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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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18 Sep 2012, 00:17
VeritasPrepKarishma wrote: Responding to a pm:
There is a difference between this question and the other one you mentioned.
In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits.
Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead.
6C1 = ways to select the one card from 6 which will form the pair 5C2 = select two different values from the 5 remaining to form the singles *This is correct * 2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits)
Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6.
Probability = 255/495 = 17/33
Awesome , this was going to be my next question, I was going to ask that the singles can be selected first and then the pair or we could have different arrangements as the question does not say that the pair should be together ( adjacent)so we could have B3A1A5A2 where A1A2 is the pair of a single suit and B3A5 are the single cards. Below explanation already clarifies that question. Thank you for hitting the bulls eye. This was exactly what was bothering me. VeritasPrepKarishma wrote: Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator.
Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct) Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct)
Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct)
Probability = 255*4!/495*4! = 17/33 This was enlightening . EvaJager wrote: one pair 6C1*5C2(4!/2!)= 720
NO Choose one pair out of 6  6C1  and you don't care for the order in which you choose the two cards Choose two pairs out of the remaining 5 pairs  5C2  and then, from each pair, you have 2 possibilities to choose one of them, therefore 5C2*2*2 Here you stop! Don't care about any order. Period.
for two pairs 6C2 * 4!/2!2! = 90
NO You choose two pairs  6C2  and you stop here. From each pair you take both cards, and don't care about any order.
Thank you Eva .. But the question I was going to ask next to you was that, " the pair can be selected first or the singles can be selected first or the pair can be in between the 2 singles etc " , the question does not say that the pair has to be together and singles together or that that the pair has to be selected first and then the singles , so this thought was confusing me. I was basically getting confused with this sum, digitcodescombination103081.html#p802805Karishma has very clearly answered that very doubt.Just wanted to know " why I was doing , what I was doing , ". Its now clear, thanks to both of you , hope I am in a better position to understand such questions now. Karishma and Eva thank you so much for this.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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29 Apr 2013, 03:41
Bunuel wrote: maheshsrini wrote: 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? 8/33 62/165 17/33 103/165 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: \(\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\). Responding to a pm: How do we obtain \(6C4*2^4\)? Think of what you have: 6 cards numbered 1 to 6 of 2 different suits. Say you have 1 to 6 of clubs and 1 to 6 of diamonds. A total of 12 cards. You want to select 4 cards such that there is no pair i.e. no two cards have the same number. This means all 4 cards will have different numbers, say, you get a 1, 2, 4 and 6. The 1 could be of clubs or diamonds. The 2 could be of clubs and diamonds and so on for all 4 cards. This means you must select 4 numbers out of the 6 numbers in 6C4 ways. Then for each number, you must select a suit out of the given two suits. That is how you get 6C4 * 2*2*2*2 This is the total number of ways in which you will have no pair i.e. two cards of same number.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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01 May 2015, 05:54
cumulonimbus wrote: Bunuel wrote: maheshsrini wrote: 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? 8/33 62/165 17/33 103/165 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  \(\frac{6}{9}\); \(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\). So \(P=1\frac{16}{33}=\frac{17}{33}\)  the same answer as above. Answer: C. Hope it helps. Hi Bunnel Can you please correct me where am I going wrong in the below approach 1P(no pair) P(no pair) = (10x8x6)/12C4 Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on. Please correct me where i am going wrong. Thanks!!



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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03 May 2015, 22:15
Radhika11 wrote: IanStewart wrote: shrouded1 wrote: This subtlety is present in a lot of probability questions and must be clarified in the Q. The above question is in MGMAT (I got it on my CAT), but it fails to tell you if the 4 cards that are picked are picked simultaneously (i.e. order does not matter) or one by one. In my opinion such questions are ambiguous and the answer is dependent on the assumption you make. Unfortunately there are more Qs like this one on the MGMAT CATs, just something to be weary of. Whether you pick the four cards simultaneously, or pick them one at a time (without replacement) doesn't actually matter if you are finding a probability; the two situations are mathematically identical. You can see this intuitively by thinking of taking hold of four cards in the deck first. If you take them all out at the same time, or if there is a nanosecond between your removing each, why would the probability that you get a pair be affected? It won't be, so the 'ambiguity' you suggest is present in such questions is no ambiguity at all. You can see that either perspective will give you the same answer, though it's easier to illustrate with a simpler example. Say you have 3 red marbles and 4 blue marbles in a bag, and you pick two (either simultaneously, or without replacement  it's the same thing), and you want to find the probability of picking two red marbles. If we look at the problem as though we are picking marbles one at a time, we have 3*2 ways of picking two reds, and 7*6 ways of picking two marbles, so the probability would be 3*2/7*6 = 1/7. If we look at the problem as though we're picking two marbles simultaneously, we have 3C2 ways of picking two red marbles and 7C2 ways of picking two marbles, so the probability would be 3C2/7C2 = 1/7. So when you stick your two hands in the bag and grab two marbles, it doesn't matter if you lift your two hands out at the same time, or take them out one at a time; the probability is the same. Note though that you need to be consistent in the calculation  if you assume order matters when you calculate the numerator, you must also assume order matters when you calculate the denominator. Hi Can you please correct me where am I going wrong in the below approach 1P(no pair) P(no pair) = (10x8x6)/12C4 Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on. Please correct me where i am going wrong. Thanks!! You are mixing concepts. You have 12 distinct cards. There are two different methods of solving this: Method 1: The "number of ways" of selecting the first card is 12, not 1. The number of ways of selecting the second card is 10. The number of ways of selecting the third card is 8 and the number of ways of selecting the fourth card is 6. Now you have the 4 cards such that there are no pairs. What are the total number of ways of selecting 4 cards? They are 12*11*10*9 Required Probability = 1  (12*10*8*6)/(12*11*10*9) = 17/33 Note that you have arranged cards in first, second, third and fourth places. It doesn't matter as long as you arrange them in the denominator as well which you did by using 12*11*10*9 and not 12C4. Method 2: 1 is the probability of selecting the first card such that there are no pairs. It can be any card so probability of picking it is 1. In this case the second card probability is 10/11 and so on... Required Probability = 1  1*(10/11)*(8/10)*(6/9) = 17/33
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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13 Jun 2015, 20:44
Hi Guys ,can someone please tell me why this method would be incorrect? To select 4 cards from 2 different suits with each having different values ,I could go the following ways a)Select 2 each from Suit A and Suit B > 6c2*4c2(To get rest 2 cards from the 4 different values of the other suit) b)3 from A and 1 from B> 6c3*3c1 c) 1 from A and 3 from B>6c1*5c3 d)All four from A > 6c4 e)All four from B>6c4 when I add them all up I get the correct answer but I'm not sure if this is the correct approach..Please guide me!! Thanks



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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14 Jun 2015, 22:21
davesinger786 wrote: Hi Guys ,can someone please tell me why this method would be incorrect? To select 4 cards from 2 different suits with each having different values ,I could go the following ways a)Select 2 each from Suit A and Suit B > 6c2*4c2(To get rest 2 cards from the 4 different values of the other suit) b)3 from A and 1 from B> 6c3*3c1 c) 1 from A and 3 from B>6c1*5c3 d)All four from A > 6c4 e)All four from B>6c4 when I add them all up I get the correct answer but I'm not sure if this is the correct approach..Please guide me!! Thanks Yes, using this method, you will get the number of ways in which you will have no pairs. When you subtract those from the total number of ways, you will get the number of ways in which you will have at least one pair. This will give you the required probability. Nothing wrong here. This method will be a bit time consuming though, hence I would suggest you to check out Bunuel's approach given above too.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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13 Jun 2016, 23:16
Bunuel wrote: maheshsrini wrote: 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? 8/33 62/165 17/33 103/165 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  \(\frac{6}{9}\); \(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\). So \(P=1\frac{16}{33}=\frac{17}{33}\)  the same answer as above. Answer: C. Hope it helps. Although I do understand your approach but not sure what's wrong with mine.! I did this as follows > P = P(one pair) + P(2 pairs) 6C1*10C1*8C1/12C4 + 6C2/12C4 6C1  no. of ways of selecting 1 pair out of 6 pairs 10C1  no. of ways of selecting 1 card out of the remaining 10 8C1  no. of ways of selecting 1 card out of the remaining 8 cards ( excluding the 9th card as that would form another pair which is considered in the second case) 12C4  total no. of ways of selecting 4 cards out of 12 cards 6C2  no. of ways of selecting 2 pairs out of 6 pairs I am not getting the right answer  not sure which point I am missing. Can you please explain? Thanks.!



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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14 Jun 2016, 21:53
MeghaP wrote: Bunuel wrote: maheshsrini wrote: 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? 8/33 62/165 17/33 103/165 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  \(\frac{6}{9}\); \(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\). So \(P=1\frac{16}{33}=\frac{17}{33}\)  the same answer as above. Answer: C. Hope it helps. Although I do understand your approach but not sure what's wrong with mine.! I did this as follows > P = P(one pair) + P(2 pairs) 6C1*10C1*8C1/12C4 + 6C2/12C4 6C1  no. of ways of selecting 1 pair out of 6 pairs 10C1  no. of ways of selecting 1 card out of the remaining 10 8C1  no. of ways of selecting 1 card out of the remaining 8 cards ( excluding the 9th card as that would form another pair which is considered in the second case) 12C4  total no. of ways of selecting 4 cards out of 12 cards 6C2  no. of ways of selecting 2 pairs out of 6 pairs I am not getting the right answer  not sure which point I am missing. Can you please explain? Thanks.! Here is your problem: 6C1*10C1*8C1 6C1  no. of ways of selecting 1 pair out of 6 pairs  fine 10C1  no. of ways of selecting 1 card out of the remaining 10  say you picked the 2 of spades 8C1  no. of ways of selecting 1 card out of the remaining 8 cards say you picked the 3 of hearts Now, in 10C1, you will pick the 3 of hearts in another case and in 8C1, you will pick the 2 of spades. Hence, there is double counting. Divide 10C1 * 8C1 by 2 to get (6C1*10C1*8C1)/(2*12C4) + 6C2/12C4 You should get the correct answer.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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17 Sep 2016, 03:21
maheshsrini wrote: 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 Another Way Total No of ways to Pick 4 cards out of 12 = \(C^4_{12}\) Prob of picking atleast 1 pair = 1  Prob of No pair. Total No. of ways to pick no Pairs = \(\frac{(12 * 10 * 8 * 6)}{4!}\) in above expression, 12 => There are 12 ways to pick 1st card 10 => There are 10 ways to pick 2nd card because 1 has already been picked and we can't pick the same number again. 8 & 6 => Same reason as that of 10. 4! => (12 * 10 * 8 * 6) gives us the Number of ways to Pick and Arrange 4 cards. So, we divide by 4! to remove the Arrangement as their sequence doesn't matter to us Therefore, Final Probability = \(1  \frac{(12 * 10 * 8 * 6)}{(4!) * C^4_{12}} = 1  \frac{(12 * 10 * 8 * 6*4!*8!)}{(4!*12!)} = 1  \frac{(12 * 10 * 8 * 6)}{(12*11*10*9)} = 1  \frac{16}{33} = \frac{17}{33}\)
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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06 May 2017, 17:15
maheshsrini wrote: 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 We can use the equation: 1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4) Let’s determine the probability of not finding a pair of cards when 4 are flipped. P(first flip) = 1 P(second flip is not a pair with the first card) = 10/11 P(third card is not a pair with the first or second card) = 8/10 P(fourth card is not a pair with the first, second, or third card) = 6/9 Thus, the probability of not selecting a pair is: 10/11 x 8/10 x 6/9 10/11 x 4/5 x 2/3 2/11 x 4 x 2/3 16/33 Since 1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4), P(at least one pair of cards in 4) = 1  P(no pairs of cards in 4) = 1  16/33 = 17/33. Answer: C
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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01 Jan 2018, 03:44
Heres a simple and cool and innovative method. Bill chooses 2 cards with at least 1 pair. Therefore the problem can be solved in the following manner: a) total possibilities = 12x11x10x9 cards b) If he choose at least 1 pair, then there are 2 Case1: Where he chooses exactly 1 pair = 12x10x8x3 (Using the Bunuel method, I simply replaced 6 by 3 because at least 1 of the 3 numbers is Case2: Where he chooses exactly 2 pairs = 12x10x2x1 (Uptil 10, then he has only 2 cards) Probability = (12x10x8x3 + 12x10x2x1)/12x11x10x9 = 26/99. But the answer is not correct. I fail to understand why? I have 25 short on the numerator.



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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31 Jan 2018, 04:00
Total no of ways = 12C4=495 No of ways of selecting 01 pair from (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) = 6 x (6C25) as one of the pair from remaining 5 could get repeated. =240 No of ways of selecting 02 pairs = 6C2 = 15 Total ways of selecting 01 or 02 pairs = 240+15=255 Prob = 255/495=17/33



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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22 Feb 2018, 13:25
Can someone please explain the 2^4 part in original solution.
If Say, I have to choose 2 numbers from 5, say the chosen numbers are 1 and 2 , then the way I can include is (1,2) (2,1) I can not include (1,1) and (2,2) how can there be four ways to get 1 and 2 from 2 suits(someone gave this ans as a solution of selecting one pair by 6c1*5c2x2^2) where I am assuming that method of getting 2^2 is same as getting 2^4.
Help will be appreciated... Thanks
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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23 Feb 2018, 01:18
Mudit27021988 wrote: Can someone please explain the 2^4 part in original solution.
If Say, I have to choose 2 numbers from 5, say the chosen numbers are 1 and 2 , then the way I can include is (1,2) (2,1) I can not include (1,1) and (2,2) how can there be four ways to get 1 and 2 from 2 suits(someone gave this ans as a solution of selecting one pair by 6c1*5c2x2^2) where I am assuming that method of getting 2^2 is same as getting 2^4.
Help will be appreciated... Thanks
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Posted from my mobile device You are looking for the ways in which you have no pair. So you will select 4 numbers out of 6 (from 1 to 6). This is done in 6C4 ways. Say we get 1, 2, 4, 5 in one case. But we have 2 suits say spades and hearts. The 1 can be of either spades or hearts (it will lead to 2 different cases). Similarly, 2 can be either spades or hearts and so on... So for each number, you have 2 options of the suit. Hence you multiply by 2 * 2 * 2 * 2 = 2^4.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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23 Feb 2018, 01:42
Hi VeritasPrepKarishmaThanks for the reply. Like in your example you mentioned 1,2,4,5.so as u rightly mentioned that each could come from different suits. So aren't we counting (1,1) here which is not required. We have already eliminates the probability of getting (3,3) (6,6) likewise should we subtract some number from 2^4. Posted from my mobile devicePosted from my mobile device



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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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24 Feb 2018, 05:23
Mudit27021988 wrote: Hi VeritasPrepKarishmaThanks for the reply. Like in your example you mentioned 1,2,4,5.so as u rightly mentioned that each could come from different suits. So aren't we counting (1,1) here which is not required. We have already eliminates the probability of getting (3,3) (6,6) likewise should we subtract some number from 2^4. Posted from my mobile devicePosted from my mobile deviceNo, we cannot have (1, 1). Note that we have 4 distinct numbers already  1, 2, 4 and 5. Each will appear once only. But we can have two cases  1 of spades, 2 of spades, 4 of hearts and 5 of spades 1 of hearts, 2 of spades, 4 of hearts and 5 of spades So by multiplying by 2, we are accounting for these 2 cases. Same goes for each number. That is why we have 2*2*2*2
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Joined: 14 Feb 2015
Posts: 1

Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6
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16 Sep 2018, 01:13
I think there is a different way to solve this Question. 12C4 = 495
at least 1 pair = 2c2 *6 * 10c2 5 = 6*(455) = 240 ( 5 for to avoid any remaining pair) At least 2 pair: 6c2 = 15
Total possibilities = 15 + 240=255
255/495 = 17/33




Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6 &nbs
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16 Sep 2018, 01:13



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