How many different necklaces can be formed wi | vedclass

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(C) The total number of beads is $n = 6 + 5 = 11$.Since the beads are arranged in a necklace,the clockwise and counter-clockwise arrangements are cons

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) The total number of beads is $n = 6 + 5 = 11$.Since the beads are arranged in a necklace,the clockwise and counter-clockwise arrangements are considered identical.The number of circular permutations of $n$ objects where $p$ are of one kind and $q$ are of another kind is given by $\frac{(n-1)!}{p!q!}$.For a necklace,we divide by $2$ to account for the reflection symmetry.Number of different necklaces $= \frac{1}{2} 	imes \frac{(11-1)!}{6!5!} = \frac{10!}{2 	imes 6! 	imes 5!}$.$= \frac{10 	imes 9 	imes 8 	imes 7 	imes 6!}{2 	imes 6! 	imes (5 	imes 4 	imes 3 	imes 2 	imes 1)}$.$= \frac{10 	imes 9 	imes 8 	imes 7}{2 	imes 120} = \frac{5040}{240} = 21$.

How many different necklaces can be formed with $6$ white and $5$ red beads?

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