Question

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

Answer 1

Number of vans $=2$

Total seats in two vans$=14$

Number of girls $=3$ and number of boys$=9$

Then, total people $=3+9=12$

Now, the number of ways arranging $12$ people on $14$ seats${=}^{14}{P}_2=\frac{14!}{(14-12)!}=\frac{14\times 13!}{2}=7\times (13!)$

Now, number of ways of arranging $3$ girls of choosing $2$ back seats $=2\times 3!$

And the number of ways arranging $9$ boys on the remaining $11$ seats${=}^{11}{P}_9=\frac{11!}{(11-9)!}=\frac{11!}{2!}$

Therefore, the complete number of ways

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

$=2\times 2\times 3\times 2\times 1\times \frac{11!}{2!}$

$=12\times 11!=12!$

Hence, the probability of three girls sitting together in aback row on adjacent seats

$=\frac{12!}{7\times 13!}$

$=\frac{12!}{7\times 13\times 12!}$

$=\frac{1}{91}$

Hence, it is complete solution.