# Probability Question (Easy)?

I have trouble with these to questions. Please show your step. Thank your for helping me. =)

How many different 5- card poker hands can be dealt that contains no aces from a regular 52-card deck?

How many different 5- card poker hands can be dealt that contains "a" aces (for a = 0 to 4) from a regular 52-card deck?

### 1 Answer

- MorewoodLv 79 years agoBest Answer
The easiest way to do this is to use "combinations" along with the "fundamental counting principle".

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Combinations (nCr on your calculator, =Combin(,) on your spreadsheet, "CHOOSE" in writing) tell you how many ways you can select a subset of a given size from a set of a given size. EXAMPLE: If I have 562 facebook friends, but I can only invite 27 of them to my birthday party, there are 562 CHOOSE 27 ways {or (562)nCr(27) or =Combin(562,27)} to select which friends to invite.{Remember to put the bigger number first or you will get an error message - how could you choose 562 friends from a list of 27?!?}

In your problem (second version, which contains the first version as a special case), how many aces are there? How many aces do you want to choose? How many ways to do that?

Also, how many NON-aces in that deck? How many NON-aces do you need to choose (for a total of 5 cards)? How many ways to do that? {This is the entire solution to the first version of your problem.}

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The fundamental counting principle tells you to multiply when you want to combine two independent choices. If every possibility for the first choice can be combined with every possible combination for the second choice, then you can make a table (choice 1 in columns, choice 2 in rows) and the number of entries is the product of the number columns and the number of rows. EXAMPLE: My mother gives me a list of 5 chaperones, exactly one of which must be "invited" to my party (none of whom are facebook friends). So now I have 5×(562CHOOSE27) ways to send out my invites.

You have to choose which aces and which non-aces to put in your 5-card poker hand, which can be done independently since all non-aces are not aces.

Source(s): Read more about the fundamental counting principle: http://www.basic-mathematics.com/fundamental-count... and about combinations: http://en.wikipedia.org/wiki/Combination