Permutations

The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.

This is a permutation question because changing the order of the selected 3 charts gives a different arrangement. Therefore, the number of arrangements is \({}^{11}P_{3}=11\times10\times9=990\). Option 165 is \({}^{11}C_{3}\), which counts selections without considering order. Exam tip: Words such as “arrange” or “in order” usually indicate permutations.

The governing concept is permutation: when distinct objects are placed in a row, their order matters, so all n objects can be arranged in n! ways. Here n = 8, and therefore the required number is 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320. Hence option A is correct. Option B, 5040, is 7!, which would arrange only seven objects. Option C, 720, is 6!, and option D, 56, is not the factorial count for a complete linear arrangement. Since every book is distinct and all eight are used, no repetition adjustment or combination formula is needed.

The relevant principle is the factorial rule for arranging distinct objects. GANIT has five letters: G, A, N, I and T, and none of them is repeated. Therefore the first position can be filled in 5 ways, the second in 4, then 3, 2 and 1 ways. The total is 5 × 4 × 3 × 2 × 1 = 5! = 120. Hence option A is correct. Option B, 60, and option D, 240, do not equal the required factorial count. Option C, 24 = 4!, would count arrangements of only four distinct letters, not all five letters. Every letter is used exactly once, so no division for repetition is needed.

This is a permutation problem because the digits are arranged in positions and changing their order creates a different number. The specified digit is fixed in the third, or middle, position, so it does not need to be chosen or arranged again. The remaining four distinct digits can be placed in the four vacant positions in 4! ways. Thus the count is 4 × 3 × 2 × 1 = 24. Therefore option C is correct. Option A, 120, would incorrectly arrange all five digits freely as 5!, while option B does not follow the factorial count. Option D also results from an incorrect restriction or multiplication.

The governing principle is permutation, because the words “in order” mean that the arrangement of the selected letters matters. For the first position, any of the 12 letters may be chosen. Once that letter is selected, 11 letters remain for the second position, since repetition is not indicated in an ordered selection of distinct letters. By the multiplication principle, the total is 12 × 11 = 132. The same result follows from the permutation formula: 12P2 = 12!/(12 − 2)! = 12!/10! = 12 × 11 = 132. Therefore option A is correct. Option B, 66, equals 12C2 and would apply if order did not matter. Options C and D do not follow from the required two successive choices.

This is an ordered arrangement problem with a fixed first digit. Since the number must start with 3, the first position has exactly one choice and 3 cannot be used again. The remaining available digits are 1, 2, 4, 6, and 8, giving five choices for the second position. After filling that position, four choices remain for the third position, and then three choices remain for the fourth. Therefore the multiplication principle gives 1 × 5 × 4 × 3 = 60 numbers. Equivalently, the last three places can be filled in 5P3 = 5 × 4 × 3 = 60 ways. Thus option B is correct. Option A incorrectly uses 5!, while C and D do not count all valid ordered selections of three remaining digits.

Because the two letters must be selected in order, exchanging their positions produces a different outcome. Therefore use a permutation. There are 15 choices for the first letter and, after it is selected, 14 choices for the second letter. The multiplication principle gives 15 × 14 = 210 ordered selections. In notation, 15P2 = 15!/(15−2)! = 15!/13! = 210. Hence option A is correct. Option B, 105, equals 15C2 and ignores order. Option C is simply 15², which would allow repetition and is not appropriate here, while option D does not count the available ordered pairs.

This problem uses permutations with a fixed first position. Because every number must start with 6, the first digit is already determined and contributes one choice. After using 6, six digits remain: 1, 2, 3, 5, 7, and 8. Three more positions must be filled without repetition. There are 6 choices for the second digit, 5 for the third, and 4 for the fourth. By the multiplication principle, the total number is 1 × 6 × 5 × 4 = 120. Equivalently, the remaining digits can be arranged in 6P3 ways, which is 6 × 5 × 4 = 120. Thus option A is correct. Option B omits one available choice, while C and D do not represent the required three-position permutation after fixing the first digit.

The three offices are distinct, so assigning the same three people to different offices produces different outcomes. The president can be selected in 7 ways, the secretary in 6 remaining ways, and the treasurer in 5 remaining ways. Thus the total is 7 × 6 × 5 = ⁷P₃ = 7!/(7−3)! = 210. Option C is therefore correct. ⁷C₃ would count only an unordered group of three people and would treat every assignment as the same, which is unsuitable for named posts. The expression 7³ incorrectly allows repetition and does not enforce different people. The value 3! counts only arrangements of an already chosen group, not the complete selection from seven people.