If 20 beads of two different colors are to be | vedclass

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(B) First,arrange the $10$ beads of the first color in a circle. The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$. Since the b

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$5(9!)^2$

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(B) First,arrange the $10$ beads of the first color in a circle. The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$. Since the beads of the same color are identical,the arrangement of $10$ beads of the first color in a circle is $\frac{(10-1)!}{2} = \frac{9!}{2}$.Now,there are $10$ gaps created between these $10$ beads. We need to place the $10$ beads of the second color in these $10$ gaps. Since the beads are now in fixed positions relative to the first set,the number of ways to arrange the $10$ beads of the second color in these $10$ gaps is $10!$.Therefore,the total number of ways is $\frac{9!}{2} 	imes 10! = \frac{9!}{2} 	imes (10 	imes 9!) = 5 	imes (9!)^2$.

If $20$ beads of two different colors are to be arranged in a necklace such that they alternate,and there are $10$ beads of each color,then what is the number of ways to arrange them?

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