# Explanation of the 'division of $52$ cards' in four groups of $13$ problem?

I have been thinking about this problem, total possible combinations of division of $52$ cards deck in $4$ groups of $13$.

I remember that the answer was ${52 \choose 13 }{39 \choose 13} {26 \choose 13} {13 \choose 13}$

What I want to know if figure represents the unique $4$ sets of $13$ cards or does this figure include the permeation as well,

Lets say $A_1,A_3,A_2,A_4$ are four combinations of $13$ cards

so does this figure count it as $1$ value or as $4!$ Can anyone help me understand that?

If you can give me some online links for similar concepts of Permutation & Combination, it wold be great?

It counts your four combinations as $4!$ different deals. The reasoning is as follows. Suppose that we label the four hands North, South, East, and West, as is standard in bridge. There are $\binom{52}{13}$ ways to choose North’s hand, then $\binom{39}{13}$ ways to choose South’s hand from the remaining $39$ cards, then $\binom{26}{13}$ ways to choose East’s hand, and finally just $\binom{13}{13}=1$ way to choose West’s hand. Since the reckoning takes into account which player gets which hand, each of the $4!$ permutations of these same four hands will be counted separately.
• @Joe: We specified which of the four sets of $13$ cards goes to North. If we switched the North and South hands, say, that would give North a different one of the $\binom{52}{13}$ possible hands, so the product above would count it as a different deal. Apr 3, 2013 at 19:02