Permutations

Correct answer: A. 60

For a three-digit number, there are 5 choices for the hundreds place. After choosing one digit, 4 choices remain for the tens place and 3 for the units place. Hence, the total number is \(5\times4\times3=60\). The value \(125=5^3\) would apply if repetition of digits were allowed. Exam tip: use \(nP_r\) for ordered selections without repetition.

Correct answer: A. (1320)

The three medals are ordered, so (^{12}P_3=1320). In exams, use permutation when medals are different.

Correct answer: A. (720)

The arrangement of (6) distinct pictures in a row is (6!=720). In exams, the word row means linear arrangement.

Correct answer: A. (5040)

The number of one-to-one placements is (7!=5040). In exams, distinct envelopes mean arranging the letters.

Correct answer: A. (60)

There are (6) letters with (3) identical and (2) identical letters, so (\frac{6!}{3!2!}=60). In exams, multiply all repeated group factorials in the denominator.

Correct answer: A. 30240

Use the complementary-counting method within permutations. First count all unrestricted arrangements of the 8 distinct students: 8! = 40320. Now count arrangements in which the two specified students sit together. Treat them as one block; this gives 7 units, which can be arranged in 7! ways, and the two students can switch places within their block in 2! ways. Thus together arrangements equal 7! × 2! = 10080. Subtracting from the total gives 40320 − 10080 = 30240. Therefore option A is correct. Option B is the unrestricted total, option C is the forbidden count, and option D results from an incorrect subtraction.

Correct answer: A. (28800)

There are two possible patterns and each has (5!\cdot5!) arrangements. The total is (2\cdot5!\cdot5!=28800).

Correct answer: A. (300)

The thousands place cannot be (0), so (5\cdot5\cdot4\cdot3=300). In exams, first choose the non-zero leading digit.

Correct answer: A. (2880)

Arrange English books first in (4!) ways, then place (4) Hindi books in (5) gaps in (^{5}P_4) ways. The total is (4!\cdot^{5}P_4=2880).

Correct answer: A. (20160)

For a necklace, rotations and reflections are both the same, so (\frac{(9-1)!}{2}=20160). In exams, distinguish necklaces from circular table arrangements.

Correct answer: A. (120)

The direct answer is A: 120. A code has three ordered positions, and six letters are available without repetition. Choose the first position in 6 ways, the second in 5 ways, and the third in 4 ways. Therefore the total is \(6\times5\times4=120\), equivalently \({}^{6}P_3=6!/(6-3)!=120\). Option A is correct. Option B, 20, resembles a combination count and ignores order. Option C, 216, allows repetition because it is \(6^3\). Option D, 36, does not represent the successive choices. For example, ABC and BAC are different codes, so order matters even when the same letters are used.

Correct answer: A. (60)

There are (5) letters with (2) identical letters, so (\frac{5!}{2!}=60). In exams, check repeated vowels carefully.

Correct answer: A. (5040)

This is (^{10}P_4=10\cdot9\cdot8\cdot7=5040). In exams, selected and arranged means use (nP_r).

Correct answer: A. (360)

For a circular garland, the number is (\frac{(7-1)!}{2}=360). In exams, count reflection as the same for garlands.

Correct answer: A. (36)

The direct answer is option A: 36 numbers. A three-digit number greater than 400 must have a hundreds digit greater than 4, because 4 in the hundreds place would give a number beginning with 4, which is not greater than 400 unless the remaining digits make it greater; here the available digits after 4 can indeed make such numbers, so we must check carefully. If the hundreds digit is 4, the tens and units can be selected in \(4\times3=12\) ways, and all are greater than 400 because the remaining two-digit part is positive. If the hundreds digit is 6 or 8, there are also \(4\times3=12\) ways for each. Thus the total is \(3\times12=36\), counting hundreds digits 4, 6, and 8. Option A is correct. Option B, 60, ignores the restriction or overcounts. Option C, 24, misses one allowed hundreds-digit case. Option D, 48, is another overcount. No digit repeats. Exam cue: for “greater than 400,” inspect the hundreds digit and remember that 4xx is greater than 400 when the last two digits are nonzero.

Correct answer: A. 10080

The governing idea is the multiplication principle combined with a position restriction. In a row of 8 seats, the two middle positions are positions 4 and 5. The specified person has 2 permitted choices. After that person is placed, the remaining 7 distinct people can be arranged freely in the remaining seats in 7! ways. Therefore the total number is 2 × 7! = 2 × 5040 = 10080, so option A is correct. Option B counts only one allowed middle position, option C is the unrestricted total 8!, and option D does not represent the product of the valid choices.

Correct answer: A. (720)

Distributing one distinct prize to each student can be done in (6!) ways. In exams, distinct prizes with one each mean factorial.

Correct answer: A. (360)

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

Correct answer: A. (14400)

Arrange boys first in (5!) ways, then place (3) girls in (6) gaps in (^{6}P_3) ways. The total is (5!\cdot^{6}P_3=14400).

Correct answer: A. 120

The last-digit condition fixes the units place as 5, so that position has only one choice. Since repetition is forbidden, the remaining three positions must be filled from the other six digits: 6 choices for the thousands place, then 5 for the hundreds place, and 4 for the tens place. By the multiplication principle, the count is 6 × 5 × 4 = 120. Thus option A is correct. There is no leading-zero issue because zero is not in the given set. Option B incorrectly uses a larger selection count, while options C and D overcount arrangements by not applying the fixed-ending and no-repetition conditions correctly.

Correct answer: A. (48)

The two particular books occupy the ends in (2!) ways and the remaining books in (4!) ways. The total is (2!\cdot4!=48).

Correct answer: A. (3600)

The particular person has (5) inner positions and the others sit in (6!) ways. The total is (5\cdot6!=3600).

Correct answer: A. (181440)

There are (9) letters with two different letters repeated twice, so (\frac{9!}{2!2!}=90720). In exams, accurate repeated-letter counting is essential.

Correct answer: A. (1680)

The speaking order is important, so (^{8}P_4=1680). In exams, order of performance means permutation.

Correct answer: A. (500)

There are (5) odd choices for the unit place, but the first digit must not be (0). The total is (5\cdot4\cdot4\cdot3=240).