Yahoo Answers: Answers and Comments for Probability Question (Easy)? [Mathematics]
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From Kung Fu Champion
enCA
Thu, 20 Jan 2011 19:05:49 +0000
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Yahoo Answers: Answers and Comments for Probability Question (Easy)? [Mathematics]
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https://ca.answers.yahoo.com/question/index?qid=20110120190549AAuLSgG
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From Morewood: The easiest way to do this is to use "com...
https://ca.answers.yahoo.com/question/index?qid=20110120190549AAuLSgG
https://ca.answers.yahoo.com/question/index?qid=20110120190549AAuLSgG
Tue, 25 Jan 2011 19:56:34 +0000
The easiest way to do this is to use "combinations" along with the "fundamental counting principle".

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 NONaces in that deck? How many NONaces 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.}

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 nonaces to put in your 5card poker hand, which can be done independently since all nonaces are not aces.