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# Probab - Deck of cards

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Senior Manager
Joined: 24 Nov 2006
Posts: 351
Followers: 1

Kudos [?]: 16 [0], given: 0

Probab - Deck of cards [#permalink]  19 Jun 2007, 14:09
00:00

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(N/A)

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
Director
Joined: 13 Mar 2007
Posts: 545
Schools: MIT Sloan
Followers: 4

Kudos [?]: 28 [0], given: 0

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 !
Manager
Joined: 28 Aug 2006
Posts: 160
Followers: 2

Kudos [?]: 13 [0], given: 0

I agree with grad_mba Should be 13^4 X24X 47
Senior Manager
Joined: 04 Jun 2007
Posts: 348
Followers: 1

Kudos [?]: 16 [0], given: 0

Re: Probab - Deck of cards [#permalink]  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]

The correct answer seems to be missing !!!

It should be (13C1)^4 * 48C2 = (13^4) * 24 * 47.
Senior Manager
Joined: 21 Jun 2006
Posts: 287
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Kudos [?]: 34 [0], given: 0

A is the answer.. I guess if it is the probability of it happening just devided the number by 52C6 (no. of combinations)
Director
Joined: 13 Mar 2007
Posts: 545
Schools: MIT Sloan
Followers: 4

Kudos [?]: 28 [0], given: 0

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 ?!
Director
Joined: 12 Jun 2006
Posts: 534
Followers: 1

Kudos [?]: 34 [0], given: 1

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
Manager
Joined: 28 Aug 2006
Posts: 160
Followers: 2

Kudos [?]: 13 [0], given: 0

Ahh, Its is a permutation question(ways) not combinations. i guess answer is A
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