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

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.

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.

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.

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) \([(10!) /(2(6 ! 5 !))]\)