before anyone starts blaming me for stating something incorrectly, I copied down the question from my book word by word so I am sorry.
I don't know how to approach the problem please help.
I was thinking of dividing 3! by 5! but that does not make any sense.
As I understand it, there is no restriction on where you can place any of the three balls. So, start with your first one. How many places can you put it? Five. And now for your second one. It's placement is not at all affected by the first, so again, you can put it in any of five places. Then third one is again five. So the number of ways to place the three balls is 5*5*5=125.
Are you told the balls are indistinguishable? That will make things more complicated.
Arrange the balls & boxes like this: $$| * ||*|*$$
This corresponds to $0$ balls in the first box, $1$ in the second, $0$ in the third, $1$ in the fourth, and $1$ in the fifth.
We can enumerate all the different ways by counting how many ways we can place $4$ bars among $7$ places. That comes out to $$\binom 74 = 35$$ ways.
In general, with $n$ balls and $k$ boxes, we have $\binom {n+k-1}{k-1}$ ways.
If the balls are indistinguishable, then "reverse" the question and ask, how many ways can we arrange three balls so that only one box holds them all? That would be five. Then, how many ways can we put two balls in one box and a single ball in another box? That would be 5*4 = 20. We might be tempted to ask how to put one ball in a box and two in another box, but we just did that, so forget that. Finally, how many ways can we put one ball in each of three boxes? That would be 5*4*3/(3*2) = 5*2 = 10. Those are all of the possible combinations. So the total number of combos is 5+20+10=35.