Number of ways of distributing 7 distinct balls into 7 distinct boxes with exactly one box with 3 balls. This is what I tried. I can distribute the balls in four ways:
1) 3,1,1,1,1
2) 3,2,1,1
3) 3,2,2
4) 3,4
For 1) I can first pick 5 boxes in ${7 \choose 5}$ ways and then pick 3 balls in ${7 \choose 3}$ ways. Moreover, I can permute this selection $5!$ ways.
For 2)-4) in a similar fashion: ${7 \choose 4}{7 \choose 3}{4 \choose 2}{2\choose 1}4!$; ${7 \choose 3}{4 \choose 2}{7 \choose 3}3!$; ${7 \choose 2}{7 \choose 3}2!$
I am not sure if I these numbers are correct (I feel like I double counted somewhere). Also, is there an alternative (or perhaps better) way to approach this problem? Thanks in advance.

We give an alternate approach that (barely) involves cases.  It will give a number to check your answer against. 
The box that gets $3$ balls can be chosen in $7$ ways, and for each way the balls it gets can be chosen in $\binom{7}{3}$ ways.
What about the rest? The remaining $4$ balls can, almost, be assigned in $6^4$ ways. Except that we do not want any of the remaining boxes to have $3$ balls.
How many bad distributions are there? The box that gets $3$ can be chosen in $6$ ways, and the balls it gets can be chosen in $\binom{4}{3}$ ways, with the remaining ball assigned in $5$ ways. Thus the total is 
$$7\binom{7}{3}\left(6^4 -(6)\binom{4}{3}(5)\right).$$
More or less equivalently, we can use Inclusion/Exclusion more explicitly. Call the boxes $1$ to $7$. For each $i$, there are $\binom{7}{3}6^4$ ways of assigning $3$  balls to Box $i$, with the rest distributed among the others.
If we add up, $i=1$ to $7$, we will have double-counted the situations where two boxes get $3$ balls. But we want to count these no times.
So from $7\cdot \binom{7}{3}\cdot 6^4$ we must subtract $2\cdot \binom{7}{2}\binom{7}{3}\binom{4}{3}\cdot 5$.  
