Permutations

Only (9) can be in the units place, and the other two places can be filled in (4\times3=12) ways. For odd numbers, check the units place first.

Using the permutation formula, \({}^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \({}^{15}P_0=\frac{15!}{15!}=1\). There is exactly one way to select and arrange zero objects: choose nothing. Option 0 is incorrect because it represents the number of objects selected, not the number of possible arrangements. Exam tip: whenever \(r=0\), \({}^{n}P_0=1\).

(5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.

(A) occurs three times and (N) twice, so (\frac{6!}{3!2!}=60). Divide by factorials of repeated groups.

Starting and ending are ordered, so ({}^{6}P_{2}=30). In a route, changing start and end changes the case.

The hundreds digit cannot be 0, so it has 3 choices: 1, 5, or 7. The tens digit then has 3 remaining choices and the units digit has 2. Thus, the total number is \(3\times3\times2=18\). Although 24 equals \(4P_3\), it includes arrangements beginning with 0, which are not 3-digit numbers. Exam tip: In number-formation questions, always check whether 0 is allowed in the first position.

The governing concept is a permutation of distinct objects in a linear order. All four direction cards are different, each card is used exactly once, and the top-to-bottom sequence distinguishes one arrangement from another. There are 4 choices for the top position, 3 remaining choices for the second position, 2 for the third, and 1 for the bottom position. Therefore, the multiplication principle gives 4 × 3 × 2 × 1 = 4! = 24. Option A is correct. Option B incorrectly uses 4², as though two independent choices were made repeatedly. Options C and D are not the complete product of the available choices and therefore do not count every possible order. This is a straight arrangement, not a circular one.

First and second speakers are ordered positions, so ({}^{14}P_{2}=182). Speaking order uses permutation.

The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.

Changing the order of books on a shelf creates a new arrangement, so it is a permutation. In selecting a committee or objects, order does not matter. Exam tip: use \({}^nP_r=\frac{n!}{(n-r)!}\) when order matters.

The governing concept is an ordered selection, or permutation, because the upper rack and lower rack are distinguishable positions. For the upper rack, any of the 8 different plates can be chosen. After that choice, 7 plates remain for the lower rack. Thus the total is 8 × 7 = 56, or equivalently 8P2 = 8!/(8−2)! = 8!/6! = 56. Therefore, option A is correct. Option C, 28, equals 8C2 and would be correct only if the two plates were selected without assigning them to separate racks. Options B and D do not follow the successive-choice process. The fact that the racks are named makes the order meaningful: plate X on the upper rack and plate Y on the lower rack differs from the reverse placement.

There are 6 available letters: P, Q, R, S, T, and U. A two-letter code is formed without repetition, so the first position has 6 choices. Once that letter is used, only 5 letters remain for the second position. By the multiplication principle, the number of codes is \\(6\times5=30\\), which is \\({}^{6}P_2\\). Therefore option B is correct.

Order matters for codes: PQ and QP are different codes. That is why combinations, which would count only selections and give 15, are not appropriate. The phrase “without repetition” prevents choices such as PP, and ensures the second position has five rather than six choices. Hence the total number of valid two-letter codes is 30.

(C) occurs twice and (O) occurs twice, so (\frac{4!}{2!2!}=6). Divide by both repeated letters.

First, second and third are distinct positions, so order matters. Hence, the number of ways is \({}^{15}P_3=15\times14\times13=2730\). The value 455 is \({}^{15}C_3\), which only selects three students and does not assign them ordered ranks. Exam tip: use permutations when ranks, posts, or positions are assigned.

The direct answer is option A, 9. A two-digit number has a tens place and a units place. Each place can contain one of the three digits 3, 6, or 9. Because repetition is allowed, the digit used first may be used again. Thus there are 3 choices for the tens place and 3 choices for the units place, giving 3×3=3^2=9 numbers. They include 33, 36, 39, 63, and so on. Option A is correct. Option B, 6, would be 3×2 and would apply when repetition is not allowed. Option C, 12, has no valid counting basis here. Option D, 3, counts only one position and forgets that the second position also has choices. None of the digits is zero, so there is no leading-zero issue. Memory cue: with repetition allowed, n positions and r choices per position give r^n.

The governing concept is permutation of distinct objects in a straight line. Since all six beads are different and every bead is used once, changing their left-to-right order creates a new arrangement. The first position has 6 choices, the second has 5 remaining choices, followed by 4, 3, 2, and 1. Therefore, the multiplication principle gives 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720. Option B is correct. Option C, 120, is 5! and omits one factor. Option A, 360, is only half of the required total, while option D, 36, does not represent all possible orders. Since the beads are placed in a straight line, no circular-arrangement adjustment such as division by 6 is appropriate.

Using the permutation formula, \({}^{n}P_r=\frac{n!}{(n-r)!}\). Thus, \({}^{5}P_0=\frac{5!}{5!}=1\) and \({}^{5}P_2=\frac{5!}{3!}=5\times4=20\). Therefore, \({}^{5}P_0+{}^{5}P_2=1+20=21\). Option 20 is only the value of \({}^{5}P_2\), not the required sum. Exam tip: \({}^{n}P_0\) is always 1.

There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.

The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.

(5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.

Four digits must be selected and arranged from the 6 given digits. Hence, the number of such numbers is \({}^6P_4=6\times5\times4\times3=360\). The value \(720\) is \(6!\), which would apply if all 6 digits were arranged, not just 4. Exam tip: use \({}^nP_r\) when order matters and repetition is not allowed.

The first, second and third positions are distinct, so order matters. Therefore, the required number is \({}^{9}P_{3}=9\times8\times7=504\). The value 84 does not represent the required permutation count. Exam tip: Use permutations when positions or order are important.

The word REFER has 5 letters, with two identical R's and two identical E's. If all letters were different, there would be \(5!\) arrangements. Interchanging the identical R's or identical E's does not create a new arrangement, so divide by \(2!\) for each repeated pair: \(\frac{5!}{2!\,2!}=30\). Option 60 accounts for only one repeated pair. Exam tip: for arrangements with repeated objects, use \(\frac{n!}{p!q!\cdots}\).

Using the permutation formula \({}^{n}P_r=\frac{n!}{(n-r)!}\), we get \({}^{12}P_0=1\) and \({}^{12}P_1=12\). Hence, \({}^{12}P_0+{}^{12}P_1=1+12=13\). Option 12 is only the value of \({}^{12}P_1\), not the required sum. Exam tip: always remember that \({}^{n}P_0=1\) for every \(n\).

The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.