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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.

Re: Probability cards [#permalink]
02 Jun 2010, 23:41

Bunuel, thank you for your answer.

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.

Re: Probability cards [#permalink]
04 Jun 2010, 11:18

nonameee wrote:

Bunuel, thank you for your answer.

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.

Re: Probability cards [#permalink]
04 Jun 2010, 12:09

Expert's post

edoy56 wrote:

nonameee wrote:

Bunuel, thank you for your answer.

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. _________________

Re: Probability cards [#permalink]
07 Jun 2010, 10:34

confused

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....?

Re: Probability cards [#permalink]
07 Jun 2010, 10:53

1

This post received KUDOS

Expert's post

ssgomz wrote:

confused

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 \(4*\frac{13*12*11*10}{52*51*50*49}\).

Re: Probability cards [#permalink]
13 Jun 2010, 02:18

Expert's post

virupaksh2010 wrote:

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).

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