Each of the coins in a collection is distinct and is either silver or

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.

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

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?

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.

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.

Hi,
I get that we don't need to actually solve, but can someone show more clearly why C is correct? From the explanations above it just looks like we know silver and gold each have the same # of permutations, but there is a restriction here. How do we account for that?

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.