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In how many ways can 52 cards be divided equally among four players(1,2,3,4) in order.

I interpreted this question as

Let there be 52 different cards(1.e each numbered with unique number from 1 to 52,like c1,c2,c3,c4....,c52).

In how many ways can cards be divided equally among four players(1,2,3,4) in order(i.e Player 1 gets card(c1,c5,c9....),Player 2 gets(c2,c6,c10...),Player 3 gets(c3,c7,c11....),Player 4 gets(c4,c8,c12....))

In my textbook it's answer is given as

For the first player we have C(52,13) choices,for the second player we have C(39,13) choices, for the third player we have C(26,13) choices and for the last player we have C(13,13) choices . Hence the total number of ways

= C(52,13)C(39,13) C(26,13)* C(13,13)

But,

According to me each player gets unique set of cards.Therefore,all will have C(13,13) ways. Hence total number of ways

=C(13,13)*C(13,13)*C(13,13)*C(13,13) =1

Where am I wrong?

jame samajoe
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  • I'm afraid you have misunderstood the question. When the book says "in order" it means that it matters which player gets which hand, so that the first player getting all the spades is different from the second player getting all the spades. – saulspatz Sep 05 '18 at 15:36
  • @fleablood according to me players(1,2,3) won't have choices from all 52,39,26 cards because all players have specific option like player1 can choose only c1,c5,c9,.... like this for all players.Is my logic wrong? – jame samajoe Sep 05 '18 at 15:39
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    To be fair it is not at all clear what the book means by "in order". However the OPs interpretation that every card is dealt in order is trivially restrictive. That means the deck itself is in order and there's only one way that could be. – fleablood Sep 05 '18 at 15:39
  • @fleablood do you have any idea what does"in order" mean ? – jame samajoe Sep 05 '18 at 15:41
  • I will cede that "in order" was a terrible choice of words. I would interpret it to mean that the cards are randomly dealt. But the diffferent possible hands order will not matter or the player puts the hand in order. That is the three card hand of c1,c2,c3 is the same as c2, c1, c3 or any other order. But the book is NOT clear. – fleablood Sep 05 '18 at 15:43
  • "In order" means that you consider the hands in the the order they are dealt; it matters which player gets which hand. If the order did not matter, you would have to divide the book's answer by $24$ since there are $4!$ ways of distributing the $4$ hands among the players. – saulspatz Sep 05 '18 at 15:44
  • "do you have any idea what does"in order" mean" No. I do not. And apparently you do not either. Asking how many ways there are to put anything in order is a trivially pointless question. There is one way to put things in order. – fleablood Sep 05 '18 at 15:45
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    saulspatz. It's not clear that the "in order" refers to the order of the players rather than the order of the hands themselves. The OPs interpertation is fair but... it makes for a trivial question. – fleablood Sep 05 '18 at 15:49

1 Answers1

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You are not wrong, you are just answering a different question than what the textbook is answering. You are counting the number of ways to achieve 13 cards to each player where the value of each card received does not matter. The textbook is counting the number of possible distributions of those cards (where the order of the cards within a player's hand does not matter, but the specific cards received does). You are treating player 1 getting all of the spades as the same as if player 1 got a random assortment of cards (since in both cases, player 1 still received 13 cards). The textbook is treating those as separate outcomes.

SlipEternal
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