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# In how many ways can one choose 6 cards from a normal deck

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Director
Joined: 19 Nov 2004
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In how many ways can one choose 6 cards from a normal deck [#permalink]  06 Jan 2005, 09:55
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In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present

A)13^4 x 48 x 47
B) 13^4 x 24 x 47
C) 48C6
D) 13^4
E) 13^4 x 48C6

Please explain your reasoning clearly.
Manager
Joined: 31 Aug 2004
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Location: Vancouver, BC, Canada
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[#permalink]  06 Jan 2005, 12:58
Hmm...I am weak at solving this kind of question. Pls help!

Director
Joined: 07 Nov 2004
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[#permalink]  07 Jan 2005, 12:12
Pick B.

We need one from each deck, and then the remaining two can be any two of the remaining 48 cards.
i.e: 13C1*13C1*13C1*13C1*48C2 = 13^4*24*47
Intern
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[#permalink]  08 Jan 2005, 05:54
gayathri wrote:
Pick B.

We need one from each deck, and then the remaining two can be any two of the remaining 48 cards.
i.e: 13C1*13C1*13C1*13C1*48C2 = 13^4*24*47

Can you please explain why it is not 13^4*48*47

Thanks
Director
Joined: 07 Nov 2004
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[#permalink]  08 Jan 2005, 08:01
Quote:
Can you please explain why it is not 13^4*48*47

Thanks

If you do 48*47, you are saying that the order in which you pick the cards matter. This is a combination problem because you just have to pick any two cards, the order in which they appear does not matter.
[#permalink] 08 Jan 2005, 08:01
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# In how many ways can one choose 6 cards from a normal deck

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