Counting: 2 dozen different donuts are given to 24 kids $2$ dozen donuts are given out to $24$ kids so that each kid gets one donut.
1. How many ways are there to distribute the donuts if the donuts are all different?
Answer: $24!$
Why couldn't we use the product rule here and say 24 donuts x 24 kids = total different outcomes. Got confused and wondering how this is different than say having $3$ shirts and $2$ pants, then you can make $6$ different outfits.
2. How many ways are here to distribute the donuts if there are 4 varieties of donuts and exactly six of each variety
Answer: $24! / 6!6!6!6!$
I've haven't seen an example like this in my textbook, is there a formal method used here? A simple explanation of the reasoning behind this appreciated.

UPDATE: For #2, I just learned the concept in class. The method used is called "Permutation with Repetitions" for anyone learning it. Common example is distinguishable balls placed into distinct bins
 A: For $(1)$ it is the product rule, but you are applying it incorrectly.  There are $24$ ways to choose the donut to give to the first kid, then $23$ ways to choose the donut to the second kid, since we can't give him the donut we've already given to the first kid, then $22$ ways to choose the donut for the third kid, and so on.  By the product rule, there are $24!$ possibilities.
For $(2)$ imagine for a moment that we can distinguish between all the donuts.  Then as before there are $24!$ ways to distribute the donuts.  But it doesn't really matter which of the chocolate donuts a particular kid gets, so long as he gets a chocolate donut, so we have  counted each distribution of the $6$ chocolate donuts $6!$ times.  The same is true of the other varieties, giving the answer shown. 
A: First part:
Consider instead the different ways of order 24 kids in line so as to distribute the donuts as they get unpacked.
Second part:
Consider the equivalent calculation of:


*

*Choosing, among 24 kids, 6 of them to give them the first kind of donuts, i.e. $24\choose 6$.

*Of the 18 kids still without donuts, we give out 6 donuts of some other kind out of the 4 kinds available to 6 other kids, i.e. $18 \choose 6$.

*Next we have $12$ kids to choose from for the 6 donuts of the third kind, i.e. $12\choose 6$.

*And we end up with one single way of giving out the last kind, i.e. $6\choose 6$.


Therefore the final calculus is:
$${24\choose 6} {18\choose 6} {12\choose 6} {6\choose 6}= \frac{24!}{ 6!6!6!6!}$$
