Originally Posted by

**Soroban** Hello, Aquafina!

We can use this fancy counting formula:

. . $\displaystyle n(A \cup B\cap C) \;=\;n(A) + n(B) + n(C)$

. . . . . . . . . . . . . . $\displaystyle - n(A \cap B) - n(B \cap C) - n(A \cap C)$

. . . . . . . . . . . . . . . . $\displaystyle + n(A \cap B\cap C)$

$\displaystyle n(2\cup3\cup5) \;=\; n(2) + n(3) + n(5) - n(2\cap3) - n(3\cap5) - n(2\cap 5) - n(2\cap3\cap5) $

. . . . . . . $\displaystyle =\; 3000 + 2000 + 1200 - 1000 - 400 - 600 + 200 $

. . . . . . . $\displaystyle = \;4400$