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P and C

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Intern
Intern
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Joined: 22 Jun 2010
Posts: 18

Kudos [?]: 10 [0], given: 1

P and C [#permalink]

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New post 26 Oct 2011, 18:45
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 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

I have one basic question. For the problem above..

13c1 * 13c1 * 13c1 * 13c1 ... we need to have 2 cards.. when we choose 2 cards is the order in which we choose the cards is important?? like taking 2 and 5 is different from 5 and 2 ??? why is it 48 * 47 instead of 48c2 ??

Kudos [?]: 10 [0], given: 1

Manager
Manager
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Joined: 19 Oct 2011
Posts: 82

Kudos [?]: 95 [0], given: 1

Re: P and C [#permalink]

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New post 30 Oct 2011, 17:13
It says how many ways, so I think that orders do matter in this case. Otherwise, the answer should be totally revised, and the order of the first four cards should be considered.

Kudos [?]: 95 [0], given: 1

Re: P and C   [#permalink] 30 Oct 2011, 17:13
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