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# Probab - Deck of cards

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Senior Manager
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Probab - Deck of cards [#permalink]

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19 Jun 2007, 14:09
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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
Director
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19 Jun 2007, 18:09
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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19 Jun 2007, 19:32
I agree with grad_mba Should be 13^4 X24X 47
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Re: Probab - Deck of cards [#permalink]

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19 Jun 2007, 22:47
[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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20 Jun 2007, 12:42
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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20 Jun 2007, 19:28
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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21 Jun 2007, 18:40
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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21 Jun 2007, 18:59
Ahh, Its is a permutation question(ways) not combinations. i guess answer is A
21 Jun 2007, 18:59
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# Probab - Deck of cards

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