Creating Unique Garlands with 3 Red & 12 Black Beads

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Discussion Overview

The discussion revolves around the combinatorial problem of determining the number of unique garlands that can be created using 3 identical red beads and 12 identical black beads. Participants explore various approaches to account for the circular arrangement of the beads and the ambiguity in the definition of "garland."

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the equation x + y + z = 12 to represent the distribution of black beads between the red beads, suggesting the number of ways is 14C2.
  • Another participant counters that black beads can also be placed before and after the red beads, leading to the equation W + X + Y + Z = 12 and suggesting the number of ways is 14C3.
  • Concerns are raised about the circular nature of garlands, noting that arrangements like $BBBBBBBBBRRR$ and $RRRBBBBBBBBB$ are equivalent.
  • One participant questions the definition of "garland," suggesting it may be culturally ambiguous and proposing that a visual representation could clarify the problem.
  • Another participant discusses partitioning the number 9 into three or fewer summands as a method to approach the problem, providing examples of possible partitions.
  • Several cases are outlined to count the arrangements based on the equality or inequality of x, y, and z, leading to a total of 19 arrangements proposed by one participant.
  • One participant expresses uncertainty about whether any cases were overlooked in their counting, while another confirms that their understanding of the cases is correct.

Areas of Agreement / Disagreement

Participants express differing views on the definition of "garland" and its implications for the counting problem. There is no consensus on the correct number of unique arrangements, with multiple approaches and interpretations presented.

Contextual Notes

The discussion highlights ambiguities in definitions and the complexity of counting arrangements in circular configurations, which may affect the accuracy of proposed solutions.

Who May Find This Useful

This discussion may be of interest to those studying combinatorial mathematics, particularly in the context of counting problems involving circular arrangements and identical objects.

grgrsanjay
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How many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?I was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z

So, x+y+z=12
no .of ways is 14C2...Was i Wrong somewhere??
 
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Black will fit not only between, but also before and after. You seem to have only one end.

W before any red
X after first red
Y after second red
Z after last red

W + X + Y + Z = 12

14C3
 
tkhunny said:
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.

How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
 
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I don't disagree with your solution. I suggest only that the definition of "garland" is ambiguous.

After a quick survey of those in my home, they ALL believed a "garland" to be more of a hanging rope, and not a wreath. I believe we have established significant ambiguity (at least cultural) in the problem statement.

Perhaps a picture was provided.
 
Plato said:
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.

How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
So,there must be less than 14C2 solutions??How do i eliminate them?

Ok,could you check whether i can do it like this

x + y + z = 12

Case 1:x=y=z
No.of.ways = (4,4,4) = 1

Case 2:two of x,y,z are equal
No.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6

Case 3: x,y,z are all unequal

To avoid double counting , i will take x<y<z

No.of ways =
(0,1,11),(0,2,10),(0,3,9),(0,4,8),(0,5,7)
(1,2,9),(1,3,8),(1,4,7),(1,5,6)
(2,3,7),(2,4,6)
(3,4,5)
= 12 ways

So,total no.of ways is 19(Wink)
 
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tkhunny said:
the definition of "garland" is ambiguous.
The definition may be ambiguous here but not in textbooks on countilng. Here is a link. The subtopic on beads is the one used in counting problems.
grgrsanjay said:
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this
So,total no.of ways is 19
YES, 19 is correct
This sort of problem has set many arguments. If you agree on what meaning of garland we use then the following will show you why. If we have three green beads and three red beads how many different garlands are possible? Well the answer is 3. All the teds together; two reds together and one separate; then all the reds are separated. There are no other possibilities.
Three can be written as 3, 2+1, or 1+1+1.

The questions using two colors have that nice solution. They quickly become a nightmare with more than two colors.
 
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I think none of my case were repeated...could you conform it whether any case is left?
 
grgrsanjay said:
I think none of my case were repeated...could you conform it whether any case is left?
Yes you are correct. At first I read it as if there were only 12 beads altogether.
 

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