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Dan
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10-card hands 11 Sep 2004, 03:44
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A deck of cards has 13 cards of each suit. How many 10-card hands having 3 cards of one kind and 2 cards of another are there? (expressed in scientific notation).



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10-card hands 11 Sep 2004, 08:06
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A deck of cards has 13 cards of each suit.
Understood.

How many 10-card hands
What is a 10-card hands?

having 3 cards of one kind and 2 cards of another are there?
When you say "kind", do you mean suit?

(expressed in scientific notation).
Do you mean Combination (nCr) notation?

Please post your questions completely and I assure you will get a ton of replies :)
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10-card hands 12 Sep 2004, 22:28
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I think you are stumbled in this question like I am, but the fact is that it's complete as is. Scienitific notation is like 7.2 x 10^3.
10-card hands I guess are hands holding 10 cards only.
I have the OA but the logic did not come to me.
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lovely
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10-card hands 13 Sep 2004, 09:35
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Dan
A deck of cards has 13 cards of each suit. How many 10-card hands having 3 cards of one kind and 2 cards of another are there? (expressed in scientific notation).
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I have the number of different 10 card hands as being:
52!/5!(47!) = 1.582 x 10^10

then you have to find the number of hands that have at least three cards of one suit and two of another...so I would say then multiplying by 10c3, and 10c2: so the final answer would be:

(1.582 x 10^10) x (10!/3!(7!)) x (10!/2!(8!)) =

8.542813 x 10^13

have no idea if this is correct
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10-card hands 14 Sep 2004, 05:52
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close! OA is

13(4C3) x 12(4C2) x 44C5 = 4.1 x 10^9.
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10-card hands 14 Sep 2004, 06:48
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There is no word 'atleast' in the query. I could not construe how 3cards of one kind and 2cards of another kind become 10 cards(Instead, it becomes 5). I believe, 'at least' or something like that must be present.


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Dan
A deck of cards has 13 cards of each suit. How many 10-card hands having 3 cards of one kind and 2 cards of another are there? (expressed in scientific notation).

I have the number of different 10 card hands as being:
52!/5!(47!) = 1.582 x 10^10

then you have to find the number of hands that have at least three cards of one suit and two of another...so I would say then multiplying by 10c3, and 10c2: so the final answer would be:

(1.582 x 10^10) x (10!/3!(7!)) x (10!/2!(8!)) =

8.542813 x 10^13

have no idea if this is correct
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10-card hands 14 Sep 2004, 08:25
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Oops, I understand the problem now :(

"A deck of cards has 13 cards of each suit. How many 10-card hands having 3 cards of one kind and 2 cards of another are there"

By suit, he means the 4 types (Diamond, Spade, etc.).
and by kind he means the 13 different cards (like 1, 2, ... Queen, King)

I initially thought they are interchangeable and my answer was: 4 (13C3) x 3 (13C2) x 26C5
This would be the answer, if the question was ""A deck of cards has 13 cards of each suit. How many 10-card hands having 3 cards of one suit and 2 cards of another are there"
Detail: There are 4 ways you can select a primary suit (say, Diamond). After selecting a suit, there are 3 ways to select 3 cards from 13.
Of the remaining three suits, there are 3 ways you can select a secondary suit (say, Spade). After selecting a suit, there are 2 ways to select 2 cards from 13. From here on, he is okay with selecting any random 5 from the remaining lot (26C5). Remember to keep the two selected suits away (52 - 2 x 13 = 26)

But since Suit and Kind are different: the answer is: 13 (4C3) x 12 (4C2) x 44C5
Detail: There are 13 ways you can select a primary kind (say, Card with value 1). After selecting a kind, there are 3 ways to select 3 cards from 4.
Of the remaining 12 kinds, there are 12 ways you can select a secondary kind (say, 2). After selecting a kind, there are 2 ways to select 2 cards from 4.
From here on, he is okay with selecting any random 5 from the remaining lot (44C5). Remember to keep the two selected kinds away (52 - 2 x 4 = 44)



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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