how many....include one card from each suit?

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Another card question..
From a deck of 52 cards, how many different four-card hands could be dealt which include one card from each suit?

The "include one card from each suit" is what I'm having difficulty putting in numbers..

Thanks for the help.
 
tkhunny said:
First Guess: (52*39*26*13)*4!
Click to expand...
Shouldn’t (52*39*26*13)/(4!)?
Note that the above equals (13)<SUP>4</SUP> or \(\displaystyle {13 \choose 1}^4\).
 
tkhunny said:
First Guess:
(52*39*26*13)*4!
Click to expand...
The point here is to take a shot at it. As you showed no work at all and provided no theory or impression, I just left the guess for you to think about. Surely you had SOME idea how to go about it. You've posted enough here to know that it is expected of you to provide better information.

Read it.
Think about it.
Come up with a plan.
Implement the plan. <== This is where I quit.
Review how you really feel about it.
Revise as necessary <== pka did the last two steps.
 
There are 52 choices for the first card; the first card can be of any suite. Because the second card must be of a different suite from the first choice, there are only 39 choices for the second card. Similarly, there are 26 choices of the third and 13 choices for the fourth card. Now divide by (4!) because there are that many ways to makes the 4 choices.
 
anna said:
like how did u get 39, how many cards are in a suit too..
Click to expand...
A common body of knowledge answers questions such as that. Most question writers assume that you know the body of knowledge. Thus you need to look the properties of a standard deck of cards.
 
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