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(B) First,arrange the $6$ distinct white roses in a circle. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. So,the number

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$43200$

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(B) First,arrange the $6$ distinct white roses in a circle. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. So,the number of ways to arrange $6$ white roses is $(6-1)! = 5! = 120$.There are $6$ gaps created between the $6$ white roses in the circle.We need to place $5$ distinct red roses in these $6$ gaps such that no two red roses are together. The number of ways to choose $5$ gaps out of $6$ is $^6C_5 = 6$.The number of ways to arrange $5$ distinct red roses in the chosen $5$ gaps is $5! = 120$.Since the garland can be flipped (clockwise and anti-clockwise arrangements are considered the same),we divide by $2$.Total number of ways $= \frac{5! 	imes ^6C_5 	imes 5!}{2} = \frac{120 	imes 6 	imes 120}{2} = \frac{86400}{2} = 43200$.

The number of different ways of preparing a garland using $6$ distinct white roses and $5$ distinct red roses such that no two red roses come together is

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