Why m + n is 85? Suppose $4$ balls are placed at random into $4$ boxes. The probability that exactly two boxes remain empty is $\frac mn$ in its simplest form.
I divided it into $2$ cases, where $2$ balls are in two boxes each; $3$ in one box and $1$ in another.
For the first case, I think there are ${4 \choose 2}$$\times$${4 \choose 1}$$\times$${3 \choose 1}$ = $72$ different ways.
For the second case, I think there are ${4 \choose 3}$$\times$${4 \choose 1}$$\times$${3 \choose 1}$ = $48$ different ways.
The total number of ways to place the balls is $4^4$ = $256$.
Thus, I calculated that $\frac mn$ = $\frac {15}{32}$
However, the answer said that m+n is $85$. Where did I do wrong?
 A: Check your first value ($72$).
You seem to count $4\choose 2$ ways to put two balls into a box, times $4\choose 1$ ways to choose the first box, times $3\choose 1$ ways to choose the second box. However, this is flawed.
For example, if the balls are $A,B,C,D$ and boxes are $1,2,3,4$, then "$A,B$ are in $1$, $C,D$ are in $3$" is counted twice: once where the chosen balls are $A,B$ and the chosen boxes are $1,3$, and again, when the chosen balls are $C,D$ and the chosen boxes are $3,1$.
Thus, the actual number is half of what you got, i.e. $36$. Added to the correct second result ($48$) this gives you the fraction $\frac{84}{256}=\frac{21}{64}$ so $m+n=21+64=85$.
A: Hard to follow your calculation.
Pick the two bins you'd like to be non-empty, $\binom 42=6$ ways to do that.
The probability that all the balls go to either of those two is $\left(\frac 12\right)^4$.  The probability that all the balls go to exactly one of those two is $2\times \left( \frac 14\right)^4$
Thus the answer is $$\binom 42\times \left(\left(\frac 12\right)^4-2\times \left (\frac 14\right)^4\right)=\frac {21}{64}$$
