<s>[INST] 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.
 [/INST] 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 $2\cdot \binom{7}{2}\binom{7}{3}\binom{4}{3}\cdot 5$.  
</s>