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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes.The number of ways of arranging one ball in each of the boxes is

$\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.
answered Jun 20, 2014 by sreemathi.v

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