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There are $20$ boxes numbered $1,2,......20$. Five boxes are selected at random and are arranged in ascending order. In how many ways this can be done so that no. $10$ is present in each selection and it comes in $3rd$ place in the arrangement.?

$\begin{array}{1 1} (A) ^{10}C_2\times ^{10}C_2 \\ (B) ^9C_2\times ^{10}C_2 \\ (C) ^9C_2\times ^{9}C_2 \\ (D) ^{19}C_4 \end{array} $

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Since no. 10 is present in each selection, only 4 boxes are to be selected.
Also since no. 10 comes in $3^{rd}$ position,
first two boxes should be numbered <10 and
next two numbers should be >10.
2 numbers are to be selected from 1,2.......9 in $^9C_2$ ways
and
2 numbers are to be selected from 11,12,......20 in $^{10}C_2$ ways.
$\therefore$ The required no. of arrangements = $^9C_2\times ^{10}C_2$

 

answered Sep 5, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p
 

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