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a probability problem, who can help? Thanks!

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a probability problem, who can help? Thanks!

by billzhao » Tue Feb 03, 2009 8:27 pm
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

my logic is: there are 52 choices for the 1st card, 39 choices for the 2nd card, 26 choices for the 3rd, 13 chocies for the 4th, 48 choices for the 5th and 47 choices for the 6th. so altogether there are 13^4*4!*48*47.

The answer for the problem is 13^4*48*47.

Which answer is correct?

Thanks!
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Re: a probability problem, who can help? Thanks!

by mrsmarthi » Tue Feb 03, 2009 9:36 pm
billzhao wrote:In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

my logic is: there are 52 choices for the 1st card, 39 choices for the 2nd card, 26 choices for the 3rd, 13 chocies for the 4th, 48 choices for the 5th and 47 choices for the 6th. so altogether there are 13^4*4!*48*47.

The answer for the problem is 13^4*48*47.

Which answer is correct?

Thanks!
I think 13 ^ 4 * 48 * 47 should be correct. Same approach as yours.
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by billzhao » Wed Feb 04, 2009 1:22 am
why no 4!? the four suits can be arranged randomly in order ya?
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by fleurdelisse » Wed Feb 04, 2009 4:48 am
order doesn't matter in this case, you need to choose 6 cards at random, regardless of which you choose first.

So, in essence you need to choose one from each suit to ensure you have the four suits, that is: 13^4
and for the last two, you've got to choose 2 out of 48. So I would have said it should be combination(48,2) and therefore (48*47)/2.
So I got this answer: 13^4*24*47

Usually, if there is nothing in the directions that specifies that order matters, I assume it doesn't.

It's weird the answer doesn't account for order in the 13^4 but then it does for the last two cards.
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by fleurdelisse » Wed Feb 04, 2009 4:58 am
can you please double check the answer you have, it probably should be 13^4*24*47, you should be using combination. please let me know. thatnks!
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by billzhao » Wed Feb 04, 2009 6:32 am
The answer is:

(a）13^6*48*47

(b) 13^6*27*47

(c) 48*C6

(d) 13^4

(e) 13^4*48*C6
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by fleurdelisse » Fri Feb 06, 2009 2:11 am
where did you get this questions from? i googled the question, and got this:
https://www.urch.com/forums/gmat-math/14 ... esent.html
(typo in answer b, as mentionned in post 7 of the above link)

so as mentioned, the correct answer should be 13^4*24*47, even though it does not appear as one of your answer choices
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by sumitkhurana » Thu Feb 19, 2009 7:25 am
Can someone find a flaw in this explaination:-

Since the order in which cards are drawn is not mentioned, we can assume that they are all drawn together. So:-

1st possibilty :-
Assuming 1 card from each and the remaining 2 cards from one of the suites.

So 13C1 * 13 C1 * 13C1 * 13 C3, there are 4 ways to find the suite from where 3 cards would be picked so :-

4*13*13*13*13C3 = 88 * 13^4 - (1)

Now, 2nd possibility:-

Assuming 2 cards extra cards are picked from 2 diff suites.

So 13C1 * 13 C1 * 13C2 * 13 C2 * 4 C2 ( ways to pick these 2 suites )

= 216 * 13^4 (2)

Total (1) + (2) = 304 * 13^4

Please explain why this logic is incorrect.
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by Rashmi1804 » Mon Mar 02, 2009 8:49 am
I too approached the problem the same way as sumit did.

but none of the options given show the answer i got!!

However, the below post seems to agree with my approach.

https://www.beatthegmat.com/telling-perm ... 30354.html

yet, Could anybody explain whats wrong with this approach or the approach given in the original file.
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