Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\); Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\)); Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:

We can choose any card for the first one - \(\frac{52}{52}\); Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left); Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left); Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T

Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T

Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.

Why wouldn't the probability be 1 * 13/51 * 13/50 * 13/49 ? We know there are 4 suits, and the probability of getting any card in one suit is 13/# of cards in the deck.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\); Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\)); Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:

We can choose any card for the first one - \(\frac{52}{52}\); Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left); Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left); Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

Hope it helps.

No posting of PS/DS questions is allowed in the main Math forum.[/warning]

In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T

Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.

Why wouldn't the probability be 1 * 13/51 * 13/50 * 13/49 ? We know there are 4 suits, and the probability of getting any card in one suit is 13/# of cards in the deck.

edoy56 wrote:

In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.

Yes you are wrong.

First of all suit and denomination are not the same: there are four suits (Clubs, Diamonds, Hearts, and Spades), and 13 different denominations (Ace, two through ten, Jack, Queen, and King).

Next, even if you are counting the probability that the 4 cards you draw are of different suits, 1 * 13/51 * 13/50 * 13/49 won't be the correct answer.

Correct answer would be \(P=\frac{(C^1_{13})^4}{C^4_{52}}\).

Or another way: \(P=4!*\frac{13}{52}*\frac{13}{51}*\frac{13}{50}*\frac{13}{49}\) (the same probability as above), we are multiplying by 4! as the favorable scenario of Clubs, Diamonds, Hearts, and Spades (CDHS) can occur in 4! # of ways, which is basically the # of permutaions of 4 letters CDHS.

Hope it helps.

P.S. This question is beyond the scope of GMAT, so I don't recommend to spend too much time on it.
_________________

why is it multiplied by 4^4 and not 4!? if the first card is taken from 1 suit the second can be from the remaining 3 because 1 suit will have only one card of one denomination....?

why is it multiplied by 4^4 and not 4!? if the first card is taken from 1 suit the second can be from the remaining 3 because 1 suit will have only one card of one denomination....?

We need 4 cards of different denomination, but these 4 cards don't have to be of the different suits.

Consider this: \(C^4_{13}\) is the # of ways to choose 4 cards of different denomination. For example one particular case from \(C^4_{13}\) would be: Jack - Queen - King - Ace.

Now, each card from this case (Jack - Queen - King - Ace) can be of 4 suits: Clubs, Diamonds, Hearts, and Spades. So taking into account the suits the case Jack - Queen - King - Ace can be formed \(4*4*4*4=4^4\) # of times. As there are \(C^4_{13}\) # of cases like Jack - Queen - King - Ace, so total # of ways to form 4 cards of different denomination is \(C^4_{13}*4^4\).

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\); Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\)); Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:

We can choose any card for the first one - \(\frac{52}{52}\); Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left); Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left); Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

How did \(\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}\) become equal to \(\frac{4^4*13*12*11*10}{52*51*50*49}\) here?

Take 4 common from the numerator and the result would be \(4*\frac{13*12*11*10}{52*51*50*49}\).

I think we we should not multiply it with 4 since we are not finding number of ways just probability so order doesnot matter..

13/52 * 13/51*13/49*13/48 should be the answer ...

Please correct me if i am wrong.

Your approach is not correct: First of all \(Probability \ of \ an \ event =\frac {# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\), so we do need to find the # of ways and if order matters for # of ways of an event, then order matters for probability too.

Answer for the original question is \(C^4_{52}=\frac{4^4*13*12*11*10}{52*51*50*49}\) ( so we are not multiplying 13/52 * 13/51*13/49*13/48 by 4 we are multiplying by 4^4), please see the solution in my first post.

If you are referring to the different question we've considered (the probability that the 4 cards you draw are of different suits), the answer for it is still not 4*13/52 * 13/51*13/49*13/48 it's \(P=\frac{(C^1_{13})^4}{C^4_{52}}=4!*\frac{13}{52}*\frac{13}{51}*\frac{13}{50}*\frac{13}{49}\), so again we are not multiplying by 4 we are multiplying by \(4!=1*2*3*4=24\), please see the solution for this question in my second post.

For more on this issue please check Probability and Combinatorics chapters of Math Book (link in my signature).

Re: Four cards are randomly drawn from a pack of 52 cards. Find [#permalink]

Show Tags

28 Nov 2015, 12:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...