Permutations

Since changing the positions of the children in the front row gives a different arrangement, order matters. Therefore, the number of arrangements is \({}^{6}P_{4}=6\times5\times4\times3=360\). Option 15 is \({}^{6}C_{4}\), which does not consider order. Exam tip: Phrases such as “stand in a row” usually indicate a permutation.

For three places, there are (5\times4\times3=60) ways. Without repetition, each next choice decreases.

Using the permutation formula \({}^{n}P_r=\frac{n!}{(n-r)!}\), we get \({}^{7}P_0=\frac{7!}{(7-0)!}=\frac{7!}{7!}=1\). There is exactly one arrangement of zero objects: choosing nothing. Option 0 confuses the number of objects with the number of possible arrangements. Exam tip: both \({}^{n}P_0\) and \({}^{n}C_0\) always equal 1.

(4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.

There are (6) letters and two identical (O)'s, so (\frac{6!}{2!}=360). Use a denominator for a repeated letter.

There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.

The direct answer is A: 18. A three-digit number cannot begin with 0. For the hundreds place, choose 1, 2 or 3: 3 choices. After that, two places remain and three unused digits remain, including 0, so there are 3 choices and then 2 choices. Therefore the total is \(3\times3\times2=18\). Option A is correct. Option B, 24, is \(4P3\) and incorrectly allows 0 in the first place. Option C, 12, misses some valid choices. Option D, 9, is also too small. The key exam check is that zero may be used in a later place, but never as the first digit of a three-digit number.

The direct answer is option B: 6 ways. The three flags are different, and all three are placed from top to bottom. For the top position there are 3 choices; after choosing one, 2 remain for the middle; the last flag goes at the bottom in 1 way. Thus the total is \(3\times2\times1=3!=6\). Option B is correct because each different top-to-bottom order is a different arrangement. Option A, 3, counts only the number of flags, not their possible orders. Option C, 9, does not follow the multiplication rule and would not correctly count arrangements. Option D, 12, is also an unsupported overcount. For example, red-blue-green and blue-red-green are different arrangements because their positions differ. Memory trick: three distinct objects in a line give \(3!=6\) orders.

The two positions are president and secretary, so they are different roles. Choosing candidate P as president and candidate Q as secretary is different from choosing Q as president and P as secretary. Therefore order matters, and a permutation rather than a combination is required.

There are 11 choices for president. After that choice is made, 10 candidates remain for secretary. By the multiplication principle, the total number of selections is \(11\times10=110\), or \({}^{11}P_2=\frac{11!}{(11-2)!}=110\). Thus option C is correct. The value 55 would come from \({}^{11}C_2\), which treats the two positions as identical and therefore does not suit distinct offices.

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

Gold, silver and bronze are distinct positions, so changing the order of the same 3 finalists changes the outcome. Hence the count is \(^{8}P_{3}\). A committee or a group of representatives has no order. Exam tip: use permutations when ranks or positions are assigned.

Right and left positions are distinct, so ({}^{6}P_{2}=30). When positions are identified, order is counted.

({}^{4}P_{3}=4\times3\times2=24) codes are possible. In a code, changing positions changes the code.

The direct answer is D: 3. EYE has three positions and E is repeated twice. Treating the two E's as temporarily different gives \(3!=6\) orders, but exchanging the identical E's does not create a new arrangement. Hence divide by \(2!\): \(3!/2!=6/2=3\). The actual arrangements are EYE, E EY? More clearly, the three patterns are EYE, E E Y, and Y E E, written as EYE, EEY and YEE. Option D is correct. Option A, 6, double-counts the identical E's. Option B, 2, misses one pattern. Option C, 1, treats all letters as if they were identical. A repeated letter is divided out, not ignored completely.

First and second speakers are ordered positions, so ({}^{12}P_{2}=132). Speaking order uses permutation.

There are (3) choices for each of the three places, so (3^3=27). With repetition allowed, choices remain the same at every place.

The governing concept is permutation: when different objects are placed in a row, their order matters. For the first position there are 5 choices, for the second 4 remaining choices, then 3, 2, and 1. Therefore, the total number of arrangements is 5 × 4 × 3 × 2 × 1 = 5! = 120. Thus option C is correct. A value of 60 would not account for all possible orders, while 25 is only 5² and does not represent placing five distinct beads without replacement. The value 10 is also unrelated to this arrangement problem. Because the beads are on a straight string rather than in a circle, rotations are not identified as the same arrangement; hence no circular-arrangement division is required.

Using the permutation formula \({}^nP_r=\frac{n!}{(n-r)!}\), we get \({}^4P_0=1\) and \({}^4P_1=4\). Therefore, \({}^4P_0+{}^4P_1=1+4=5\), so option D is correct. Option A is only the value of \({}^4P_1\). Exam tip: \({}^nP_0\) is always equal to 1.

There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.

This is a permutation of all the letters because different orders produce different words. BIHAR contains five letters: B, I, H, A, and R. Each letter occurs only once, so there is no repetition to divide out. The first position can be filled in 5 ways, the second in 4 ways, and so on, giving 5 × 4 × 3 × 2 × 1 = 5! = 120. Therefore, option B is correct. The value 60 is only half of the required factorial count, 25 does not represent arranging five distinct letters, and 10 is also not obtained from the permutation rule. Since every letter must be used exactly once, the appropriate formula is n!, not a combination formula.

(4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.

For four places, there are (5\times4\times3\times2=120) ways. Without repetition, choices keep decreasing.

The prizes are distinct, so ({}^{8}P_{3}=8\times7\times6=336). Order is counted for distinct prizes.

The direct answer is B: 6. NOON contains four letters, with N repeated twice and O repeated twice. If all positions were different there would be \(4!=24\) orders. Swapping the two N's changes nothing, so divide by \(2!\); swapping the two O's also changes nothing, so divide by another \(2!\). Thus \(4!/(2!2!)=24/4=6\). Option B is correct. Option A, 12, divides by only one repeated pair. Option C, 4, divides too much or uses an incorrect count. Option D, 24, ignores both repetitions. The rule is to divide by the factorial of every repeated group.

({}^{10}P_{0}=1) and ({}^{10}P_{1}=10), so the sum is (11). In simple sums, find the values separately first.