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I tried it as

(number of ways selecting 4 girls out of 5)(arranging them on their seats )(ways we can find four consecutive sear for these girls)(selecting 10 seats from the remaining 12 seats )(arranging students in these 10 seats)

$=(^5 _4)(4!)(4)(^{12}_{10})(10!)$ = $^{11}P_6(6!)(2)$

But the answer is not this .

Answer is given as http://i.stack.imgur.com/MWKVF.jpg

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    Mind telling us what the answer is? – 2017-02-24
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    @CompulsiveMathurbator answer is $^{11}P_9 (5!)(46)$ – 2017-02-24

1 Answers 1

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Let us start with selecting the places for the girls. We have the following possibilities:

  • $5$ places at the back of one of the $2$ vans. Evidently there are $2$ possibilities.
  • $4$ consecutive places at the back of one of the vans ($2\times2=4$ possibilities) and another place such that exactly $4$ girls sit together ($11$ are available). There are $4\times11=44$ possibilities.

So there are $46$ possibilities in total, and each of them provides in $5!$ possibilities to place the girls. Then there are $11$ places left for the $9$ boys, leading to $P_9^{11}$ possibilities. So our final result is: $$46\times5!\times P_9^{11}\text{ possibilities}$$