P&C q6

[Saumyojit]

In how many ways can a pack of 52 cards be divided Equally among 4 players in *order*?

In order means what?

There are 4 suites , 13 cards of each suite .

52c13 --> selecting any 13 cards from 52 and giving to the first person.

39c13 --2ns person

26c13 --- 3rd person

13c13--- 4 th person .

Multiplying each other will it give me the answer?

I didn't see the Answer.

@JeffM @Subhotosh Khan

 

[Saumyojit]

okay this is right.
No issues

 

[Steven G]

Here is how I would think about it.
There are 52! ways of arranging the deck. The 1st 13 cards goes to the 1st player, the 2nd 13 cards goes to the 2nd player and so on.
So the division of the 52 cards to the 4 players is done.

Now the order of the 13 cards which a player gets does not matter. I therefore claim the answer is (52!)(13!)^4.
Is that what you got?

 

[pka]

There is a standard way(order) to deal a deck of well shuffled cards to four players. The dealer begins with the person on her/his left. Going clockwise until the last card dealt goes to the dealer.

 

[Dr.Peterson]

In how many ways can a pack of 52 cards be divided Equally among 4 players in *order*?

In order means what?

I'm not sure what "in order" means; the English is not clear. It probably means that the 4 players are distinguished; it may or may not mean that the order in which the cards are dealt to each player matters, though it probably doesn't because of the phrase "divided equally".

There are 4 suites , 13 cards of each suite .

52c13 --> selecting any 13 cards from 52 and giving to the first person.

39c13 --2ns person

26c13 --- 3rd person

13c13--- 4 th person .

Multiplying each other will it give me the answer?

I think you're right; I think Jomo's answer is wrong. Tell me why.

Can you see a way to use pka's idea (looking at the order in which the cards are dealt) to find an answer (by then accounting for different orders for any given player)? You should get the same numerical answer.

 

[Steven G]

I'm not sure what "in order" means; the English is not clear. It probably means that the 4 players are distinguished; it may or may not mean that the order in which the cards are dealt to each player matters, though it probably doesn't because of the phrase "divided equally".

I think you're right; I think Jomo's answer is wrong. Tell me why.

They both give the same value so one can't be wrong and the other correct.

 

[Dr.Peterson]

Here is how I would think about it.
There are 52! ways of arranging the deck. The 1st 13 cards goes to the 1st player, the 2nd 13 cards goes to the 2nd player and so on.
So the division of the 52 cards to the 4 players is done.

Now the order of the 13 cards which a player gets does not matter. I therefore claim the answer is (52!)(13!)^4.
Is that what you got?

Did you mean it the way it came out, or did something get lost? (I actually misread it at first as if you said (52C13)^4, because what you wrote made no sense; it's presumably either a typo or a failure in the new version of Xenforo.)

 

[Steven G]

Did you mean it the way it came out, or did something get lost? (I actually misread it at first as if you said (52C13)^4, because what you wrote made no sense; it's presumably either a typo or a failure in the new version of Xenforo.)

(52!)/(13!)^4

15 factorial divided by 13! to the 4th power.

The division sign initially was missing.

The OP's product and my result equal the same value.

 

[Steven G]

There is a standard way(order) to deal a deck of well shuffled cards to four players. The dealer begins with the person on her/his left. Going clockwise until the last card dealt goes to the dealer.

That is correct. However, giving the top 13 cards to each player in order is equivalent to what I wrote as far as how many different 13 card hands each player can get. You know this is true better than I do!

 

[Dr.Peterson]

(52!)/(13!)^4

15 factorial divided by 13! to the 4th power.

The division sign initially was missing.

The OP's product and my result equal the same value.

Yes, and your method (now that I've read it carefully) is what I was hinting at when I asked,

Can you see a way to use pka's idea (looking at the order in which the cards are dealt) to find an answer (by then accounting for different orders for any given player)? You should get the same numerical answer.