We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
224
5
avatar+36 

In how many ways can seven beads of distinct colors be put on the hexagonal grid shown, if reflections and rotations of an arrangement are considered equivalent?

 Jan 8, 2019
 #1
avatar+700 
+1

\(\dfrac{7\cdot(5!)}{6} = \boxed{140}\)

 

- PM

 #2
avatar+700 
+1

There are 7 points, so there are 7! = 5040 ways to arrange the points without considering overcounting.

 

Let's first consider rotations. If we number the 6 beads on the outside as ABCDEF and the center bead as G, the arrangement of ABCDEFG is considered the same as BCDEFGA. Since there are 6 different locations the outer bead A can be, (A cannot be in the center) we divide 5040 by 6 to get 840.

 

Let's now consider reflections. We can draw six axes for the beads to reflect on, and ABC G(center) DEF is considered the same as DEF G(center) ABC. We divide 840 by 6 to get our answer, \(\boxed{140}\).

 

Hope this helps, 

 

- PM cool

 #3
avatar+36 
+1

Thank you! smiley

Starz  Jan 9, 2019
 #4
avatar+36 
+2

Just found the correct answer. Thank you for the help! smiley

 

There are 7! ways to put the beads on the grid, not considering rotations and reflections. Arrangements can be reflected or not reflected and can be rotated by 0, 60, 120, 180, 240, or 300 degrees, so they come in groups of twelve equivalent arrangements. Correcting for the symmetries, we find that there are 7!/12 or 420 distinct arrangements.

Starz  Jan 9, 2019
 #5
avatar+700 
+1

You are very welcome! laugh

PartialMathematician  Jan 12, 2019

15 Online Users

avatar