1. counting

A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead. How many such garlands(distinguishable) are possible?plz discuss.

2. Originally Posted by Navesh
A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead. How many such garlands(distinguishable) are possible?plz discuss.
Hello,

you've got exactly 5 places where you can put a bead.

For the first red bead you can choice out of 5 places, for the second red bead there are only 4 places left. Similar to the two yellow beads and the blue on:

That means you've got 5*4*3*2*1 = 120 possibilities. BUT : Because you can't distinguish between the first or the second red bead, and you you can't distinguish between the first or the second yellow bead, the amount of possibilities is 4 times to great. So finally:

$\frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{2\cdot 2}=30$

Greetings

EB

3. Originally Posted by earboth
Hello,

you've got exactly 5 places where you can put a bead.

For the first red bead you can choice out of 5 places, for the second red bead there are only 4 places left. Similar to the two yellow beads and the blue on:

That means you've got 5*4*3*2*1 = 120 possibilities. BUT : Because you can't distinguish between the first or the second red bead, and you you can't distinguish between the first or the second yellow bead, the amount of possibilities is 4 times to great. So finally:

$\frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{2\cdot 2}=30$

Greetings

EB
It seems that the problem has been solved as if the beads are being arranged along a row. Won't the circular geometry affect the result?

4. Hello, Navesh!

You're right . . . EB didn't consider the circular arrangement.

How many such garlands (distinguishable) are possible?

I've approached this a number of ways,
. . and I've finally come up with the right answer $(4).$

For the moment, imagine that the beads are of five different colors.

Since they are arranged in a circle, the first bead can be placed anywhere.
Then the four remaining beads can arranged in $4! = 24$ ways.

But we do not have five different colors; there is duplication.
The two red beads can be switched without creating a new arrangement.
The two yellow beads can be switched without creating a new arrangment.
Hence, there are: $\frac{24}{2\cdot2} = 6$ ways.

Now, the garland can be "flipped".
That is:
. . . . . + - - * - - - - - - -
. . . . . * - - - - - * - - - -
. . . . . + - - - - - - - - - -
. . . . . + - - - - - - - - - -
. . . . . + - * - * - - - - - -

5. Hello, Navesh!

You're right . . . EB didn't consider the circular arrangement.

How many such garlands (distinguishable) are possible?

I've approached this a number of ways,
. . and I've finally come up with the right answer $(4).$

For the moment, imagine that the beads are of five different colors.

Since they are arranged in a circle, the first bead can be placed anywhere.
Then the four remaining beads can arranged in $4! = 24$ ways.

But we do not have five different colors; there is duplication.
The two red beads can be switched without creating a new arrangement.
The two yellow beads can be switched without creating a new arrangment.
Hence, there are: $\frac{24}{2\cdot2} = 6$ ways.

Now, the garland can be "flipped over"
. . say, left to right over a vertical axis of symmetry.

. . . . . . . . . . .R . . . . . . . . . . . . . . . . . . . . R
. . . . . . . B . . . . . R. . . . . . . . . . . . . . .R . . . . . B
That is:. . . . . . . . . . . is the same as:

. . . . . . . . .Y . . Y . . . . . . . . . . . . . . . . . Y . . Y

So I assumed that I would divide the remaining $6$ arrangements by 2.
. . But the answer not $6 \div 2 = 3$, but $4.$
It seems that the six arrangements are not two sets of mirror-images.

I found this quite difficult to justify . . . then I found a baby-talk approach.

The two Reds and two Yellows can be placed in two arrangements:

. . . . R . . R . . . . . . . R . . Y
. . . . . . . . . . . and . . . . . . . . disregarding rotations and reflections.
. . . . Y . . Y . . . . . . . Y . . R

. . . . R .*. R . . . In the first arrangement, the Blue can be placed
. . . . . . . .* . . . .in three positions: between the Reds, between
. . . . Y .*. Y . . . . the Yellows, or between a Red and a Yellow.

. . . . R .*. Y . . In the second arrangement, the Blue can be placed in
. . . . . . . . . . . . exactly one position: between a Red and a Yellow.
. . . . Y . . R . . . . . . . (The rest are rotations and reflections.)

And there are the four possible arrangements.