How do the answers to combinatorial problems change if instead of 4 different objects we have 4 identical ones? I think I did the first parts of these correctly, but I really don't know about the last part? Could I just divide all my previous answers by $4!$
If you have $4$ children, $8$ unique fruit, and $8$ identical candy bars.
1: how many ways can $8$ fruit be distributed among $4$ kids. There are no restrictions. (is this $4$^8)
2: how many ways can you distribute the fruit if you have to give at least $1$ to each kid. (Is this $4$!{$8$C$4$})
3: how many ways can you give candy bars to the kids. there are no restrictions. (Is this (($8$+$4$-$1$)C($4$-$1$))
4: how many ways can you give out the candy bars if each kid has to get $1$. (give each kid $1$ bar, and do the same as problem 3?)
How would 1 - 4 change if the kids were replaced by $4$ identical bowls?
 A: Part 2 is the hard one, it is an inclusion-exclusion problem.  From part 1,there are $4^8$ ways to hand out the candy bars.  Now we exclude the ways some child doesn't get one.  There are ${4 \choose 1}3^8$ way to give all the bars to three kids, so we subtract those.  Now if we gave them all to two kids we have subtracted them twice, so add them back in.  That is ${4 \choose 2}2^8$.  If we gave them all to one kid, they were counted once in the first batch, subtracted three times in the second, added three times in the third, so we are good.   Final answer is $4^8-4\cdot 3^8+6\cdot 2^8$
A: 2:
This result has a generating function of 
$(x+x^2/2!+x^3/3!+...)^4$
The answer is the same as $8!$ times the coefficient of $x^8$ in the above expression.
3:
Put the candy bars in a line, and divide them with 3 sticks. However, the sticks can stay next to each other. Thus this is same as arranging 8 candies and 3 sticks
Answer = $(8+3)!/(8!3!)$
4:
This is simply a specialization of 2, with 1 in place of $8!$
Answer = $\binom 7 3$
