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,Originally Posted by Navesh
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:
$\displaystyle \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{2\cdot 2}=30$
Greetings
EB
Hello, Navesh!
You're right . . . EB didn't consider the circular arrangement.
A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead.
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 $\displaystyle (4).$
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 $\displaystyle 4! = 24$ 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: $\displaystyle \frac{24}{2\cdot2} = 6$ ways.
Now, the garland can be "flipped".
That is:
. . . . . + - - * - - - - - - -
. . . . . * - - - - - * - - - -
. . . . . + - - - - - - - - - -
. . . . . + - - - - - - - - - -
. . . . . + - * - * - - - - - -
Hello, Navesh!
You're right . . . EB didn't consider the circular arrangement.
A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead.
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 $\displaystyle (4).$
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 $\displaystyle 4! = 24$ 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: $\displaystyle \frac{24}{2\cdot2} = 6$ 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 $\displaystyle 6$ arrangements by 2.
. . But the answer not $\displaystyle 6 \div 2 = 3$, but $\displaystyle 4.$
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.