Permutations

Correct answer: A. 210

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.

Correct answer: B. 30

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.

Correct answer: A. (90)

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

Correct answer: A. (60)

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

Correct answer: A. (210)

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

Correct answer: A. (360)

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

Correct answer: A. (60)

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

Correct answer: A. (48)

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

Correct answer: A. (360)

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

Correct answer: A. (3024)

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

Correct answer: A. (17280)

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

Correct answer: A. 120

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.

Correct answer: A. (240)

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.

Correct answer: A. (360)

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

Correct answer: A. (180)

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.

Correct answer: A. (180)

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

Correct answer: A. (1440)

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

Correct answer: A. 6720

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.

Correct answer: A. (990)

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

Correct answer: A. (576)

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

Correct answer: A. (48)

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

Correct answer: A. (120)

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

Correct answer: A. (504)

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

Correct answer: A. (720)

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

Correct answer: A. (181440)

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.