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) $\dfrac{1}{91}$
(B) $\dfrac{1}{92}$
(C) $\dfrac{1}{81}$
(D) $\dfrac{1}{82}$

Answer 1

Hint: For answering this question we will find the total number of possible arrangements and the possible number of when 3 girls sitting together in a back row on adjacent seats. Then we will use them and find their ratio which will be equal to the probability of 3 girls sitting together in a back row on adjacent seats.

Complete step by step answer:
Now considering from the question we have 2 vans having 7 seats in each and 9 boys and 3 girls.
The number of ways of arranging 12 people on 14 seats without restriction is ${}^{14}{{P}_{12}}=\dfrac{14!}{2!}$ .
Now the number of ways of choosing back seats in 2 vans is 2. And the number of ways arranging 3 girls on adjacent seats is $2\left( 3! \right)$. And the number of ways of arranging 9 boys in the remaining seats is $^{11}{{P}_{9}}$ .
Therefore, the number of ways of arranging if three girls should sit together in a back row on adjacent seats is $2\times 2\left( 3! \right){{\times }^{11}}{{P}_{9}}$ .
The probability of 3 girls sitting together in a back row on adjacent seats is $\dfrac{2\times 2\left( 3! \right){{\times }^{11}}{{P}_{9}}}{^{14}{{P}_{12}}}$ .
After simplifying this we will have $\dfrac{4!{{\times }^{11}}{{P}_{9}}}{^{14}{{P}_{12}}}\Rightarrow \dfrac{4!\times 11!\times 2!}{14!\times 2!}=\dfrac{4\times 2\times 3}{14\times 13\times 12}=\dfrac{1}{13\times 7}$.
Hence the answer is $\dfrac{1}{91}$ .

So, the correct answer is “Option A”.

Note: While answering this question we should make a note that the value of $^{n}{{C}_{r}}$ is different from the value of $^{n}{{P}_{r}}$ which are defined as follows $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . If we use the other one here we will have the answer as $\dfrac{4!{{\times }^{11}}{{C}_{9}}}{^{14}{{C}_{12}}}\Rightarrow \dfrac{4!\times 11!\times 2!\times 12!}{9!\times 14!\times 2!}=\dfrac{4!\times 11\times 10}{14\times 13}=\dfrac{1320}{91}$ which is a wrong answer.