Expert: Daniel Mazur - 7/2/2009

Question
Hi Dan,

I'm back with math again.

There's a question I recently went through which is as follows:

" Q.) In how many ways can a pack of 52 cards be divided among four players equally? "

The way I figured it out is this:

(^52 C _13)  x  (^39 C _13)  x  (^26 C _13)  x  (^13 C _13)

But when I checked the answer, it also has a factor '4!' in the denominator of the expression I wrote above.


In general it says, that
"the number of ways in which mn different items can be divided equally into 'm' groups, each containing n objects and THE ORDER OF THE GROUPS IS NOT IMPORTANT, is given by
{(mn)!/(n1)^m}/m!"

Now in the above expression I don't understand why the factor m! has come into the denominator just like 4! has come in our question.
If the order of distributing the pack of cards to the four guys is not important, then the answer should be the expression I wrote "multiplied" by 4!, not divided by it, since we can distribute the four packs of cards among the four guys in 4! ways after selecting them in
(^52 C _13)  x  (^39 C _13)  x  (^26 C _13)  x  (^13 C _13) ways
, and if the order is important, then the cards can be distributed in just one way, the expression I wrote.

But this reasoning doesn't agree with the general formula I wrote above in which the expression is divided by this factor (4! in our case) and not multiplied, as I reasoned out above.

Please solve this confusion.

Thanks,
Shikhin

Answer
Dear Shikhin,

in the problem "In how many ways can a pack of 52 cards be divided among four players equally?" you don't say, how they define <equally>. I will take it that you mean <so that each player has the same number of cards>, because that's how you are solving the problem. But there are other ways, how to define <equal> in cards:-)...

I agree with the official solution, the 4! belongs to the denominator. Your expression
(^52 C _13)x(^39 C _13)x(^26 C _13)x(^13 C _13)
is a good start, but then we must DIVIDE by the number of permutations in the group of 4 players.

Example: You deal one round, where (for convenience) player A gets all the hearts, player B gets all clubs and the rest is somehow distributed among the other two players. Then you deal a second round, where player A gets all clubs and player B gets all hearts, while the rest of the cards are distributed among players C and D exactly the same way as before. These two DEALINGS, while different in who has what cards, are considered THE SAME SITUATION by the problem definition.

In some sense, by saying "THE ORDER OF THE GROUPS IS NOT IMPORTANT" the problems treats players as indistinguishable... So the number of distinct SITUATIONS in the game is smaller than the number of distinct DEALINGS.

This is always the case in statistical physics when you consider side-by-side similar systems, where in one the particles (players) are considered distinguishable and in the other they are indistinguishable. The systems of indistinguishable particles always offer SMALLER number of distinct realizations than systems of distinguishable particles - and the difference is a factor of N! in denominator just like you observe in your card problem.

Cheers,
Daniel