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Suppose that four persons enter a lift on the ground floor of a building. There are seven floors above the ground floor and each person independently chooses her exit floor as one of these seven floors. If each of them chooses the topmost floor with probability $\frac{1}{3}$ and each of the remaining floors with an equal probability, then
the probability that no two of them exit at the same floor equals.

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$\frac{200}{729}$

$\frac{220}{729}$

$\frac{240}{729}$

$\frac{180}{729}$

answer is A.

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Detailed Solution

$^{6} C_{4} \cdot 4! {(\frac{1}{9})}^{4} +^{6} C_{3} \cdot 3! \cdot^{4} C_{1} \cdot {(\frac{1}{9})}^{3} (\frac{1}{3})$

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