There are $n^m$ ways to place $m$ labelled balls into $n$ labelled boxes with no constraints.
When it comes to the problem such that at least one box has more than $k$ balls.
My thought was
Step 1. choose $k+1$ balls to form a big ball gives $\binom{m}{k+1}$ ways
step 2. solve the unconstraint problem to place $m-k$ labelled balls(1 big ball + $m-(k+1)$ original balls) into $n$ labelled boxes, which comes out $n^{m-k}$ ways
put thogether using product rule we have $$\binom{m}{k+1}n^{m-k}$$ ways.
Validate with an example below the above was not correct.
Example with $m=4; n=2, k=2;$ based on the above ,$\binom{4}{3}2^{2} =16$ but acutually we only have 10 diffent ways as $$(1234,0)(0,1234)(123,4)(4,123),(124,3)(3,124)(134,2)(2,134)(234,1)(1,234)$$
Could anybody help find where was wrong and give a correct way to solve this problem?