In a game, each player receives 5 cards from a deck of 52 different cards. How many different groupings of?

Question

cards are possible in this game??

how to do this?

Answer 1

Tough Question...

If I remember my probability right, it should be something like this...

Player 1 chooses 5 cards from the 52-card deck. This is written as: (52 choose 5)

Player 2 chooses 5 cards froim the remaining 47-cards. This is written as: (47 choose 5)

Player 3 chooses 5 cards froim the remaining 42-cards. This is written as: (42 choose 5)

Let's say there's only 3 players (just repeat the above process for more).

For the first player, there is (52 choose 5) different card combinations. For the second player, there is (47 choose 5) different card combinations. And for the third, (42 choose 5) combinations...

Thus, the total # of combinations is:
(52 choose 5)*(47 choose 5)*(42 choose 5)

Since:
(Y choose X) = Y! / (X! * (Y-X)!) (note: not as bad as it looks typed out...)

where X! is: 1*2*3*4...*X, called X factorial...

Thus:
(52 choose 5)*(47 choose 5)*(42 choose 5)

= 52! * 47! * 42!
--------------------------------------------
5! * (52-5)! * 5! * (47-2)! * 5! * (42-5)!

= 52! * 47! * 42!
-------------------------------------------
47! * 42! * 37! * 5! * 5! * 5!

= 52!
----------------------------
37! * 120 * 120 * 120

= (52 * 51 * ... * 2 * 1)
--------------------------------
(37 * 36 * ... * 2 *1) * 120 * 120 * 120

= (52 * 51 * ... * 39 * 38)
--------------------------------
120^3

= 3.391 * 10^18

Of course, there are different ways of distributing the cards (ie. one per person), but, if my understanding is correct, it should work out to the same result... (based on five unique cards for X people from a 52-card deck...)

Answer 2

Since the order of the cards does not matter, this is a combination, and not a permutation set.

It would be C(52,5)

As an example, the number of five-card hands possible from a standard fifty-two card deck is:

n!/k!(n-k)!

52!/5!(52-5)! = 2598960

Answer 3

the first card can be any of the 52, the second can only be 51 possibilities and so on. thus the number of options are:

52*51*50*49*48

Answer 4

Let n = number of cards
and r = number of cards in a group

number of groups = n!/(r!(n-r)!)

= 52!/(5!(47!)) = 2,598,960

Answer 5

1, it is C(52, 5)
2, C(52, 5)*C(47, 5)
3, C(52, 5)*C(47, 5)*C(42, 5)
............................................
10, C(52,5)*C(47,5)*upto C(7,5)