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January 19th, 2008, 09:40 AM  #1 
Newbie Joined: Jan 2008 Posts: 3 Thanks: 0  Possibilities with a standard deck of cards?
First of all forgive me for the mistakes I'm gonna make. My english is in a very bad condition but wanted to try for an answer to my problem. I have a standard deck of cards and I lay them all face up. There are 52! possible combinations that we can see. 52!=52*51*50*...2*1 But if the suit and colour dont matter, how many different combinations are? I mean that is we change place 2 cards with the same number (ex 8 of hearts and 8 of diamonds) we dont get a new combination. Hope you understood my question Thanks in advance 
January 19th, 2008, 12:07 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
(52!)/(4!)^13 = 92024242230271040357108320801872044844750000000000

January 19th, 2008, 12:38 PM  #3 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
There is a general formula. If you have a group of element's with n elements and k groups, each consisting of ki, 1>i>n, then you can rearrange the starting group in: n!/(k1!*k2!*k3!*...*ki!) ways. Therefore, if you have 52 cards and the color doesn't matter, you can rearrange it in: 52!/(4!)^13 ways just like CRGreathouse sad. This is because you have 13 groups each composed of 4 elements. Each group represents a card type (Jack, 10's..) and there are 4 of the cards in each group because there are four colors. You could solve the problem where the color does matter as: 52!/(1!)^52 Stupid, but true. 
January 19th, 2008, 07:52 PM  #4 
Newbie Joined: Jan 2008 Posts: 3 Thanks: 0 
I thank you both for your answer. Milin I understood the way solving this through your post. Thanx for your explanation. 

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