Perms and Coms - what am I doing wrong??

Question

The question is:

Find the number of ways that 6 different coloured balls can be placed in 3 non-identical urns so that no urn is empty.

Here's my working:

Let $A$ be the event Urn $A$ is empty

$B$: Urn $B$ is empty

$C$: Urn $C$ is empty

$n(A^C \cap B^C \cap C^C)$ \begin{align} &=n(A\cup B\cup C)^C\\ &=\text{Total}-n(A\cup B\cup C)\\ &=3^6-\left\{n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\right\}\\ &=3^6-\left\{(2^6-2)+(2^6-2)+(2^6-2)-1-1-1+0\right\}\\ &=546 \end{align}

But the correct answer is $540$.

I don't know what I've done wrong.

Can someone please explain this to me?

Thanks a lot.

Answer 1

Your $n(A), n(B), n(C)$ are not correct.

"$A$ is empty" means that all balls are assigned to the $2$ other boxes. This corresponds to assigning each of the $6$ balls a label $B$ or $C$. There are obviously $2^6$ ways to do this.

Similarly for $n(B)$ and $n(C)$.

So, at the end you have $3^6- (3\cdot 2^6 -3)$.