Permutations

The governing concept is permutation because the three prizes are distinct. Choosing student X for the first prize and student Y for the second is different from reversing those assignments, so order of assignment matters. There are 7 choices for the first prize, then 6 remaining choices for the second, and finally 5 remaining choices for the third. Thus the number of assignments is 7 × 6 × 5 = 210, which is also 7P3 = 7!/(7−3)!. Therefore, option A is correct. Option C, 35, would count only an unordered group of three students using 7C3. Option B is not the relevant permutation value, and option D incorrectly treats the choices as independent with repetition allowed.

The governing concept is permutation with repeated objects. CIVIC contains five letters: C, I, V, I, C. If all five positions were treated as distinct, there would be 5! arrangements. However, the two C letters are identical, so interchanging them does not produce a new visible arrangement; the same applies to the two I letters. We therefore divide by 2! for the repeated C’s and by 2! for the repeated I’s: 5!/(2! × 2!) = 120/4 = 30. Hence option B is correct. Option D, 120, ignores repeated letters. Options A and C do not apply the required duplicate correction and therefore undercount or miscount the distinct arrangements.

For different posts, order matters, so (^{10}P_2=90). In exams, do not use combinations when posts are distinct.

There are (5) letters with (2) identical letters, so the number is (\frac{5!}{2!}=60). In exams, remember to divide by identical letters.

For three ordered positions, (^{7}P_3=7\cdot6\cdot5=210). In exams, ranks always mean order matters.

The chairs are distinct and (4) students are seated, so (^{6}P_4=360). In exams, account for unused chairs carefully.

There are (5,4,3) choices for the three places, so (5\cdot4\cdot3=60). In exams, choices reduce without repetition.

The hundreds place cannot be (0), so (4\cdot4\cdot3=48). In exams, always check the zero restriction on the first digit.

There are (6) letters with (2) identical letters, so (\frac{6!}{2!}=360). In exams, put factorials of repeated letters in the denominator.

Order is important in a signal, so (^{9}P_4=3024). In exams, words like signal and code usually indicate permutation.

Treat the four girls as one block, then (6!\cdot4!=17280). In exams, use the block method for togetherness conditions.

For seating distinct people around a circular table, rotations are considered identical because there is no fixed first seat. The governing formula is therefore (n − 1)!, not n!. With n = 6, fix one person as a reference and arrange the remaining five people in the five available relative positions: (6 − 1)! = 5! = 5 × 4 × 3 × 2 × 1 = 120. Thus option A is correct. Option B, 720, is 6! and incorrectly treats a circular arrangement like a row. Option C, 60, and option D, 360, do not result from the circular-permutation formula.

Treat the two people as one block, giving (6) units in a circle, so (5!\cdot2!=240). In exams, also multiply by the internal arrangement of the block.

Among (6) letters, (2) are identical, so (\frac{6!}{2!}=360). In exams, count only the repeated identical letters.

There are (3) even choices for the units place and (5\cdot4\cdot3) ways for the rest, so (180). In exams, apply the last-digit condition first.

There are (3) odd choices for the units place and the remaining places are filled in (5\cdot4\cdot3) ways. The total is (180).

Treat the two subjects as two blocks, so (2!\cdot5!\cdot3!=1440). In exams, arrange inside each block separately.

The governing concept is an arrangement without repetition. Because a password is ordered, ABCDE and BACDE are different passwords; hence combinations cannot be used. Fill the five positions successively: 8 choices for the first, 7 for the second, 6 for the third, 5 for the fourth, and 4 for the fifth. Thus the count is 8P5 = 8 × 7 × 6 × 5 × 4 = 6720, so option A is correct. Option B, 56, is 8C5 and ignores order. Option C, 336, omits one factor, while option D, 8^5, incorrectly permits repetition.

Selection with order is (^{11}P_3=11\cdot10\cdot9=990). In exams, the word arranged suggests permutation.

Treat the four red balls as one block, then (4!\cdot4!=576). In exams, arrange the block and outside units separately.

All possible last digits are even, but the hundreds place cannot be (0). Total numbers are (4\cdot4\cdot3=48).

This is a one-to-one arrangement of (5) persons and (5) rooms, so (5!=120). In exams, assignments to distinct rooms are permutations.

There are three different posts, so (^{9}P_3=504). In exams, different named posts make order important.

The arrangement of six distinct objects in six distinct positions is (6!=720). In exams, use factorial when matching positions are distinct.

The English word has (9) letters with (A) twice and (Y) twice, so (\frac{9!}{2!2!}=90720). In exams, count repeated letters from the exact word given.