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In how many ways can 3 girls and 9 boys can be seated in two vans having 3 numbered seats in the front and 4 numbered seats at the back so that the 3 girls are to be seated together in the back row?

$\begin{array}{1 1} 6.^{11}P_9 \\ 12.^{11}P_9 \\ 24.^{11}P_9 \\ 30.^{11}P_9 \end{array}$

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There are 14 seats, 3 girls and 9 boys.
3 girls can be seated in backrow in the seats numbered 1,2,3 or 2,3,4. (in 2 ways in $1^{st}$ van.)
Similarly in two ways they can be seated in other van.
Girls among themselves can be arranged in $3!$ ways.
After fixing girls remaining 11 seats are to be occupied by 9 boys in
$^{11}P_9$ ways.
$\therefore$ The required no. of arrangements = $4.3!.^{11}P_9=24.^{11}P_9$
answered Aug 20, 2013 by rvidyagovindarajan_1
 

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