Question

In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats?

Answer 1

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