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Seven people enter a lift. The lift stops at three (unspecified) floor. At each of the three. no one enters the lift, but at leat one person leaves the lift. After the three floor stops, the lift is empty. In how many ways can this happen?

Why did my attempts fail?

let $x_{i's}$ be the number of persons leaving lift at once (any floor). Then problem becomes finding integer solution of: $$x_{1}+x_{2}+x_{3}=7\ \text{where,}\ \ 1\le x_{i}\le7$$

$\implies x'_{1}+x'_{2}+x'_{3}=4\ \text{where}\ \ x_i=1+x'_{i}$

The number of solutions: $\displaystyle{{4+3-1}\choose{4}}=15$, but this didn't work, please explain where I went wrong and give your solutions too.

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My read is that your answer is correct provided that the seven passengers are indistinguishable.

If you can distinguish the passenger an inclusion-exclusion argument yields $3^7-3\cdot 2^7 + 3\cdot 1^7 =1806$ possibilities.

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    if you don't mind please elaborate how you got last line (all possibilities). – 2017-02-18
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    As Rolf said, using the [inclusion-exclusion principle](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), due to the stated restriction that "at least one person leaves the lift" on each of the three floors. The first term $3^7$ gives the options without that restriction. The second term subtracts off the cases where each of the floors in turn has zero leavers. The final term adds back the cases where $2$ floors have no leavers to compensate for the over-subtraction of such cases in the second part. – 2017-02-18