# Garland trouble

#### grgrsanjay

##### New member
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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#### tkhunny

##### Well-known member
MHB Math Helper
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

#### Plato

##### Well-known member
MHB Math Helper
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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#### tkhunny

##### Well-known member
MHB Math Helper
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.

#### grgrsanjay

##### New member
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 Last edited:

#### Plato

##### Well-known member
MHB Math Helper
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.
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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#### grgrsanjay

##### New member
I think none of my case were repeated...could you conform it whether any case is left?

#### Plato

##### Well-known member
MHB Math Helper
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.