In how many ways can 4 beads be selected and | vedclass

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(A) The number of ways to select $4$ beads from $6$ distinct beads is given by $^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$. For a necklace,the clockw

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$45$

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(A) The number of ways to select $4$ beads from $6$ distinct beads is given by $^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$. For a necklace,the clockwise and anti-clockwise arrangements are considered identical. The number of circular arrangements of $n$ objects is $(n-1)!$. For a necklace of $4$ beads,the number of arrangements is $\frac{(4-1)!}{2} = \frac{3!}{2} = \frac{6}{2} = 3$. Therefore,the total number of ways to form the necklace is $15 	imes 3 = 45$.

In how many ways can $4$ beads be selected and arranged in a necklace from $6$ distinct beads?

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