There are 8 different beads in a box with the colors red, orange, yellow, green, blue, indigo, violet, and white. The red, orange, yellow, green, and blue beads should be together,... There are 8 different beads in a box with the colors red, orange, yellow, green, blue, indigo, violet, and white. The red, orange, yellow, green, and blue beads should be together, but the orange and green beads should not be beside each other. How many ways can you arrange the beads to form a bracelet?
Understand the Problem
The question asks for the number of ways to arrange 8 beads in a bracelet, where 5 specific beads must be together, and two of those beads cannot be next to each other. This is a combinatorics problem that combines permutation and restriction.
Answer
$216$
Answer for screen readers
$216$
Steps to Solve
- Consider the 5 beads as a single unit
Since the red, orange, yellow, green, and blue beads must be together, we can treat them as a single unit. This leaves us with effectively 4 units: the group of 5 beads, and the indigo, violet, and white beads.
- Arrange the 4 units in a circle
The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. Here, $n=4$, so there are $(4-1)! = 3! = 6$ ways to arrange the 4 units in a circle.
- Arrange the 5 beads within their unit
The 5 beads (red, orange, yellow, green, and blue) can be arranged in $5!$ ways. However, we have the restriction that the orange and green beads cannot be next to each other.
- Calculate arrangements where orange and green are together
To find the arrangements where orange and green are together, treat orange and green (OG) as a single unit. Then we arrange the 4 units (OG, red, yellow, blue) which gives us $4!$ arrangements. Since orange and green can be arranged as OG or GO, we multiply by 2. So there are $2 \cdot 4! = 2 \cdot 24 = 48$ arrangements where orange and green are together.
- Calculate arrangements where orange and green are not together
The total number of arrangements of the 5 beads is $5! = 120$. Subtract the arrangements where orange and green are together to get the arrangements where they are not together: $120 - 48 = 72$.
- Account for bracelet flips
Since a bracelet can be flipped, we must divide by 2 to account for arrangements that are mirror images of each other. Thus we divide $6$ by $2$ giving 3. Multiplying by $72/2$ is not correct because the internal arrangement of the 5 beads is not symmetrical in general. Flipping the bracelet does not necessarily result in a valid arrangement since orange and green can't be next to each other.
- Combine the arrangements
Multiply the number of circular arrangements of the 4 units by the number of valid arrangements of the 5 beads together: $6 \cdot 72 = 432$.
- Divide by two for bracelet symmetry For a bracelet, we note that rotations and reflections are considered the same arrangement. The arrangements found include both clockwise and counter-clockwise arrangements. Therefore, we need to divide by 2 to account for the symmetry. However, we can only divide by 2 if flipping the bracelet gives a valid arrangement where orange and green are still not together.
Since we account for orange and green not being next to each other in every arrangement of the 5 beads, we should divide by two to account for the bracelet nature: $432/2 = 216$.
$216$
More Information
The problem combines circular permutation concepts with restricted arrangements. The fact that it is a bracelet is also important because it reduces the number of arrangements.
Tips
A common mistake is forgetting to account for the bracelet being flipped. Also, it is easy to mess up the counting when taking into account the restrictions.
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