Beautiful Necklaces
How many necklaces can be formed by joining together 12 green beads and 3 red beads?
Note: All beads have to be used. Rotations and reflections are considered identical.
The answer is 19.
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2 solutions
I didn't know where to place this rant on the question wording. Necklaces are not supposed to be able to be reflected, the ones which are supposed to be reflected are bracelets. You can check it on wikipedia.
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Thanks for sharing that fact. I've updated the question accordingly.
In future, if you spot any errors with a problem, you can “report” it by selecting the “dot dot dot” menu in the lower right corner. You will get a more timely response that way.
For god's sake, what rotation are we talking about? Nothing is stated explicitly in the problem and I don't see anything to rotate here.
wow nice solution, but i couldn't visualize what the 7 symmetric configurations look like. can you help me out? thanks!!!
Does your method work if we increase the number of red beads to more than 3. If it does then it will be an absolutely BRILLIANT way to enumerate number of partitions of an integer.
This is the same as the number of (unordered) partitions of 1 2 into at most 3 parts. More information on this general problem is given here .
Since there are only 1 9 such partitions in this case, I will list them all out so that the reader can see the process involved. I will list them as triplets in (not necessarily strictly) descending order with 0 's as "fillers":
(
1
2
,
0
,
0
)
,
(
1
1
,
1
,
0
)
,
(
1
0
,
2
,
0
)
,
(
9
,
3
,
0
)
,
(
8
,
4
,
0
)
,
(
7
,
5
,
0
)
,
(
6
,
6
,
0
)
(
1
0
,
1
,
1
)
,
(
9
,
2
,
1
)
,
(
8
,
3
,
1
)
,
(
8
,
2
,
2
)
,
(
7
,
4
,
1
)
,
(
7
,
3
,
2
)
,
(
6
,
5
,
1
)
,
(
6
,
4
,
2
)
,
(
6
,
3
,
3
)
,
(
5
,
5
,
2
)
,
(
5
,
4
,
3
)
,
(
4
,
4
,
4
)
.
Exactly how I did it too! Thank you for the interesting link as well.
why can't we do it this way that there are (12+1) places for the 3 red beads to go so it can be done in C(13,3) and we divide it by two as symmetric necklaces would be double counted......Hence, our answer should be 286/2 = 143
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This would still result in over-counting since it does not take into account the (assumed) fact that the necklace can be flipped over as well.
Thought it mattered that the necklace can be rotated and flipped/turned over
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It does, and this counting process accounts for that. Any triplet ( a , b , c ) accounts for any bead configuration where there the red beads are separated by a , b and c white beads in any orientation. Thus all rotations/reflections of the "base" triplet are accounted for, as they are cyclic in nature.
Thanks. I've updated the question accordingly.
In future, if you spot any errors with a problem, you can “report” it by selecting the “dot dot dot” menu in the lower right corner. You will get a more timely response that way.
Let's count the number of ordered combinations first.
1 2 ! 3 ! 1 5 ! = 4 5 5
Now let's take into account rotations. There are in total 12+3=15 rotations. It is important to consider the case where the red beads are equally spaced. This case only occurs 5 times rather than 15.
1 5 4 5 5 − 5 + 5 5 = 3 1
Finally we take into consideration reflections. There are 7 symmetric configurations.
2 3 1 − 7 + 7 = 1 9