Three horses H_{1}, H_{2}, H_{3} entered a fi | vedclass

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(D) The horse $H_{1}$ can choose any of the $7$ portions ($P_{1}$ to $P_{7}$),so there are $7$ ways for $H_{1}$.Since no $2$ horses can enter the same

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$210$

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(D) The horse $H_{1}$ can choose any of the $7$ portions ($P_{1}$ to $P_{7}$),so there are $7$ ways for $H_{1}$.Since no $2$ horses can enter the same portion,after $H_{1}$ has chosen a portion,there are $6$ portions remaining for $H_{2}$. Thus,there are $6$ ways for $H_{2}$.After $H_{1}$ and $H_{2}$ have occupied their respective portions,there are $5$ portions remaining for $H_{3}$. Thus,there are $5$ ways for $H_{3}$.By the fundamental principle of counting,the total number of ways the $3$ horses can graze the field is $7 	imes 6 	imes 5 = 210$ ways.

Three horses $H_{1}, H_{2}, H_{3}$ entered a field which has $7$ portions marked $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}$ and $P_{7}$. If no $2$ horses are allowed to enter the same portion of the field,in how many ways can the horses graze the grass of the field?

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