beads... of three colors: red, yellow, and green.
>
seven-bead bracelets
three beads of any one color; two of a second color; two of a third."
And my count is:
(1) 3 choices for tripled color.
(2) EITHER the three beads are together, in which case the other 2 pairs
go into four adjacent spaces: 00**, 0*0*, 0**0, *00* (4 ways)
OR they are two-and-one with a single gap, and the other four go
0 0**, 0 *0*, 0 *00, * *00, * 0*0, or * 00* (6 ways)
OR they are two-and-one with a double gap and the other four go
00 ** , 0* *0, *0 0*, or *0 *0 (4 ways)
OR they are single with gaps of size 1,1 and 2 and the other four go
0 0 **, 0 * *0, 0 * 0*, or * * 00 (4 ways)
Total 18 patterns x 3 color choices = 54
The Polya-Burnside method could presumably also be used here.
-Robert Dawson
> "And what was the experiment exactly?" I asked.
>
> "Why," said Miss Tryon, "I wanted to see how many different bracelets were
> produced by these haphazard instructions. There were thirty-two, which is
> about what might have been expected. It would have been possible-but only
> just possible-for every girl to construct a bracelet different in appearance
> from all the others."
> How many girls are there in the class?
>
> -end of problem---
>
> Now, the answer to this, according to the newsletter is 54.
>
> My count is like this:
> 1. there are 3 colors to choose the three beads from.
> 2. For each of the above, on the bracelet there can be either three beads
> together, or two together one in-between the rest (3 ways to do this), so
> total 4 ways for the 3 beads
> 3. For each of the above, there's 6 ways to place the remaining 4 beads -[4
> choose 2].
>
> so, I am getting 6*4*3=72.
>
> am I counting wrong or what?
>
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