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How many ways are there to choose 3 cards from a standard deck of 52 cards, if all three cards must be of different suits? (Assume that the order of the cards does not matter.)

 Sep 16, 2016

WARNING: I am not toooooo good at this probability stuff....but here is my answer

you draw one card....it is a ' given ' it WILL be ONE of the suits

probability:  1


Now you have 51 cards left, 39 of which are a 'different' suit. SO you have a 39/51 chance of getting a different suit.

1 x 39/51


Now you have 50 cards left....  only 26 are a different suit than the first two drawn, so you have a 26/50 chance of a different suit


1 x 39/ 51  x 26/50 = .3976    or  39.76% chance of getting THREE different suits (using a standard poker deck of 52)



Anyone else get the same answer???

 Sep 16, 2016
edited by ElectricPavlov  Sep 16, 2016

Sorry....I now see your question reads 'how many different ways'   NOT "what is the probability"  ......I'll leave the answer to someone who KNOWS how to do it!     

ElectricPavlov  Sep 16, 2016

For the first card, we can pick any card = 52 choices


For the second card, we have 39 ways to pick a card of a different suite from the first one.


And for the last card, we have 26 ways to pick a card that is different from the first two suites.


So....the total number of ways =   52 * 39 * 26  =  52,728  ways




cool cool cool

 Sep 16, 2016

There are 4C3 ways to choose which suit goes first and stuff so thats 4. 


There are 13^3 = 2197 ways to  choose the cards. 


2197*4 = 8788



 Dec 29, 2016

First, we choose the suits. There are (4C3) ways to do this. Then, we choose one of 13 cards from each of the chosen suits. There are (13^3 = 2197)  ways to do this. The total number of ways to choose 3 cards of different suits is therefore (4*2197 = 8788)

 Mar 15, 2017

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