In how many ways can eight distinct balls be distributed to three distinct boxes so that each box must contain at least one ball?
So in order to approach this problem I was first thinking of assigning one ball to each box. Since there are 8 total balls I have 8C3 ways of picking 3 of them. Out of those 3 that I picked they can be arranged in 3 different ways since there are 3 boxes. Now I have to arrange the remaining 5 balls. This can be done in 5x5x5 ways. Then I multipled 3(8C3)(5x5x5) to get my final solution. However my textbook says the solution should be 5796. What is wrong with my approach/what adjustments can be made to make my approach work?
When you pick $3$ balls to put in boxes, and then assign the rest, you're doing some over counting. Suppose you put balls $A$, $B$ and $C$ in the three boxes, and then dump $D$-$H$ all in box $1$. That's the same as if you had started the boxes with $D$, $B$ and $C$, and then put everything else in box $1$, but you're counting those as separate cases.
To avoid overcounting, try a different approach: For example, there are $3^8$ ways of putting balls in boxes, total. How many of those ways leave at least one box empty? If you can subtract those, you'll have your answer.
If you're familiar with the inclusion/exclusion principle, you might find it handy here.