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The man wants to paint $15$ benches in red, green and blue colours. How many ways he can paint the benches, such that he must paint at least one bench in each colour. We can't rearrange benches after painting.

My solution: There are $3^{15}$ ways to paint all benches, using 3 colours. There are $3 \cdot 2^{15}$ ways to paint all benches, using 2 colours. There are $3$ ways to paint all benches, using 1 colour.

So the answer is $3^{15} - 3\cdot 2^{15} + 3 = 14 250 606$, because we first must subtract all ways to paint using two colours, which also contain the ways to paint using one colour, so we also have to add $3$ at the end.

Another explanation can be done using Euler's diagrams.

I know this problem can be tackled using the Inclusion–exclusion principle.

Namely, the formula $|A \cap B \cap C| = |A| + |B| + |C| - |A \cup B| - |A \cup C| - |B \cup C| + |A \cup B \cup C|$.

The problem is I don't really understand the physical sense of sets $A, B, C$. What do they represent? What are the elements they consist of?

But the most absurd thing for me is that it seems like $|A| = |B| = |C| = 1$, but in the same time $|A \cup B|$ and $ |A \cup C|$ are significantly bigger then $|A| = |B| = |C| = 1$.

I know it's wrong, but I don't have any other ideas.

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    The statements in your second paragraph are incorrect. You'll be less likely to make a mistake if you consider sets $A_{rgb}, A_{rg}, A_{rb}, A_{gb}, A_r, A_g, A_b$, where, for instance, $A_{rg}$ is the set of all ways to colour the benches without using any colours besides red and green. (However, using only red or only green is allowed.) – 2017-02-20
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    `the physical sense of sets |A|...` Those are not sets, those are natural numbers. $|A|$ is the number of elements in set $|A|\,$. – 2017-02-20
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    @dxiv It was my typo, sorry. I'm not talking about cardinalities. – 2017-02-20
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    The question is incomplete. Can we rearrange the benches after painting or are they fixed so order matters? – 2017-02-20
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    @mathreadler Clarified. – 2017-02-20

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$|A| \ne 1$. Sense is the following. Let $A$ be a set of all paints with at least one red bench, $B$ be a set of all paints with at least one blue bench and $C$ be a set of all paints with at least one green bench. Then we have $$|A \cap B \cap C| = |A| + |B| + |C| - |A \cup B| - |A \cup C| - |B \cup C| + |A \cup B \cup C|$$ that is number of paints with all three properties equals to sum of numbers of paints with any one properties minus sum of numbers of paints with at least two properties, since all these paints were added twice plus numbere of paints with all three properties since they where added $3 - 3 = 0$ times.

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    Let $C = \{\text{red}, \text{blue}, \text{green}\}$. Then $A \subset C^{15}$, such $\forall (a_1, \dotso, a_{15}) \in A \ \exists i \text{ such that } a_i = \text{blue}$? – 2017-02-20
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    Yes, you are right. – 2017-02-20
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    Unfortunately, it's hard to calculate (at least for me), for example $|A|$. – 2017-02-20
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    In this sense $|A| = 3^{15} - 2^{15}$ because we need to exclude from all $3^{15}$ paints those $2^{15}$ without blue color. – 2017-02-21