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Each of the coins in a collection is distinct and is either silver or
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27 Dec 2015, 08:54
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76% (01:18) correct 24% (01:25) wrong based on 301 sessions
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Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent? (1) The display contains an equal number of gold and silver coins. (2) If only the silver coins were displayed, 5,040 different arrangements of the silver coins would be possible.
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Re: Each of the coins in a collection is distinct and is either silver or
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29 Dec 2015, 20:25
Definitely not my strongest area. But here is my rationale:
Given: Coins are either gold or silver. Each of the coins is distinct.
To find: Number of ways that the coins can be displayed in a row if no 2 coins are adjacent
To answer this question we will need the number of distinct gold and silver coins.
Statement 1:
Although we are told the number of gold and silver coins are equal the exact number of each of the coins are still unknown. Hence is insufficient.
Statement 2:
Possible ways of arranging the silver coins alone = 5040. This is 7! So we know that there are 7 distinct silver coins. No info about the number of gold coins so insufficient.
Both together  We have the number of silver coins from statement 2 and number of gold coins from statement 1.
This is sufficient information to answer the question.
Hence answer is C.
Please correct me if i am wrong. Feedback appreciated.



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Re: Each of the coins in a collection is distinct and is either silver or
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03 Jan 2016, 05:15
Statement 1: tells us there are equal number of gold and silver coins, i. e number of coins is even . But the number is not specified. Insufficient Statement 2: Number of ways to arrange silver coins = 5040 = 7! ( by factoring) Thus there are 7 silver coins, but there is no information on gold coins. Thus Insufficient
1+ 2 Since number of silver coins = number of gold coins we know there are 7 silver and 7 gold coins Thus Sufficient
Ans: C



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Re: Each of the coins in a collection is distinct and is either silver or
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04 Jan 2016, 04:37
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent? (1) The display contains an equal number of gold and silver coins. (2) If only the silver coins were displayed, 5,040 different arrangements of the silver coins would be possible. In the original condition, there are 2 variables(the number of silver coins and gold coins), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), s=g and s=g=7 is derived from s!=5040, which is unique and sufficient. Therefore the answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Each of the coins in a collection is distinct and is either silver or
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08 Feb 2016, 00:34
nipunjain14 wrote: Statement 1: tells us there are equal number of gold and silver coins, i. e number of coins is even . But the number is not specified. Insufficient Statement 2: Number of ways to arrange silver coins = 5040 = 7! ( by factoring) Thus there are 7 silver coins, but there is no information on gold coins. Thus Insufficient
1+ 2 Since number of silver coins = number of gold coins we know there are 7 silver and 7 gold coins Thus Sufficient
Ans: C Number of ways to arrange silver coins = 5040 = 7! ( by factoring) Could you explain how exactly do you factor the following and were able to find that it was 7! When I factor it I don't always get the same result and wanted to see if there was some trick to this that I am not aware of. I mean it's easy to memorize each value, but I'am all ears for a new trick that I could use should the need for it ever arise.



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Re: Each of the coins in a collection is distinct and is either silver or
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08 Feb 2016, 04:21
angoae wrote: nipunjain14 wrote: Statement 1: tells us there are equal number of gold and silver coins, i. e number of coins is even . But the number is not specified. Insufficient Statement 2: Number of ways to arrange silver coins = 5040 = 7! ( by factoring) Thus there are 7 silver coins, but there is no information on gold coins. Thus Insufficient
1+ 2 Since number of silver coins = number of gold coins we know there are 7 silver and 7 gold coins Thus Sufficient
Ans: C Number of ways to arrange silver coins = 5040 = 7! ( by factoring) Could you explain how exactly do you factor the following and were able to find that it was 7! When I factor it I don't always get the same result and wanted to see if there was some trick to this that I am not aware of. I mean it's easy to memorize each value, but I'am all ears for a new trick that I could use should the need for it ever arise. If you factor 5040, it can be written as : 2*2*2*2*3*3*5*7 It can be further be written as > 2*3*4*5*6*7 which is equal to 7! Hope this is clear



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Re: Each of the coins in a collection is distinct and is either silver or
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08 Feb 2016, 07:19
I agree answer should be C. But I am wondering if the promt didn't restrain the coins to being next to each other how would this could change the logic. Any thouhgts on this? Bunuel? MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent?
(1) The display contains an equal number of gold and silver coins. (2) If only the silver coins were displayed, 5,040 different arrangements of the silver coins would be possible. In the original condition, there are 2 variables(the number of silver coins and gold coins), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), s=g and s=g=7 is derived from s!=5040, which is unique and sufficient. Therefore the answer is C.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Each of the coins in a collection is distinct and is either silver or
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08 Feb 2016, 13:00
If there are no constraints on the arrangement of the coins (gold and silver don't have to alternate), then the answer would still be the same (the answer to whether the information is sufficient or not). The information we need to answer the question is:  Number of gold coins  Number of silver coins If we CAN find those two numbers, we know we will have enough information to answer the question. We don't actually need to solve the problem of how many different arrangements there are. Statement 1 tells us that the numbers are equal, and statement 2 tells us that we can calculate the number of silver coins. Once we have that information we know we can answer the question, whether the additional constraint of alternating colours is there or not. ***IMPORTANT NOTE*** You do not actually need to know that 5040=7! Solving this will not contribute to answering the question. We are told that there are 5040 arrangements of silver coins. That means that we can figure out the actual number of silver coins if we wanted to, but we don't need to know the actual number. All we need to know is that we CAN get that number. I.e. the data is SUFFICIENT.
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Re: Each of the coins in a collection is distinct and is either silver or
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08 Feb 2016, 18:59
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent? (1) The display contains an equal number of gold and silver coins. (2) If only the silver coins were displayed, 5,040 different arrangements of the silver coins would be possible. In the original condition, you need to figure out the number of silver coins and gold coins, which makes 2 variables. In order to match with the number of equations, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become silver=gold=20, which is unique and sufficient. Therefore, the answer is C. > For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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