300)

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.

Correct answer: A. (80)

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

Correct answer: B. (300)

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

Correct answer: C. 20

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.

Correct answer: B. (48)

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

Correct answer: B. (2784)

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

Correct answer: C. (100)

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

Correct answer: A. (1728)

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.

Correct answer: A. (35)

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

Correct answer: A. (486)

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.

Correct answer: A. (225)

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

Correct answer: B. (360)

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

Correct answer: B. (1440)

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.

Correct answer: A. (56)

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

Correct answer: A. (140)

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