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

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???

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

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

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)