Permutations

Correct answer: B. (4320)

Subtract the (6!) arrangements where the specified flag is at the top from the total (7!). For one forbidden position, subtraction is quick.

Correct answer: B. (144)

The last digit can be (1,3,5) in (3) ways; the first place then has (4) nonzero choices, followed by (4\cdot 3). For odd numbers, fix the unit digit first.

Correct answer: A. (50400)

There are (10) letters with (S) three times, (T) three times and (I) twice, so the count is (10!/(3!3!2!)). For long words, make a frequency table.

Correct answer: C. (120)

The direct answer is C: 120. The three prizes are different, so assigning prize 1, prize 2 and prize 3 is an ordered selection of three different students. Choose the recipient of the first prize in 6 ways, the second in 5 ways, and the third in 4 ways. The total is \(6\times5\times4={}^{6}P_3=120\). Option C is correct. Option A, 18, is too small. Option B, 60, does not count all assignments. Option D, 216, is \(6^3\) and would allow one student to receive multiple prizes, which the question forbids. Since prizes are distinct, order matters even though recipients cannot repeat.

Correct answer: B. (6720)

We must choose and arrange (5) persons, so the number is (^{8}P_{5}). Seating includes both selection and order.

Correct answer: B. (48)

(N) and (R) can occupy the ends in (2!) ways, and the remaining (4) letters in (4!) ways. With end restrictions, fill the ends first.

Correct answer: C. (240)

Treat (1) and (2) as a block, giving (5!) arrangements and (2!) internal orders. For adjacent digits, the block method is direct.

Correct answer: A. (3024)

This is an ordered selection of (4) books from (9), so the number is (^{9}P_{4}). Arrangement on a shelf requires order.

Correct answer: C. (84)

Adding arrangements starting with smaller available letters at each position and then adding (1) gives rank (84). In rank problems, count position by position.

Correct answer: A. (3600)

Total circular arrangements are (7!), and adjacent cases are (2\cdot 6!), so subtraction gives (3600). In circular arrangements, start with ((n-1)!).

Correct answer: A. (10080)

There are (8) letters with (C,L,U) each repeated twice, so the count is (8!/(2!2!2!)). Identifying repetitions is the key step.

Correct answer: B. (960)

There are (4) position pairs for (A,B) and (2) orders, then (5!) ways for the rest. For a fixed gap, count position pairs.

Correct answer: A. (600)

If first is (6), count (^{6}P_{4}); if last is (6), subtract leading-zero cases from choices; then subtract the intersection. Use inclusion-exclusion for 'or'.

Correct answer: A. (360)

For (4) fingers, choose and assign rings in (^{6}P_{4}) ways. Placing objects in distinct positions is an ordered selection.

Correct answer: A. (9979200)

There are (11) letters with (I,N,A) each repeated twice, so the count is (11!/(2!2!2!)). In long words, note repeated letters separately.

Correct answer: A. (720)

Treat the three women as one block, so (6) units have (5!) circular arrangements and the women have (3!) internal ways. In circular block problems, count units carefully.

Correct answer: A. (90)

Divisibility by (4) depends on the last two digits; there are (15) valid ordered ending pairs and the first two places fill in (^{5}P_{2}) ways. For divisibility by (4), check the last two digits.

Correct answer: B. (48)

The two specified pictures occupy the two middle positions in (2!) ways, and the remaining (4) pictures in (4!) ways. Fill restricted positions first.

Correct answer: C. (60)

This is an ordered selection of (3) from (5), so it is (^{5}P_{3}). When there is no repetition and order matters, use permutation.

Correct answer: B. (3024)

Both selection and order are involved, so the count is (^{9}P_{4}). When order matters, do not use combination.

Correct answer: A. (96)

With (P) fixed first, subtract the (4!) arrangements where (L) is last from the remaining (5!). In mixed conditions, apply fixed conditions first.

Correct answer: D. (3600)

First fill the two ends with ordinary students in (^{5}P_{2}) ways, then arrange the remaining (5) students in the middle (5) places. For end restrictions, fill the ends first.

Correct answer: C. (480)

Arrange the four consonants in (4!) ways and place the two vowels in (^{5}P_{2}) ways in the gaps. Use the gap method for non-adjacent vowels.

Correct answer: A. (2520)

Include (1,2), choose (3) more digits from the remaining (7), then arrange the (5) digits in (5!) ways. For compulsory objects, include them first and choose the rest.

Correct answer: A. (600)

Subtract the (5!) arrangements where that person sits on the fixed chair from the total (6!). With one forbidden seat, subtract the invalid cases from total.