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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

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by shashank.ism » Tue Feb 09, 2010 1:16 pm

In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

(13^4) x 48 x 47
(13^4) x 27 x 47
48C6
(13^4)
(13^4) x 48C6

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sanjayism: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Sat Oct 31, 2009 10:52 am; Location: rourkela; Thanked: 1 times

by sanjayism » Wed Feb 10, 2010 10:58 pm

I am confused with answer choices;

answer may be
13^4*48C2

kumar sanjay

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Wed Feb 10, 2010 11:13 pm

sanjayism wrote:I am confused with answer choices;

answer may be
13^4*48C2

52 cards in a deck -13 cards per suit
First card - let us say from suit hearts = 13C1 =13
Second card - let us say from suit diamonds = 13C1 =13
Third card - let us say from suit spade = 13C1 =13
Fourth card - let us say from suit clubs = 13C1 =13
Remaining cards in the deck= 52 -4 = 48
Fifth card - any card in the deck = 48C1
Sixth card - any card in the deck = 47C1
Total number of ways = 13 * 13 * 13 * 13 * 48 * 47 = 134 *48*47

so ANS is A

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Ian Stewart: GMAT Instructor; Posts: 2621; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Thu Feb 11, 2010 12:22 am

I'm not sure where this question is from, but it has been posted on this forum before, and it is a mess; none of the answer choices are correct. It is a much harder question than it may appear at first (and is *way* out of scope for the GMAT), and most of the 'solutions' I've seen posted to it, including the one above, dramatically overcount the number of possibilities. In another thread, Brent Hanneson has given an excellent explanation of this:

www.beatthegmat.com/choose-6-cards-from ... 45411.html

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Thu Feb 11, 2010 1:25 am

Ian Stewart wrote:I'm not sure where this question is from, but it has been posted on this forum before, and it is a mess; none of the answer choices are correct. It is a much harder question than it may appear at first (and is *way* out of scope for the GMAT), and most of the 'solutions' I've seen posted to it, including the one above, dramatically overcount the number of possibilities. In another thread, Brent Hanneson has given an excellent explanation of this:

www.beatthegmat.com/choose-6-cards-from ... 45411.html

Thanks Ian, Brent has given very good explanation of the question.

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komal: Legendary Member; Posts: 777; Joined: Fri Jan 01, 2010 4:02 am; Location: Mumbai, India; Thanked: 117 times; Followed by:47 members

by komal » Tue Feb 16, 2010 10:32 am

shashank.ism wrote:In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

(13^4) x 48 x 47
(13^4) x 27 x 47
48C6
(13^4)
(13^4) x 48C6

4 cards from 4 suits = 13^4

Now the 5th card can be drawn in 48 ways and the 6th one in 47 ways

So the number of ways = 13^4 * 48 * 47

(A) is correct

https://www.e-gmat.com/

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Ian Stewart: GMAT Instructor; Posts: 2621; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Tue Feb 16, 2010 12:11 pm

komal wrote:
shashank.ism wrote:In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

(13^4) x 48 x 47
(13^4) x 27 x 47
48C6
(13^4)
(13^4) x 48C6
4 cards from 4 suits = 13^4

Now the 5th card can be drawn in 48 ways and the 6th one in 47 ways

So the number of ways = 13^4 * 48 * 47

(A) is correct

That is the solution I've seen posted in several threads in response to this question, and it is not correct, though it may appear to be at first. There are several problems with it:

* your first four cards do not need to be of different suits. We could pick, for example, three Spades first, then a Club, Heart and Diamond, and end up with all four suits

* the order of the selection does not matter here; that needs to be accounted for. Otherwise you are counting the same selections multiple times. See Brent's explanation at the link I posted above.

The correct answer to this question is not among the answer choices provided, and in any case, it is well beyond the scope of the GMAT.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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jeffedwards: Master | Next Rank: 500 Posts; Posts: 109; Joined: Mon Oct 12, 2009 4:58 am; Thanked: 12 times; Followed by:1 members; GMAT Score:720

by jeffedwards » Tue Feb 16, 2010 12:17 pm

I agree, I got A[/b]

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komal: Legendary Member; Posts: 777; Joined: Fri Jan 01, 2010 4:02 am; Location: Mumbai, India; Thanked: 117 times; Followed by:47 members

by komal » Tue Feb 16, 2010 7:14 pm

Ian Stewart wrote:
That is the solution I've seen posted in several threads in response to this question, and it is not correct, though it may appear to be at first. There are several problems with it:

* your first four cards do not need to be of different suits. We could pick, for example, three Spades first, then a Club, Heart and Diamond, and end up with all four suits

* the order of the selection does not matter here; that needs to be accounted for. Otherwise you are counting the same selections multiple times. See Brent's explanation at the link I posted above.

The correct answer to this question is not among the answer choices provided, and in any case, it is well beyond the scope of the GMAT.

Yes i learnt this method from one of ur posts. Dint find any problem with it until u mentioned it above. Thank you very much.

https://www.e-gmat.com/

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