I want to distribute $ 6 $ distinct balls in $ 10 $ distinct boxes such that at least one box contains more than $ 1 $ ball. Did I calculate correct?
There are $ \binom{10}1 = 10 $ ways to choose a box.
We will choose $ 2 $ balls and put them in the box we have chosen - $ \binom{6}2 = 15 $
We will now distribute the rest of the balls - $ 10^4 $.
Therefore, by the rule of product, the answer is $ 10\cdot15\cdot10^4 $.
Lets solve the question using complement rule as @lulu mentioned such that all possible distributions without restriction - the number of distributions where no box contain more than one balls.
All possible distributions without restriction = $10^6$
the number of distributions where no box contain more than one balls: Select $6$ boxes among $10$ boxes to put the balls , and distribute these $6$ distinct balls into $6$ distinct boxes by $6!$ . Hence , $$\binom{10}{6}6!$$
So , $$10^6 -\binom{10}{6}6!=848,800$$