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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
 #1
avatar+19857 
+5

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
 #2
avatar+19857 
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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
 #3
avatar+106532 
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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
 #4
avatar+349 
0

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

 

cool

 Dec 29, 2016
 #5
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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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