Question

The number of ways in which in a necklace can be formed by using 5 identical red beads and 6 identical black beads is (a) \([(11 !) /(6 ! 4 !)]\) (b) \({ }^{11} \mathrm{P}_{6}\) (c) \([(10 !) /(2(6 ! 5 !)]\) (d) None of these

Short answer

The number of ways to form a necklace using 5 identical red beads and 6 identical black beads is (c) \([(10!) /(2(6 ! 5 !))]\).

Step-by-step solution

Identify the type of arrangement

Since the problem is about arranging beads in a necklace, it is considered a circular arrangement rather than linear arrangement. In circular arrangements, the position of objects relative to each other matters, rather than their exact position.

Apply the circular permutation formula

For circular permutations, (n-1)! ways are used to arrange n objects in a circle, instead of n! ways in a linear arrangement. Here, we have 5 red identical beads and 6 black identical beads, a total of 11 beads. However, since the beads are identical within their color, we don't need the full (n-1)! formula. Instead, we need to find a way to combine them into a unique arrangement.

Group the beads by color

Let's group the beads by color: 5 red (R) beads and 6 black (B) beads. We need to find the number of different ways these groups can be combined to form a necklace in such a way that their relative positions stay the same.

Apply the combination formula

We have a total of 11 beads, out of which 6 are black, so we need to find the number of ways to choose 6 black beads out of 11 beads. The combination formula can be written as - \[ C(n, r) = \frac{n!}{r!(n-r)!} \] Where n is the total number of objects and r is the number of objects to be selected. For our problem, n = 11 and r = 6. Plugging these values into the formula, we get - \[ C(11,6) = \frac{11!}{6!(11-6)!} = \frac{11!}{6!5!} \] Since it's a circular arrangement, one of the identical objects (e.g. one of the black beads) can be fixed as a reference point. That leaves us with 10 remaining beads. So, we need to divide our combination by 2. \[ \frac{11!}{2(6!5!)} \] The correct answer is: (c) \([(10!) /(2(6 ! 5 !))]\)

Key concepts

Understanding the Combination Formula

When we deal with selecting a subset of items from a larger set, where the order of selection does not matter, we use the **combination formula**. This formula helps us find how many ways we can choose 'r' objects from 'n' total objects. The formula is expressed as: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] In this formula,

• \( n! \) (n factorial) means the product of all positive integers up to \( n \).
• \( r! \) is the factorial of the number of objects chosen, and \((n-r)!\) is the factorial of the difference between the total and chosen objects.

For example, if you want to choose 6 beads from a total of 11, you would calculate \( C(11, 6) \), resulting in \(\frac{11!}{6!5!}\). Notice that the order does not matter, so using combinations is the best approach in scenarios like arranging beads in a circular necklace where only the relative arrangement matters.

Exploring Circular Arrangement

Unlike the linear arrangements where the order of objects matters entirely, the **circular arrangement** makes the position of each object relative to others the focus. In a circular setup, the starting point is not fixed, meaning that rotating the circle doesn't create a new arrangement. Instead of using \( n! \) (for linear arrangements), we use \( (n-1)! \) for circular arrangements, as the circle can be rotated.
This is because, by fixing one object as a reference point, we remove the redundant arrangements caused by rotations. For example, arranging 11 beads in a circle translates to \( 10! \) possible unique arrangements if we fix one bead and arrange the remaining."
In our problem, fixing one bead helps reduce repetition, aligning with the number of distinct patterns possible in circular setups. This subtle change streamlines understanding without needing to consider rotated duplicates as different possibilities.

Concept of Identical Objects

Handling **identical objects** like the beads in our necklace problem can simplify complex permutations. When multiple objects are identical, swapping them doesn't result in a new arrangement. Thus, we need a modified approach to avoid overcounting duplicate arrangements.

• For arranging 'r' identical objects in 'n' total objects, consider that many permutations will look alike.
• For example, changing two identical red beads doesn't change the overall appearance of your arrangement.

In such cases, we divide total permutations by the factorial of identical counts. For our problem:

• 5 identical red beads and 6 identical black beads.
• Calculate unique arrangements over the total, corrected by \( \frac{1}{5!} \) for red and \( \frac{1}{6!} \) for black duplicates.

This approach makes calculations manageable by ensuring each counted permutation is truly unique, enhancing accuracy in problems involving repeats or identical items.

Applying Permutation Formula

The **permutation formula** helps us calculate arrangements where order matters, especially in linearly organized objects. It describes how many ways to arrange 'n' items in a particular order: \[ P(n, r) = \frac{n!}{(n-r)!} \]Although not directly used in our necklace exercise, understanding of permutations helps highlight how it differs from combinations. Permutation emphasizes sequence, useful when the exact order or rank of items concerns us, unlike our bead scenario. For a necklace...

• Adjustments account for rotation. Thus, instead of full permutations, combine them with circular rules and identical adjustments.
• Permutation-related skills ensure each unique sequence gets considered, but here avoids unwanted duplicates by focusing on combinations.

By correctly applying logic depending on circular or linear nature and object sameness, you can effectively calculate correct answers, critical when designing solutions for complex permutation problems.
Remember, permutations address the order while combinations value selection without worrying about arrangement.