In how many ways can 15 guests be accommodated in four rooms if no room is to be left empty?

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Without the specified restriction, there are $4^{15}$ ways. In $3^{15}$ of them, the first (second, third, fourth) room will be empty. In $2^{15}$ ways, a fixed pair of rooms will be empty, and in one way, when three fixed rooms will be empty. By the inclusion-exclusion principle (see problem $\underline{60435}$), there are $4^{15}-4 \cdot 3^{15}+6 \cdot 2^{15}-4$ ways when all rooms are non-empty.

## Answer

$4^{15}-4 \cdot 3^{15}+6 \cdot 2^{15}-4$ ways