I could use some help figuring out how to solve this

I can't remember

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May 30, 2008
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I have a question that I do not know exactly how to go about solving.

Eight cards are drawn from a standard deck of 52 cards.
How many 8-card hands, having 5 cards of one suit and 3 cards
of another suit can be formed?

I am not exactly sure how to go about solving this problem. Could someone please take me through the steps of solving this problem?
 
\(\displaystyle \left[ {4\binom {13}{5}} \right]\left[ {3\binom {13}{3}} \right]\)
 
Let me make sure I understand.

If I have C(52,5)=52!/(5!(52-5)!) = 52!/(5!*47!)

and

if I have C(52,3)=52!/(3!(52-3)! = 52!/(3!*49!)

Then I simply multiply the answers to those two combinations?

Do the 13's represent the number of cards in each suit?

Where did you get the 4 and 3 from?
 
I can't remember said:
Do the 13's represent the number of cards in each suit?
Where did you get the 4 and 3 from?
There are 13 different cards in any suit.
There are 4 ways to choose the five-card suit.
There are then 3 ways to choose the three-card suit.
 
The first response I recieved had a kind of notation I am not used to. Could you explain if the 4 is supposed to multiplied by 13/5 or is it 4!(13-5)!...
 
These are simple combinations.
\(\displaystyle \binom {13}{5} = \frac{{13!}}{{5!\left( {13 - 5} \right)!}} = \frac{{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}}{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}\).
 
I got 4,416,984 ways, is this correct. I multiplied the number of ways (4) by the first combination to get 5,148, then the number of ways (3) times the second combination to get 858, then I multiplied them together. Is that correct?
 
I can't remember said:
I got 4,416,984 ways, is this correct. I multiplied the number of ways (4) by the first combination to get 5,148, then the number of ways (3) times the second combination to get 858, then I multiplied them together. Is that correct?
Yes, I also got 4416984
 
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