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Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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29 Jul 2012, 10:11

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Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240 B. 960 C. 120 D. 40 E. 5760

I tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}

However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. } Using permutations - 8* 1*6*4= 48*4...(crash )

I see that in both the examples, the "Combinations" number is "Permutations/4"...Is there a rule to quickly relate Permutations with Combinations?

I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240 B. 960 C. 120 D. 40 E. 5760

I tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}

However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. } Using permutations - 8* 1*6*4= 48*4...(crash )

I see that in both the examples, the "Combinations" number is "Permutations/4"...Is there a rule to quickly relate Permutations with Combinations?

I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value? A. 8/33 A. 62/165 C. 17/33 D. 103/165 E. 25/33

Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \(\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\).

\(C^4_6\) - # of ways to choose 4 different cards out of 6 different values; \(2^4\) - as each of 4 cards chosen can be of 2 different suits; \(C^4_{12}\) - total # of ways to choose 4 cards out of 12.

So \(P=1-\frac{16}{33}=\frac{17}{33}\).

Or another way:

We can choose any card for the first one - \(\frac{12}{12}\); Next card can be any card but 1 of the value we'v already chosen - \(\frac{10}{11}\) (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left); Next card can be any card but 2 of the values we'v already chosen - \(\frac{8}{10}\) (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left); Last card can be any card but 3 of the value we'v already chosen - \(\frac{6}{9}\);

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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29 Jul 2012, 15:39

Bunuel, Thanks for finding a similar question. However, I am not sure how to calculate the number of ways to obtain "a pair" using PErmutations. The question that you have posted is different - it asks us to calculate probability. This one is different because it asks us to calculate "combinations." I see that the author has just modified the same question. The source of this question is 700-800 series by MGMAT.....

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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31 Jul 2012, 19:49

1) how many possible pairs? ==> 6

2) for each pair we need to find the no. of combinations of the other two cards. Since 2 cards are already chosen from the deck (the pair), there are 10 remaining cards.

C(10,2) = 45

3) 6*45 = 270 => this is no. of combinations of 4 cards that contain at least one pair. But this also includes the combinations with 2 pairs that should be eliminated.

4) how many combinations with 2 pairs? 6* C(5,1) = 30

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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05 Aug 2012, 04:22

manishgeorge wrote:

Thanks for the quick response, how ever how did you arrive are step 4?

I understand the point that there are combinations with 2 pairs that needs to be eliminated.But I didnt understand the formula

4) how many combinations with 2 pairs? 6* C(5,1) = 30

Consider it like this-

We have to eliminate the case where two pairs are selected.

So C(6,1) is how u chose the first pair. Remaining number of pairs are 5. So if u choose 2 cards that are of same number, then C(5,1) are the number of ways to do it. Therefore C(6,1)*C(5,1)=30

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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05 Aug 2012, 06:18

voodoochild wrote:

Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240 B. 960 C. 120 D. 40 E. 5760

I tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}

However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. } Using permutations - 8* 1*6*4= 48*4...(crash )

I see that in both the examples, the "Combinations" number is "Permutations/4"...Is there a rule to quickly relate Permutations with Combinations?

I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

Correct?

If "How many such combinations are possible?" refers to 4 cards containing at least one pair of cards with the same value, then the correct answer is not listed.

The total number of possibilities to chose any 4 cards, regardless the order in which they were chosen, is given by \(12C4=\frac{12*11*10*9}{4!}=\frac{990}{2}=495\). The number of possibilities to chose a set of 4 cards containing no pair of equal numbers, regardless the order in which they were chosen, is given by \(12*10*8*6/4!=240\). First card 12 possibilities, second any card with a number different from the first one, so \(12-2 =10\) possibilities, for the third one \(12-4=8\) possibilities, and for the fourth card \(12-6=6\) possibilities. Divide by \(4!\), since order doesn't matter.

So, the number of possibilities to get at least one pair of cards with equal numbers among the chosen 4 is \(495 - 240 = 255\), which is not listed as an answer!

In its present form, this cannot be a real GMAT test question, the wording is really poor.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible? (A) 240 (B) 960 (C) 120 (D) 40 (E) 5760

I would use the FCP to solve this problem.

First two cards = the pair = there are six ways to get a pair. Then two cards from the remaining 10 ---> 10C2 = 45 6*45 = 270

BUT, now we have to eliminate things that have been counted twice. What has been counted twice? The two pair combos. If I pick, say, two 4's as my first two cards, and then I happen to get two 2's as my last two cards, that will result in the same hand, the same combination, as if I pick the pair of 2's first and then the pair of 4s. How many pairs of pairs are there?

(6 choices for the first pair)*(5 for the remaining pair) = 30.

So, the 30 two-pair combinations have been counted twice, so subtract that number once.

270 - 30 = 240, the OA.

voodoochild wrote:

I tried the above problem using Combinations, and I got the correct answer which is A.

However, I crashed while using Permutations.

You know, from my perspective, this is analogous to someone saying --- When I translated the French sentence with a French dictionary, it was no problem, but when I translated the French sentence with a German dictionary, I had all kinds of problems!

Combinations and permutations are two very different things. If a problem, such as this one, is about combinations --- that is to say, only the final grouping matters, and order does not matter in the least --- then a combinations approach is correct and permutations approach is not correct, unless you do something to the permutations, like nPr/r!, to make it a combinations approach in disguise.

Here's my advice. If the problem is about combinations, don't use permutations. Period.

voodoochild wrote:

Is there a rule to quickly relate Permutations with Combinations? I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

nCr = nPr/r! is always true, absolutely true, 100% of the time. I believe you calculated the permutations incorrectly --- to be honest, I have no idea what you are doing in your permutations calculations: I do not recognize them either as permutations or as anything related to this problem. The underlying problem I believe is, again, this problem is by its very nature of combination problem. It's not at all clear to me how permutations would be the least bit helpful in solving it. It's like trying to use German to translate a French sentence.

Does all this make sense?

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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07 Aug 2012, 11:29

In my previous post I interpreted the question as the number of possibilities to get at least one pair, which in fact means either one or two pairs. I got 255 possibilities.

The number of possibilities to get two pairs is 6C2=6*5/2=15, we just have to choose two pairs out of 6. Obviously, the number of possibilities for exactly one pair is 255 - 15 = 240.

Here, I am presenting an approach to obtain directly this number:

For the one pair, we have 6 possibilities to choose from. For the other two cards, not to form a pair: we have for the first card 10 possibilities, then for the second 8 possibilities (one pair already chosen, and the card cannot be the pair of the previous one). This would give 10*8=80 possibilities, but we have to divide by 2, as order doesn't matter, and we are left we 80/2 = 40.

Thus, the final number of possibilities is 6*40 = 240.

Answer A
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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08 Aug 2012, 09:04

mikemcgarry wrote:

You know, from my perspective, this is analogous to someone saying --- When I translated the French sentence with a French dictionary, it was no problem, but when I translated the French sentence with a German dictionary, I had all kinds of problems!

Combinations and permutations are two very different things. If a problem, such as this one, is about combinations --- that is to say, only the final grouping matters, and order does not matter in the least --- then a combinations approach is correct and permutations approach is not correct, unless you do something to the permutations, like nPr/r!, to make it a combinations approach in disguise.

Mike, Thanks for your elaborate reply. I believe that the analogy drawn by you is not 100% accurate. Mathematics and language are completely different. 2+2 = 4 everywhere in the Universe. Language is fluid. I see your point about Permutations and Combinations. The Combinations solution is computable. However, I wanted to "test" my permutations skills by trying the same problem with Permutations. I would really appreciate if you could help me with an alternate solution.

Here's what I could get : Total number of permutations = 12*1*10*8=960 ways to choose a-pair. Now, these cards could be arranged in 4!/(2!*2!) = 6.

hence, total number of permutations = 960*6. Now the correct answer = 960*6/24. I am not sure why we are supposed to divide by 4! = 24.

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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08 Aug 2012, 10:31

voodoochild wrote:

mikemcgarry wrote:

You know, from my perspective, this is analogous to someone saying --- When I translated the French sentence with a French dictionary, it was no problem, but when I translated the French sentence with a German dictionary, I had all kinds of problems!

Combinations and permutations are two very different things. If a problem, such as this one, is about combinations --- that is to say, only the final grouping matters, and order does not matter in the least --- then a combinations approach is correct and permutations approach is not correct, unless you do something to the permutations, like nPr/r!, to make it a combinations approach in disguise.

Mike, Thanks for your elaborate reply. I believe that the analogy drawn by you is not 100% accurate. Mathematics and language are completely different. 2+2 = 4 everywhere in the Universe. Language is fluid. I see your point about Permutations and Combinations. The Combinations solution is computable. However, I wanted to "test" my permutations skills by trying the same problem with Permutations. I would really appreciate if you could help me with an alternate solution.

Here's what I could get : Total number of permutations = 12*1*10*8=960 ways to choose a-pair. Now, these cards could be arranged in 4!/(2!*2!) = 6.

hence, total number of permutations = 960*6. Now the correct answer = 960*6/24. I am not sure why we are supposed to divide by 4! = 24.

Help?

Because when you have chosen your cards, you took into account order: first card, second card,... So, for a given set of 4 cards ABCD, you counted all possible permutations ACBD, ADBC, ... That's why you have to divide by 4! which is exactly the number of different arrangements of the 4 cards.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

-1- For first two places select a number out of 6 to form a pair = 6C1=6 >> e.g 1,1 2,2 3,3 4,4 5,5 6,6 Lets say 1,1 is selected for first pair.

-2- For remaining two places select two cards from remaining 10 = 10C2 = 45 Suit 1 : 2 | 3 | 4 | 5 | 6 Suit 2 : 2 | 3 | 4 | 5 | 6

Because question restrict two pairs in 4 drawn cards, thus we will have to subtract those pairs selected in 10C2 i.e 2,2 3,3 4,4 5,5 6,6 =45 - 5 = 40

Thus total combinations are 6 * 40 = 240 --> A
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

-1- For first two places select a number out of 6 to form a pair = 6C1=6 >> e.g 1,1 2,2 3,3 4,4 5,5 6,6 Lets say 1,1 is selected for first pair.

-2- For remaining two places select two cards from remaining 10 = 10C2 = 45 Suit 1 : 2 | 3 | 4 | 5 | 6 Suit 2 : 2 | 3 | 4 | 5 | 6

Because question restrict two pairs in 4 drawn cards, thus we will have to subtract those pairs selected in 10C2 i.e 2,2 3,3 4,4 5,5 6,6 =45 - 5 = 40

Thus total combinations are 6 * 40 = 240 --> A

Hey I think you got this a bit wrong.. The question never restricts two pairs..

45*6= 270 Subtract the cases such as 2,2,4,4 and 4,4,2,2 or 3,3,5,5 & 5,5,3,3 ..this 2 are the same combinations and have been double counted.. Number of such cases: 6C1 * 5C1=30 240... Your method works coz u rejected the dual pairs..But it works for a different reason altogether
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Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

-1- For first two places select a number out of 6 to form a pair = 6C1=6 >> e.g 1,1 2,2 3,3 4,4 5,5 6,6 Lets say 1,1 is selected for first pair.

-2- For remaining two places select two cards from remaining 10 = 10C2 = 45 Suit 1 : 2 | 3 | 4 | 5 | 6 Suit 2 : 2 | 3 | 4 | 5 | 6

Because question restrict two pairs in 4 drawn cards, thus we will have to subtract those pairs selected in 10C2 i.e 2,2 3,3 4,4 5,5 6,6 =45 - 5 = 40

Thus total combinations are 6 * 40 = 240 --> A

Hey I think you got this a bit wrong.. The question never restricts two pairs..

45*6= 270 Subtract the cases such as 2,2,4,4 and 4,4,2,2 or 3,3,5,5 & 5,5,3,3 ..this 2 are the same combinations and have been double counted.. Number of such cases: 6C1 * 5C1=30 240... Your method works coz u rejected the dual pairs..But it works for a different reason altogether

but question clearly says a pair of cards and that does not include 2 pairs.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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27 Jun 2014, 12:02

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The above explanations are not clear to me. I have tried hard to make myself understand but I am failed in doing that. Kindly make me understand in another possible way.

The above explanations are not clear to me. I have tried hard to make myself understand but I am failed in doing that. Kindly make me understand in another possible way.

Thank You in advance

Dear deya, I'm happy to respond. My friend, with all due respect, this is not a very good question. Make you understand? I understand the problem, but how can I "make you understand" if I know nothing about you? Please explain, in as much detail as possible: (1) what you do understand about this problem? (2) what steps or statements in the above solutions do you not understand? (3) what did you think of and what did you try when you approached the problem? Part of being a successful GMAT student is learning to ask excellent questions. It takes a great deal more work, but if you put in the effort to ask excellent questions, it will actually prime your mind to understand more deeply. Learning is not just about getting the information: much more importantly, learning is about preparing you mind to absorb the information. Does all this make sense? Mike
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Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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02 Jul 2014, 23:41

Dear mikemcgarry, Thanks for the quick repy and I am apologizing for killing you time. I went through the question again and this time it is clear to me. Previously I couldnt make out that why we need to subtract 30 from 270. Actually here we just need to find the combination of choices if there are only one pair of same value. I was reading it atleast. So, my mistake. Thanks again.

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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08 Oct 2014, 19:24

this is how i saw it .... out of 4 cards chosen there has to be only one pair

so pair can be chosen (1,1) or (2,2) ...or (6,6)

this means slecting 1 from first group and 1 from second group how can this be done

from the first group one card can be chosen in ( 6 ways..1or 2 or 3 ....6 ) now to make a pair we need the similar card from grp 2 ( which can be done only in one way ) ............... so now we have got our one pair ) and we have chosen ( 2 cards) now remaining cards( 10 ) ....we still need to pick up two cards.. how can this be done ......

possiblity one both the cards are chosen from gr 1 i.e 5c2 = 10 ( only 5 cards are left bcz we have already chosen one card from each gr to make the pair)

possiblity two both cards are chosen from 2nd grp 5c2 =10

possiblity 3 one card is chosen from gr 1 and second from grp 2

now out of 5 cards ( 1 can be chosen in 5 ways) lets say we chose ( 4)

now from the second grp we can not chose 4 bcz that will make a pair ( so no of ways we can chose 2nd card from gr2 = 4ways)

so no of ways of chosing 2 cards( ne from each such that no ar is formed = 5*4 =20

so final ans possible ways of chosing 1 pair * ( possinlity 1 OR possiblity 2 ORpossiblity 3)

Re: Randolph has a deck of 12 playing cards made up of only 2 [#permalink]

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09 Oct 2014, 14:08

Hello, I'm stuck here:

I did 12 * 1*10*8 Number of ways of picking any card out of 12 (first blank), then picking it's pair for the second blank (that's one). Then picking any card from the ten remaining, and then picking any other card making sure that it's not the 3rd card's pair.

I got 960, and I understand that in this method position is imp, whereas in the question it is not. I'm wondering how do I factor this out? Thanks!