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(D) The number of circular permutations of $n$ distinct things taken $r$ at a time is given by $P(n, r) / r = \frac{n!}{r(n-r)!}$.For $n_1$,we have $n

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$\frac{9}{5}$

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(D) The number of circular permutations of $n$ distinct things taken $r$ at a time is given by $P(n, r) / r = \frac{n!}{r(n-r)!}$.For $n_1$,we have $n=9$ and $r=5$:$n_1 = \frac{9!}{5(9-5)!} = \frac{9!}{5 \cdot 4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5 \cdot 4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{5} = 604.8$ (Wait,let us use the formula $n_1 = { }^9 C_5 \cdot (5-1)! = \frac{9!}{5!4!} \cdot 4! = \frac{9!}{5!}$).$n_1 = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5!} = 9 \cdot 8 \cdot 7 \cdot 6 = 3024$.For $n_2$,the number of linear permutations of $8$ distinct things taken $4$ at a time is $P(8, 4) = \frac{8!}{(8-4)!} = \frac{8!}{4!} = 8 \cdot 7 \cdot 6 \cdot 5 = 1680$.Therefore,$\frac{n_1}{n_2} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{8 \cdot 7 \cdot 6 \cdot 5} = \frac{9}{5}$.

If the number of circular permutations of $9$ distinct things taken $5$ at a time is $n_1$ and the number of linear permutations of $8$ distinct things taken $4$ at a time is $n_2$,then $\frac{n_1}{n_2}=$

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