How many ways can a deck of 52 cards be broken up into a collection of unordered piles How many ways can a deck of 52 cards be broken up into a collection of unordered piles of sizes:  four piles of 13 cards. 
Soln: This is where my problem lies.  I knew the solution had to be something of the form $$\frac{52!}{(13!)^4}$$ But the only reason I figured that out was the way that the phrasing of the values was made.  When they used the term "collection of unordered piles of sizes... " I was under the impression that some sort of selection with repetition process of the form C(r + n - 1, r)  would have been used.  Yet with the solution using the form it does it indicates that order does indeed matter since it is a permutation. Also the real solution is actually $$\frac{\frac{52!}{(13!)^4}}{4!}$$
why are we dividing by another 4! ? is that to take into account that any of the four subgroups have repetition of some fashion? 
 A: The extra factor of $4!$ comes from the ordering of the piles themselves.
When we counted $\frac{52!}{(13!)^4}$ ways of distributing the cards into
piles, we counted each of the desired collections of cards several times.
For example, the configuration in which we have one pile containing all thirteen spades, one with all the hearts, one with all the diamonds, and one with all the clubs, is a single collection of unordered piles of cards and should only be
counted once. But when counting $\frac{52!}{(13!)^4}$ ways of distributing the cards, we count each
of the following as separate configurations:
Spades in the first pile, hearts in the second, diamonds in the third, clubs in the fourth.
Hearts in the first pile, spades in the second, diamonds in the third, clubs in the fourth.
Diamonds in the first pile, hearts in the second, spades in the third, clubs in the fourth.
And so forth, just permuting the order in which we list the piles of cards.
There are $4!$ permutations of the four piles, so the $\frac{52!}{(13!)^4}$ configurations included $4!$ copies each possible collection
of four piles of thirteen unordered cards.
Therefore there are actually only $\frac{52!}{4!(13!)^4}$ such collections.
EDIT: I was calling some of these unordered (or partially unordered)
configurations arrangements, but to many people "arrangement" implies order (despite the fact that people also often feel compelled to write "ordered arrangement"), so I have used other terms instead.
SECOND EDIT: I see that the question asked not only where the term
$4!$ came from, but why the expression $\frac{52!}{(13!)^4}$
(which resembles the answer to a permutations-with-repetitions problem)
occurred in the result rather than, say, $\binom{55}{52}$.
Perhaps the best way to understand this is to really understand where
the "permutations-with-repetitions" formula comes from in the first place.
The thing to remember about this formula is that it removes information
about order. 
It is exactly the same idea that gives us the different formulas for
permutations and combinations. The number of permutations of $4$ distinguishable items from a set of $9$ items (order matters!), without repetition, is ${}_9P_4 = 9\times8\times7\times6 = 9!/5!.$
The number of combinations of $4$ distinguishable items from a set of $9$ items (order does not matters), without repetition, is 
${}_9C_4 = 9!/(4!5!) = {}_9P_4/4!.$ 
Why can we divide ${}_9P_4$ by $4!$ and the result is ${}_9C_4$? 
Because each of the ${}_9P_4$ permutations counted $4!$ different ways
of arranging a given combination of $4$ items in order, and we only
want to count that combination once.
If you simply had $r$ items, all different,
there would be $r!$ ways to put them in a unique sequence; but if
$r_1$ of the items are identical, within any ordered arrangement of the items
there are $r_1!$ ways to permute these $r_1$ items among the $r_1$ places
in the arrangement where those items occur.
The answer $r!$ is based on the assumption that each of these permuted
arrangements is distinguishable from all the others, but actually we
consider them all to be the same.
Therefore the answer $r!$ overcounted the number of distinct ordered
arrangements by a factor of $r_1!$.
If there are another $r_2$ items that are identical to each other,
the answer $r!$ overcounted by an additional factor of $r_2!$, and so forth.
In the case of four piles of $13$ cards each, if the order of the piles
mattered and the order of cards in the piles also mattered, then we could
name a sequence of $52$ distinct places where cards can be placed
(the names might be phrases such as "fifth card in the third pile"),
and there would be $52!$ possible arrangements of the cards,
each one deemed to be different from all others.
But in the stated problem, we do not care about the order of the $13$
cards within the first pile; we can permute them among each other without
producing any "different" outcome, so the answer $52!$ overcounts
the number of outcomes by a factor of $13!$.
There is an additional factor of $13!$ for each of the other piles as well.
In all of these arguments, the same principle applies: 
correct the way the simple "permutation" formula overcounts the number of outcomes.
In all cases, the reason the "permutation" formula overcounts outcomes
is that it distinguishes all $k!$ permutations of a $k$-item
subset of the things being permuted, whereas we consider all of these
outcomes to be the same.
The factor of $4!$ comes from a similar idea, except that it refers to
how many ways we can permute the piles, rather than permutations
of subsets of individual cards.
