# 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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