Combinatorics - How many unique ways to put marbles on vertices

Mark Batten-Carew12 · 2014-08-08 14:26:45

I have already read and enjoyed the Combination and Permutations page at www.mathsisfun.com/combinatorics/combinations-permutations.html
Very clear!

But I have an issue that goes one step further that I would love to know the answer to...

The question is, how many unique combinations (excluding duplicate arrangements) are there in the following situation...

Simple 2D version: Imagine you have a hexagon, and you can put zero or one marble on each vertex of the hexagon. How many possible arrangements of marbles are there, excluding duplicates by either rotation or mirroring?

The maximum is 2**6 (there could be one marble or not on each of 6 spots) which is 64, but that number is too large because it does not eliminate all the duplicate arrangements with rotation or flipping of the locations of the marbles. Eg, any pattern of marbles has 6 possible orientations due to rotation. After that how many arrangements are duplicated by mirror imaging? By trial and error, I have determined the number is 12 unique combinations of marbles on a hex ring.

Now the real question is for the actual 3D version: How many unique combinations are there of the 12 possible positions surrounding any cannon ball in a 3D FCC spherical packing arrangement, which is the 3D equivalent of a 2D hex ring? (If it would help, see this wikipedia page for info about FCC sphere packing: http://en.wikipedia.org/wiki/Close-pack … al_spheres )

In the hex case above, there are 3 axes, each of which can have 1 or 0 marbles at either end. In FCC, there are 6 axes each of which can have 1 or 0 marbles at either end. This gives a maximum of 2**12 arrangements, but how many are left after eliminating duplicates, similar to 64 being reduced to 12?

Thanks for any help.

PS. FYI, this is in support me trying to help people understand a new theory of particle physics called QGD (see www.QuantumGeometryDynamics.com). Answering this question will simplify my demonstration of how particles move.

John E. Franklin · 2014-08-13 13:12:09

i think the best bet is to get some marbles and superglue to make a 13 piece solid. i.m getting the feeling that there are triangles and rectsngles in this irregular platonic solid. funny since its the densest.

John E. Franklin · 2014-08-13 13:31:08

i think there are four center planes with the benzene ring around the center. when u can visualize these planes, then you can reduce permutations by mentally checking all four planes. also a savvy algorithm might be programmable for 2^12 reduction...

John E. Franklin · 2014-08-13 14:43:07

i think i got the platonic solid.
its a cube with the eight corners
cut off exactly halfway from edge
to edge!!!! so u get eight triangles
and six square sides. My brother
tipped me off on this solution
because at UPS where he works
he spins boxes by opposite corners
to see the six sides. but we also
see the six edges cut in half. our
benzenr six ring!!!!

John E. Franklin · 2014-08-13 14:58:53

used pythagoreans to verify center ball is right size and its the same as other 12.
so now all u need for your model is a
rubiks cube or any nice cube shape.
all u have to do is realize that the
6 ball ring is simply at a 45 degree
angle on a rubiks cube so just use
the twelve edge pieces to work
out the problem. and now computers
can be used even easier!

John E. Franklin · 2014-08-13 15:07:35

so we have inadvertently changed
from hex angles to our familiar
orthogonal angles. And now that
i recall, the articles on fcc try to
show a cube on an angle!
so to do this problem of how many
combiations, we can just concentrate
on the geometry of the edges on a cube
and forget any dimensions and lengths
as it is perfectly symmetrical and easy
now...

John E. Franklin · 2014-08-19 10:46:11

i'm getting 144 ways to light up the 12 bulbs.

John E. Franklin · 2014-08-19 11:22:12

REM Start of BASIC! Program

dim maxesindexup1[4097]
xx=0

gosub zeroarraylbl3

for jqa=0 to 1
for jqb=0 to 1
for jqc=0 to 1
for jqd=0 to 1
for jqe=0 to 1
for jqf=0 to 1
for jqg=0 to 1
for jqh=0 to 1
for jqi=0 to 1
for jqj=0 to 1
for jqk=0 to 1
for jql=0 to 1
print "new shape computation"
jqvalmax=0
jqval=0
gosub incxxlbl1

gosub tmpvarlbl2
gosub getvaluelbl4
gosub replacemaxlbl6

gosub zquarteryyyylbl11
gosub zquarteryyyylbl11
gosub zquarteryyyylbl11
gosub zquarteryyyylbl11

gosub xquarteryyyylbl12
gosub xquarteryyyylbl12
gosub xquarteryyyylbl12
gosub xquarteryyyylbl12

gosub lefttorightlbl5
gosub getvaluelbl4
gosub replacemaxlbl6

gosub zquarteryyyylbl11
gosub zquarteryyyylbl11
gosub zquarteryyyylbl11
gosub zquarteryyyylbl11

gosub xquarteryyyylbl12
gosub xquarteryyyylbl12
gosub xquarteryyyylbl12
gosub xquarteryyyylbl12

maxesindexup1[1+jqvalmax]=77

next jql
next jqk
next jqj
next jqi
next jqh
next jqg
next jqf
next jqe
next jqd
next jqc
next jqb
next jqa

! print xx

gosub countmaxeslbl7

end

incxxlbl1:
xx=xx+1
return

tmpvarlbl2:
aa=jqa
bb=jqb
cc=jqc
dd=jqd
ee=jqe
ff=jqf
gg=jqg
hh=jqh
ii=jqi
jj=jqj
kk=jqk
ll=jql
return

zeroarraylbl3:
for jqlbl3 = 1 to 4096
maxesindexup1[jqlbl3]=0
next jqlbl3
return

getvaluelbl4:

jqval=aa
jqval=jqval+jqval
jqval=jqval+bb
jqval=jqval+jqval
jqval=jqval+cc
jqval=jqval+jqval
jqval=jqval+dd
jqval=jqval+jqval
jqval=jqval+ee
jqval=jqval+jqval
jqval=jqval+ff
jqval=jqval+jqval
jqval=jqval+gg
jqval=jqval+jqval
jqval=jqval+hh
jqval=jqval+jqval
jqval=jqval+ii
jqval=jqval+jqval
jqval=jqval+jj
jqval=jqval+jqval
jqval=jqval+kk
jqval=jqval+jqval
jqval=jqval+ll

! print jqval

return

lefttorightlbl5:
aa=jqa
dd=jqb
cc=jqc
bb=jqd
ff=jqe
ee=jqf
hh=jqg
gg=jqh
ii=jqi
ll=jqj
kk=jqk
jj=jql
return

replacemaxlbl6:

if (jqval > jqvalmax ) then

jqvalmax = jqval

print jqvalmax

endif

return

countmaxeslbl7:
jqshapecount=0

for xxlbl7=1 to 4096

if (77= maxesindexup1[xxlbl7]) then
jqshapecount=jqshapecount+1
endif

next xxlbl7
print "shape count"
print jqshapecount
print "**********"

return

turnonzaxislbl8:

ccc=cc
aaa=aa
bbb=bb

cc=hh
aa=ee
bb=dd

hh=kk
ee=ii
dd=ll

kk=gg
ii=ff
ll=jj

gg=ccc
ff=aaa
jj=bbb

return

turnonyaxislbl9:

aaa=aa
eee=ee
iii=ii

aa=dd
dd=cc
cc=bb
bb=aaa

ee=hh
hh=gg
gg=ff
ff=eee

ii=ll
ll=kk
kk=jj
jj=iii

return

computelbl10:

gosub getvaluelbl4
gosub replacemaxlbl6

return

zquarteryyyylbl11:

gosub turnonzaxislbl8
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10

return

xquarteryyyylbl12:

gosub turnonxaxislbl13
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10
gosub turnonyaxislbl9
gosub computelbl10

return

turnonxaxislbl13:

aaa=aa
bbb=bb
ddd=dd

aa=cc
cc=kk
kk=ii
ii=aaa

bb=gg
gg=jj
jj=ff
ff=bbb

dd=hh
hh=ll
ll=ee
ee=ddd

return

anonimnystefy · 2014-08-19 18:19:46

Hi Mark

I'm getting 13 for the 2D case.

allan1085 · 2014-08-21 07:35:21

20 couples go to 5 progressive dinners, 4 courses and 4 couples at each dinner. What is the formula so that no 2 couples are together more than once, and no couple serves the same course more than once?

Last edited by allan1085 (2014-08-22 07:31:40)

bobbym · 2014-08-21 07:38:10

Hi;

Question has been answered for you before.

http://www.mathisfunforum.com/viewtopic … 81#p294481 post #28.

If you need more help post there. This thread is for another problem and is still active,

John E. Franklin · 2014-08-26 10:20:40

the dot product div length product = cosine(angle) in 3d. angles between edges of cube from center are 60 degrees to closest edge. because of the hex ring on an angle, multiples of 60 are also there.

Last edited by John E. Franklin (2014-08-26 10:57:35)

Math Is Fun Forum

#1 2014-08-08 14:26:45

Combinatorics - How many unique ways to put marbles on vertices

#2 2014-08-13 13:12:09

Re: Combinatorics - How many unique ways to put marbles on vertices

#3 2014-08-13 13:31:08

Re: Combinatorics - How many unique ways to put marbles on vertices

#4 2014-08-13 14:43:07

Re: Combinatorics - How many unique ways to put marbles on vertices

#5 2014-08-13 14:58:53

Re: Combinatorics - How many unique ways to put marbles on vertices

#6 2014-08-13 15:07:35

Re: Combinatorics - How many unique ways to put marbles on vertices

#7 2014-08-19 10:46:11

Re: Combinatorics - How many unique ways to put marbles on vertices

#8 2014-08-19 11:22:12

Re: Combinatorics - How many unique ways to put marbles on vertices

#9 2014-08-19 18:19:46

Re: Combinatorics - How many unique ways to put marbles on vertices

#10 2014-08-21 07:35:21

Re: Combinatorics - How many unique ways to put marbles on vertices

#11 2014-08-21 07:38:10

Re: Combinatorics - How many unique ways to put marbles on vertices

#12 2014-08-26 10:20:40

Re: Combinatorics - How many unique ways to put marbles on vertices

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