$\begin{array}{1 1}(A)\;18720\\(B)\;18270\\(C)\;17280\\(D)\;12780\end{array} $

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- $P(n,r)=\large\frac{n!}{(n-r)!}$
- $n!=n(n-1)(n-2)(n-3)....(3)(2)(1)$

Total no. of balls =9

5 ball cannot fit 3 small box

Required number of arrangements =$4P_3\times 6!$

$4P_3=\large\frac{4!}{1!}$$=4\times 3\times 2$

$\Rightarrow 24$

$6!=6\times 5\times 4\times 3\times 2\times 1$

$\Rightarrow 720$

$4P_3\times 6!=24\times 720$

$\Rightarrow 17280$

Hence (C) is the correct answer.

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