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Senior Manager
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Probab - Deck of cards [#permalink]
 19 Jun 2007, 15:09
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
How many ways can one choose 6 cards from a normal deck of cards (13*4) in order 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
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Director
Joined: 13 Mar 2007
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Hmm I am getting - 13^4 x 24 x 47
Four suits - 13c1 x 13c1 x 13c1 x 13c1
Remaining two cards - 48c2
=> 13^4 x 24 x 47 !
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Manager
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I agree with grad_mba Should be 13^4 X24X 47
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Senior Manager
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Re: Probab - Deck of cards [#permalink]
19 Jun 2007, 23:47
[quote="Andr359"]How many ways can one choose 6 cards from a normal deck of cards (13*4) in order 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[/quote]
Agree with grad_mba and vijay2001.
The correct answer seems to be missing !!!
It should be (13C1)^4 * 48C2 = (13^4) * 24 * 47.
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Senior Manager
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A is the answer.. I guess if it is the probability of it happening just devided the number by 52C6 (no. of combinations)
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Director
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i dont understand ..can someone explain ?
If it is veiwed as -
#ways for picking 1st card = 13
2nd - 13
3rd - 13
4th - 13
5th - 48
6th - 47
yes, choice A seems to work.
But what was wrong in my 1st approach ?!
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Director
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There are 52 cards in a deck and 13 cards in a suit.
imagine 6 slots:
Slot 1: 13 different cards for clubs
51 cards left to pick from - Slot 2: 13 different cards for hearts
50 cards left to pick from - Slot 3: 13 different cards for aces
49 cards left to pick from - Slot 4: 13 different cards for spades
***at this point we have 1 card from each suit, now there are no constraints and we can choose whatever cards we like***
Slot 5: 48 cards left to pick from
Slot 6: 47 cards left to pick from
13^4(48)47
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Manager
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Ahh, Its is a permutation question(ways) not combinations. i guess answer is A
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close
