Question

The number of ways in which a necklace can be formed by using 5 identical red beads
and 6 identical black beads is?
(a) $\dfrac{11!}{6!5!}$
(b) ${}^{11}{{P}_{6}}$
(c) $\dfrac{10!}{2(6!5!)}$
(d) none of these

Answer 1

Hint: Given question is based on formation of necklace. As the necklace has a circular shape so this Question is based on circular permutation. The number of ways to add the n distinct objects around a fixed circle is circular permutation. The arrangement can be either clockwise or in an anticlockwise direction. In this question, we use the condition of circular permutation and solve the question.

Formula Used
Number of ways =$\dfrac{(n-1)!}{2}$ ways

Complete step by step Solution:
Given the Number of red beads = 5
Number of black beads = 6
Total number of beads = 6 + 5 = 11
Now we use the condition of circular permutation to solve the question.
We know the number of circular permutations of n distinct things can be arranged in (n – 1)! Ways
Then 11 beads can be arranged in ( 11 – 1 )! Ways
i.e. in 10! Ways
The arrangement can be either in the direction of clockwise or anticlockwise.
But in the question, the arrangement is identical.
As there is no dependency on the position of beads in a clockwise or anticlockwise manner.
As there are 5 red and 6 black beads in the necklace
so the number of ways beads can be arranged in $\dfrac{(n-1)!}{2.x!.y!}$ ways
The number of ways beads can be arranged in$\dfrac{11-1!}{2(5!)(6!)}$ ways
Hence, the number of ways a necklace can be arranged = $\dfrac{10!}{2(5!)(6!)}$

Hence, the correct option is (c).

Note: Students make mistake in arranging the beads. While solving this type of question, students must take care of whether it is in a clockwise or an anticlockwise manner, and while formula is used to solve the question. Students must take care of these types of things while solving the question.