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In arrangements of (7) distinct letters, how many have (A) not in the first position, (B) not in the second position, and (C) not in the third position?

Answer and explanation

Correct answer: (3216)

By inclusion-exclusion, the count is (7!-3\cdot 6!+3\cdot 5!-4!). For multiple forbidden positions, this method is safe.

Related tags

PermutationsForbidden PositionsInclusion ExclusionPermutations And CombinationsMathematicsClass 11 Mcq

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