A jeweler wants to create a circular bracelet that contains 6

Question

https://i.imgur.com/rTmxuVz.png

A jeweler wants to create a circular bracelet that contains 6 different gemstones: diamond, emerald, ruby, sapphire, tanzanite and opal. Two bracelets are considered different only when the positions of the gemstones are different relative to each other. How many different bracelets can the jewel create?

(A) 5
(B) 120
(C) 240
(D) 364
(E) 720

(A) 5
(B) 120
(C) 240
(D) 360
(E) 720

BrentGMATPrepNow · Accepted Answer

Although we can quickly apply the circular arrangement formula (i.e., number of ways to arrange n objects in a circle = (n - 1)!), we can also solve the question using the Fundamental Counting Principle (FPC, aka the slot method). In the process of doing so, you'll also learn WHY the circular arrangement formula works

First label the six places for gemstones as follows: 
 https://i.imgur.com/3NF2kYd.png 
We can place the diamond in one of the   6   available spots.
We can place the emerald in one of the   5   available spots.
We can place the ruby in one of the   4   remaining spots.
We can place the sapphire in one of the   3   remaining spots.
We can place the tanzanite in one of the   2   remaining spots.
We can place the opal in the   1   remaining spot.
So, the total number of ways to place the gemstone = (  6  )(  5  )(  4  )(  3  )(  2  )(  1  ) =   720   ways

The answer, however, is  not  E, because we have inadvertently counted every possible arrangement 6 times.

For example, the six bracelet shown here...
 https://i.imgur.com/ZcaUxlm.png 
... are all the  same , because the  relative positions  of the six gemstones are the same in each case.

Since we have counted each unique arrangement 6 times, we must divide   720   by 6 to get  120  possible arrangements

Answer: B

ASIDE: In general, if we have n objects to arrange in a circle (where arrangements are considered different only when the positions of the objects are different relative to each other), then the total number of arrangements = n!/n, which can be simplified to get  (n-1)!  although I prefer  n!/n , because it helps us see why the formula works. 
That is,  n!  represents the number of ways to arrange the n objects (if we ignore the relative position proviso), and  n  represents the number of times each unique arrangement has been counted (when we observe the relative position proviso)

Archit3110 · Answer

this is circular combination ; number of ways to arrange 6 different gemstons ( n-1)! ; (6-1)! ; 5! ; 120
option B

BrentGMATPrepNow · Answer

Archit3110  has provided a super-fast solution. 
Does anyone have a solution that doesn’t involve the circular arrangement formula?

PyjamaScientist · Answer

One way to do it in case one doesn't remember the (n-1)! Formula is to assume one position fixed and then placing rest of the items relative to that one fixed position. Since that one fixed position takes away one spot, we are left with (n-1) places left to be occupied. And thus (n-1)! arrangements are possible in those (n-1) places.

Posted from my mobile device

Posted from my mobile device

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A jeweler wants to create a circular bracelet that contains 6

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A jeweler wants to create a circular bracelet that contains 6

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards