We have 14 seats in two vans. And there are 9 boys and 3 girls. The number of ways of arranging 12 people on 14 seats without restriction is
14P12 = 14!/2! = 7(13!)
Now, the number of ways of choosing back seats is 2.
And the number of ways of arranging 3 girls on adjacent seats is 2(3!). And the number of ways of arranging 9 boys on the remaining 11 seats is 11P9 ways. Therefore, the required number of ways
= 2.(2.3!). 11P9
= (4.3! 11!)/2! = 12!
Hence, the probability of the required event
= 12!/(7. 13!) = 1/91