Just give it a think

There are \(2015\) seats numbered \(1,2,3,\cdots,2015\).

There are also \(2015\) persons who are holding tickets \(1,2,3,\cdots,2015\).

They have seats at random.

Then the probability that exactly \(3\) persons take seats having same numbers as that on their tickets is \(P\).

Find \(\lfloor10^{6}P\rfloor\).

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