Permutations

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

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.

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.

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.

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

(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.

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

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

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

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

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.

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

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'.

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

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.

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.

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.

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

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.

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

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

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.

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.

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.

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.