Permutations

The permutation formula is \(^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \(^{10}P_1=\frac{10!}{9!}=10\). Hence, 10 is the correct option. Although 100 is \(10\times10\), arranging just one item gives only 10 possible choices. Exam tip: remember directly that \(^{n}P_1=n\).

There are (6) choices for the first position and (5) for the second, so (6\times5=30). Count order when positions are distinct.

While arranging books, ABC and BAC are different arrangements, so order matters and it is a permutation. In a committee, only the members matter, not their order. Exam tip: if changing order changes the outcome, use permutation.

Because three cards are selected and placed in order, this is a permutation rather than a combination. For the first position there are 5 choices, for the second there are 4 remaining choices, and for the third there are 3 choices. By the multiplication principle, the number of ordered arrangements is 5×4×3=60. Equivalently, 5P3=5!/(5-3)!=5!/2!=60. Therefore option B is correct. The value 10 is 5C3 and counts selections where order is ignored; multiplying 10 by 3! gives 60. The value 15 does not represent the required ordered selection, and 125 incorrectly allows replacement at every stage.

Because two (O)'s are identical, arrangements are (\frac{4!}{2!}=12). For identical objects, divide by their factorial.

The order matters when 3 people are placed in a row, so we use permutations: \({}^{4}P_{3}=\frac{4!}{(4-3)!}=4\times3\times2=24\). Hence, the correct answer is 24. The value 16 does not represent the correct selection-and-arrangement count. Exam tip: Words such as “in a row” or “arrange” usually indicate a permutation.

The direct answer is A: 720. There are six different balls, so every ball is distinguishable and the order matters. Fill the six places from left to right: 6 choices for the first place, then 5, 4, 3, 2 and 1. Thus the number is \(6\times5\times4\times3\times2\times1=6!=720\). Option A is correct because it gives 720. Option B, 36, is not the product of the decreasing choices. Option C, 120, equals \(5!\), as if only five objects were arranged. Option D, 60, also does not count all six-place arrangements. Remember: arranging all \(n\) different objects in a row gives \(n!\).

Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.

The permutation formula is \({}^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \({}^{8}P_0=\frac{8!}{(8-0)!}=\frac{8!}{8!}=1\). There is exactly one way to arrange zero selected objects: choose nothing. Option 0 represents the number of objects selected, not the number of permutations. Exam tip: remember directly that \({}^{n}P_0=1\) and \({}^{n}P_n=n!\).

The governing concept is permutation in a linear arrangement: when distinct objects are placed in a row, their order matters. For the first position, any of the 3 different coins may be chosen. Once it is placed, 2 coins remain for the second position, and the last coin must occupy the third position. By the multiplication principle, the number of arrangements is 3 × 2 × 1 = 6. Equivalently, arranging all 3 distinct objects gives 3! = 6. Thus option D is correct. Option A counts only the choices for the first position. Option B incorrectly uses 3², as though repetition were allowed, and option C has no valid permutation calculation. Each coin is used exactly once.

There are two distinct posts, so ({}^{10}P_{2}=10\times9=90). Order matters for different designations.

This is a permutation of distinct letters because R, A, V and I are four different letters. In a complete arrangement, the first position has 4 choices, the second has 3 remaining choices, the third has 2, and the last has 1. Thus the total is 4 × 3 × 2 × 1 = 4! = 24. Hence option B is correct. The value 16 does not represent the successive-choice product, 12 omits one factor or uses an unsuitable partial arrangement, and 4 merely counts the letters rather than arranging them. Since no letter is repeated, no division for repetition is required.

There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.

The permutation formula is \({}^{n}P_r=\frac{n!}{(n-r)!}\). Thus, \({}^{4}P_4=\frac{4!}{(4-4)!}=\frac{24}{0!}=24\), since \(0!=1\). Therefore, 24 is the correct answer. A value such as 16 does not result from applying the permutation formula correctly. Exam tip: when \(r=n\), use \({}^{n}P_n=n!\) directly.

For three places, there are (4\times3\times2=24) ways. Without repetition, choices decrease at each next place.

The order of files on a shelf matters, so use permutations. The number of ways is \({}^{7}P_{3}=7\times6\times5=210\). The value 35 is \({}^{7}C_{3}\), which counts selections only and ignores order. Exam tip: use \(nP_r\) when changing the positions or order creates a new arrangement.

The word LEVEL contains five positions, but L is repeated twice and E is repeated twice. If all five letters were temporarily treated as different, there would be \(5!\) arrangements. However, interchanging the two Ls does not create a new visible word, and the same is true for the two Es. We must therefore divide by \(2!\) for each repeated letter.

The number of distinct arrangements is \(\frac{5!}{2!2!}=\frac{120}{4}=30\). The repeated letter V occurs only once and needs no division. Therefore choice C, (30), is correct. Counting \(5!=120\) would count identical-letter interchanges repeatedly.

The three ranks are ordered, so ({}^{5}P_{3}=60). In rank questions, use permutation, not combination.

The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.

The governing concept is permutation because three children are selected and then placed in ordered positions on the stage. The first position can be filled by any of 9 children, the second by any of the remaining 8, and the third by any of the remaining 7. Therefore the multiplication principle gives 9 × 8 × 7 = 504. In notation, this is 9P3 = 9!/(9−3)! = 9!/6! = 504. Hence option B is correct. The value 84 equals 9C3, which counts only the group of three and ignores their order. The values 81 and 729 treat the choices incorrectly, such as using repeated or constant choices, even though one child cannot occupy more than one position.

There are 6 choices for gold, then 5 for silver and 4 for bronze. Thus, the total is \(6\times5\times4=120\). \(\binom{6}{3}=20\) is incorrect because medal order matters. Exam tip: use permutations when positions or ranks are distinct.

The governing concept is permutation of all distinct objects in a linear arrangement. For the first place in the list there are 5 choices. After one key is used, 4 choices remain for the second place, followed by 3, 2, and finally 1 choice. Thus the total number of orders is 5 × 4 × 3 × 2 × 1 = 5! = 120. Therefore option D is correct. A circular-arrangement formula is not suitable because the keys are written in a straight list, so the two ends are distinct and rotations are not identified. Option B equals 5P2 and counts only two positions. The values 25 and 100 arise from incorrectly allowing repeated choices or using an unrelated square-based calculation.

The four pictures are distinct and are arranged in a left-to-right row, so order matters. Hence, the number of arrangements is \(4! = 4 \times 3 \times 2 \times 1 = 24\). A value such as \(16\) is not the correct permutation count. Exam tip: for arranging \(n\) distinct objects in a row, use \(n!\).

For (2) ordered positions from (5) letters, ({}^{5}P_{2}=20). In ordered pairs, ((P,Q)) and ((Q,P)) are different.

There are (4) letters and two identical (O)'s, so (\frac{4!}{2!}=12). Use a denominator for repeated letters.