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 three 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?

• A. $\frac{1}{91}$
• B. $\frac{1}{92}$
• C. $\frac{1}{81}$
• D. $\frac{1}{82}$

Answer 1

The correct option is A $\frac{1}{91}$

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 $14P_{12}=\frac{14!}{12!}=7(13!)$

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 $11P_9$.

Therefore, the required number of ways

$2\times 2(3!)\times 11P_9=4(3!)\times \frac{11!}{2!}=12!$

Hence, the probability of $3$ girls sitting together in a back row on adjacent seats $=\frac{12!}{7(13!)}=\frac{1}{91}$