# Making a bangle from $6$ green and $6$ blue beads

How many different ways are there to make a bangle by selecting $6$ beads from $6$ green beads and $6$ blue beads?

(The bangle is a closed loop, and rotations of the same bangle are to be counted only once. The upside-down version of a bangle is not considered to be the same bangle.)

• I'm beginning to wonder about ggbbgb and bbggbg. Are they really different? If you took the first bangle off and put it back on backwards (I.e. flipping it upside down) - wouldn't it be identical to the second as it's now reversed? Feb 28, 2016 at 19:11
• imagine the beads have a top and a base (V) - they ARE NOT the same bangle; @PaulEvans
– JMP
Feb 28, 2016 at 19:33
• But surely the beads can be spun around if that's the case; @JonMarkPerry Feb 28, 2016 at 19:35
• @PaulEvans; might as well have 2 bangles...
– JMP
Feb 28, 2016 at 19:37
• @PaulEvans; mind you, if the beads are orientable its a different Q
– JMP
Feb 28, 2016 at 19:38

There are:

13 different bangles.

Because:

Imagine a 6 digit binary number where the 1s are green and the 0s are blue.
This number can rotated to give the same bangle.
Doubling for swaping blue with green when there's an unequal number gives:
2 for 111111 - 2 numbers.
1 for 101010 - 2 numbers.
2 for 110110 - 6 numbers.
1 for 111000 - 6 numbers.
2 for 111100 - 12 numbers.
2 for 111110 - 12 numbers.
1 for 110010 - 12 numbers.
2 for 111010 - 12 numbers.
---------------------------------------------
13 bangles for all 64 numbers.

14

Argument:

• If we cut the bangle up, we get a green-blue string with six symbols; altogether, there are $2\cdot2\cdot2\cdot2\cdot2\cdot2=64$ such strings. However, now we are over-counting, as many rotations yield rotated versions of the same bangle.
• First, there are the two strings $bbbbbb$ and $gggggg$ of period length 1; they both are counted only once. (This case yields two strings and two bangles.)
• There is one string $bgbgbg$ of period length 2; this string yields the same bangle as $gbgbgb$. (This case yields two strings and one bangle.)
• There are two strings of period 3: $bbgbbg$ and $ggbggb$. These strings yield the same bangles as $gbbgbb$ and $bgbbgb$, respectively as $bggbgg$ and $gbggbg$. (This case yields six strings and two bangles.)
• The remaining $54$ strings correspond to nine bangles, that each is counted in six rotated versions. (This case yields 54 strings and nine bangles.)

Altogether, this yields $2+1+2+9=14$ bangles.

• Actually it seems that the answer is $13$, not $14$. Refer the Paul's answer. Mar 24, 2016 at 1:27

We can consider $4$ cases:

All green: $gggggg$

One blue: $gggggb$

Two blues: $ggggbb$ $gggbgb$ $ggbggb$

Three blues: $gggbbb$ $ggbgbb$ $gbgbgb$ $gbggbb$

By symmetry on green/blue, this gives a total of $14$ bangles.

Note that $ggbgbb$ and $gbggbb$ are different because of their orientations.

• I now think ggbgbb and gbggbb are the same as turning one upside down gives the other. Feb 28, 2016 at 19:30