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8 students are to be seated in a row, but two particular students must not sit together. How many arrangements are possible?

Answer and explanation

Correct answer: 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.

Related tags

PermutationsNot TogetherSeatingPermutations And CombinationsMathematicsClass 11 Mcq

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