Last visit was: 19 Nov 2025, 05:03 It is currently 19 Nov 2025, 05:03
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,385
Own Kudos:
778,202
 [19]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,385
Kudos: 778,202
 [19]
Each of the coins in a collection is distinct and is either silver or 27 Dec 2015, 08:54
4
Kudos
Add Kudos
14
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

15% (low)

Question Stats:

78% (01:14) correct 22% (01:23) wrong based on 866 sessions

History

Date
Time
Result
Not Attempted Yet
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.
ShowHide Answer
Official Answer
C
Most Helpful Reply
avatar
CV7585
avatar
Joined: 29 Jul 2015
Last visit: 14 Jun 2020
Posts: 53
Own Kudos:
42
 [9]
Given Kudos: 68
Location: Australia
Schools: McCombs '19 (D) Darden '19 (D) ISB '18 (A) Tepper '19 (WL) Duke '19 (D) HEC Dec"18 (A)
GMAT 1: 680 Q49 V33
GPA: 3.25
WE:Business Development (Energy)
Schools: McCombs '19 (D) Darden '19 (D) ISB '18 (A) Tepper '19 (WL) Duke '19 (D) HEC Dec"18 (A)
GMAT 1: 680 Q49 V33
Posts: 53
Kudos: 42
 [9]
Each of the coins in a collection is distinct and is either silver or 29 Dec 2015, 20:25
7
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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.
General Discussion
User avatar
nipunjain14
User avatar
Joined: 07 May 2015
Last visit: 12 Aug 2017
Posts: 151
Own Kudos:
83
 [3]
Given Kudos: 21
Location: India
Schools: ISB '19 Kellogg '20 Yale '20 HEC Dec"18
GMAT 1: 660 Q48 V31
GPA: 3
Schools: ISB '19 Kellogg '20 Yale '20 HEC Dec"18
GMAT 1: 660 Q48 V31
Posts: 151
Kudos: 83
 [3]
Each of the coins in a collection is distinct and is either silver or 03 Jan 2016, 05:15
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,070
Own Kudos:
19,390
 [4]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,070
Kudos: 19,390
 [4]
Each of the coins in a collection is distinct and is either silver or 04 Jan 2016, 04:37
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
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.
User avatar
angoae
User avatar
Joined: 19 Apr 2015
Last visit: 02 Nov 2019
Posts: 11
Own Kudos:
148
Given Kudos: 748
Concentration: Technology, Marketing
Schools: Mendoza '20 Georgia Tech '19 Madison '19 Boston U '19 (D) Tepper '19 Tippie '19 Carroll '19 (WL) Kelley '19 (WD) Simon '19 Krannert '20 Iowa State '19 (S) Katz '19 (WD) Leeds '19 (A)
GMAT 1: 540 Q38 V27
GMAT 2: 620 Q42 V32
GPA: 3.55
WE:Marketing (Manufacturing)
Schools: Mendoza '20 Georgia Tech '19 Madison '19 Boston U '19 (D) Tepper '19 Tippie '19 Carroll '19 (WL) Kelley '19 (WD) Simon '19 Krannert '20 Iowa State '19 (S) Katz '19 (WD) Leeds '19 (A)
GMAT 2: 620 Q42 V32
Posts: 11
Kudos: 148
Each of the coins in a collection is distinct and is either silver or 08 Feb 2016, 00:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nipunjain14
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
Show more



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.
User avatar
nipunjain14
User avatar
Joined: 07 May 2015
Last visit: 12 Aug 2017
Posts: 151
Own Kudos:
83
Given Kudos: 21
Location: India
Schools: ISB '19 Kellogg '20 Yale '20 HEC Dec"18
GMAT 1: 660 Q48 V31
GPA: 3
Schools: ISB '19 Kellogg '20 Yale '20 HEC Dec"18
GMAT 1: 660 Q48 V31
Posts: 151
Kudos: 83
Each of the coins in a collection is distinct and is either silver or 08 Feb 2016, 04:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
angoae
nipunjain14
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.
Show more



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
User avatar
davedekoos
User avatar
Joined: 09 Jul 2013
Last visit: 07 Nov 2025
Posts: 97
Own Kudos:
340
 [2]
Given Kudos: 11
Posts: 97
Kudos: 340
 [2]
Each of the coins in a collection is distinct and is either silver or 08 Feb 2016, 13:00
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,070
Own Kudos:
19,390
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,070
Kudos: 19,390
Each of the coins in a collection is distinct and is either silver or 08 Feb 2016, 18:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Amrita01
User avatar
Joined: 25 May 2024
Last visit: 24 May 2025
Posts: 60
Own Kudos:
24
Given Kudos: 23
Location: India
AMRITHA: P
Concentration: Sustainability, Entrepreneurship
Schools: ESSEC MiM "27 (II)
GPA: 8.7
Products:
My Reviews
Schools: ESSEC MiM "27 (II)
Posts: 60
Kudos: 24
Each of the coins in a collection is distinct and is either silver or 18 Jun 2024, 19:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Can anybody explain how to actually find the answer to this question? Is it 7! * 8 ?

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,385
Own Kudos:
778,202
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,385
Kudos: 778,202
Each of the coins in a collection is distinct and is either silver or Updated on: 18 Aug 2025, 00:27
Originally posted by Bunuel on 18 Jun 2024, 23:48.
Last edited by Bunuel on 18 Aug 2025, 00:27, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Amrithapallath
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.

Can anybody explain how to actually find the answer to this question? Is it 7! * 8 ?

Posted from my mobile device
Show more

­When combining the statements, we find that there are 7 distinct gold coins and 7 distinct silver coins in the collection. They can be arranged in a row such that no two coins of the same color are adjacent in 7! * 7! * 2 ways. Consider the following two scenarios:

GSGSGSGSGSGSGS
SGSGSGSGSGSGSG

For each of these scenarios, the number of ways to arrange the coins is 7! * 7!. Therefore, the total number of ways is 7! * 7! * 2.­
User avatar
ishita_sud
User avatar
Joined: 14 May 2023
Last visit: 11 Sep 2025
Posts: 9
Given Kudos: 36
Posts: 9
Kudos: 0
Each of the coins in a collection is distinct and is either silver or Updated on: 18 Aug 2025, 00:29
Originally posted by ishita_sud on 17 Aug 2025, 14:44.
Last edited by Bunuel on 18 Aug 2025, 00:29, edited 2 times in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If the coins would have been identical, then then we should have divided it by 7X 7 ? Right?
Bunuel

Amrithapallath
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.

Can anybody explain how to actually find the answer to this question? Is it 7! * 8 ?

Posted from my mobile device
­When combining the statements, we find that there are 7 distinct gold coins and 7 distinct silver coins in the collection. They can be arranged in a row such that no two coins of the same color are adjacent in 7! * 7! * 2 ways. Consider the following two scenarios:


GSGSGSGSGSGSGS
SGSGSGSGSGSGSG

For each of these scenarios, the number of ways to arrange the coins is 7! * 7!. Therefore, the total number of ways is 7! * 7! * 2.­
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,385
Own Kudos:
778,202
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,385
Kudos: 778,202
Each of the coins in a collection is distinct and is either silver or 18 Aug 2025, 00:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ishita_sud
If the coins would have been identical, then then we should have divided it by 7X 7 ? Right?
Bunuel

Amrithapallath
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.

Can anybody explain how to actually find the answer to this question? Is it 7! * 8 ?

Posted from my mobile device
­When combining the statements, we find that there are 7 distinct gold coins and 7 distinct silver coins in the collection. They can be arranged in a row such that no two coins of the same color are adjacent in 7! * 7! * 2 ways. Consider the following two scenarios:


GSGSGSGSGSGSGS
SGSGSGSGSGSGSG

For each of these scenarios, the number of ways to arrange the coins is 7! * 7!. Therefore, the total number of ways is 7! * 7! * 2.­
Show more

If the coins were identical within each color, then there would be no 7! * 7! factor. You’d only have 2 possible alternating patterns: starting with gold or starting with silver.

GSGSGSGSGSGSGS
SGSGSGSGSGSGSG
Moderators:
Bunuel
Math Expert
105383 posts
kingbucky
496 posts