In how many ways a pack of 52 cards be formed into 4 group of 13 cards? According to the question the way I solved was 
$$52_{\mathbf C_{13}}×39_{\mathbf C_{13}}×26_{\mathbf C_{13}}×13_{\mathbf C_{13}}×4!$$
But in the solution it is given as
$$52_{\mathbf C_{13}}×39_{\mathbf C_{13}}×26_{\mathbf C_{13}}×13_{\mathbf C_{13}}×\frac{1}{4!}$$
Now my question is why it is $\frac{1}{4!}$ not $4!$
 If there are 4 groups say $$\mathbf A, \mathbf B,\mathbf C, \mathbf D$$
Then they should interchange between them in $4!$ ways. Please explain
 A: My short answer is that breaking the cards into groups 
A,B,C,D and then permuting the groups (which there are 4! ways to do) does not give a new grouping. In this case, once I have chosen my four groupings A,B,C,D, the grouping B,A,C,D isn't a different grouping. It is the same grouping just in a different order. We would multiply by 4! if permuting gave new groupings. We divide by 4! because permuting actually gives repetitions, not new groupings. I can separate the cards into A= spades, B= diamonds, C= clubs, and D= hearts, or I could separate the cards into A= diamonds, B=spades, C= clubs, and D=hearts. If I care about which of the four groups is A, which one is B, which one is C, and which one is D, then there are $52 C_{13}\times 39 C_{13} \times 26 C_{13}\times 13 C_{13}$ ways to separate the cards into four groups. But if I don't care about which group is A, which one is B, etc. (which we don't in this case), then A= spades, B= diamonds, C= clubs, and D= hearts and A= diamonds, B=spades, C= clubs, and D=hearts are the same group, just in a different order. Therefore using $52 C_{13}\times 39 C_{13} \times 26 C_{13}\times 13 C_{13}$ overcounts. This is because it includes repetitions (for example, $52 C_{13}\times 39 C_{13} \times 26 C_{13}\times 13 C_{13}$ counts A= spades, B= diamonds, C= clubs, and D= hearts and A= diamonds, B=spades, C= clubs, and D=hearts as two separate things, which we do not want. Let's understand why dividing by 4! is the right thing to do. Let $P$ be the set of all 4-tuples (A,B,C,D), where A,B,C,D are 4 groups of 13 cards. Here, order matters. Now let $Q$ be the set of all sets $\{A,B,C,D\}$, where $A,B,C,D$ are 4 groups of 13 cards. Here, order does not matter. The formula you have, $52 C_{13}\times 39 C_{13} \times 26 C_{13}\times 13 C_{13}$, is counting how many members are in $P$. This is because which group of 13 you choose first matters, which group you choose second matters, etc. But what the question is asking is how many members are in $Q$. So now the task is to figure out how many things are in $Q$ once you know how many things are in $P$. We define the function $f:P\to Q$ by $f((A,B,C,D))=\{A,B,C,D\}$. This is a surjection, but not an injection, because $$f((A,B,C,D))=\{A,B,C,D\}=\{B,A,C,D\}=f((B,A,C,D)).$$ Here, $(A,B,C,D)\neq (B,A,C,D)$, because in the tuples, order matters. But $\{A,B,C,D\}=\{B,A,C,D\}$, because in the sets, order does not matter. Now we look at a set $S=\{A,B,C,D\}$ in $Q$ and ask what is the cardinality of $f^{-1}(S)$? How many members of $P$ does $f$ send to $Q$ (in other words, how many different 4-tuples can we make from a set of four elements?) This is 4!, because every permutation of those four elements gives a different 4-tuple. Therefore for every $S\in Q$, $f^{-1}(S)$ has $4!$ elements. This means that for every element of $Q$, there are $4!$ elements of $P$. Therefore $|P|=4! |Q|$. Dividing by $4!$ gives $$|Q|= |P|/4!.$$ Now substitute $|P|=52 C_{13}\times 39 C_{13} \times 26 C_{13}\times 13 C_{13}$ to find $|Q|$, which is what you want.
A: This is usually solved using multinomial coefficients, see also Multinomial_theorem. The number of possibilities to give $13$ bridge cards to each of the players $A,B,C,D$ (and the labels count) is
$$
\binom{52}{13\ 13\ 13\ 13}
=
\frac{52!}{13!\ 13!\ 13!\ 13!}
=
\frac{52!}{13!^4}\ ,
$$
which corresponds to the more pedestrian product
$$
\binom{52}{13}\cdot
\binom{39}{13}\cdot
\binom{26}{13}\cdot
\binom{13}{13}\ .
$$
If now the players do not count (and the situation has nothing to do with the bridge game), then we have to divide the number by $4!$, since the permutation group of the letters / players $A,B,C,D$ act freely on each "bridge game configuration". So the final answer is
$$
\frac 1{4!}\cdot\frac {52!}{13!^4}
=
2235197406895366368301560000\ ,
$$
with the computer check, here sage:
sage: multinomial(13, 13, 13, 13) / factorial(4)
2235197406895366368301560000
sage: factorial(52) / factorial(13)^4 / factorial(4)
2235197406895366368301560000

A: There are $\binom{52}{13,13,13,13}=\frac{52!}{13!^4}$ ways to split up in $4$ distinguishable groups of $13$ cards.
The groups are not distinguishable and every combination of non-distinguishable groups is counted exactly $4!$ times.
To find the number of combination of non-distinguishable groups we must divide by $4!$, resulting in: $$\frac{52!}{13!^44!}$$ 
