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.)

tertre Sep 16, 2016

#1**+4 **

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???

ElectricPavlov Sep 16, 2016

#2**0 **

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**0 **

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

CPhill Sep 16, 2016

#4**+2 **

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

NinjaAnswer Dec 29, 2016