Question

Four boys and eight girls sit at random in a row. Find the probability that each boy sits between two girls?
(a) $\dfrac{7! \times 8!}{3! \times 12!}$
(b) $\dfrac{{}^{7}{{C}_{3}}}{12!}$
(c) $\dfrac{{}^{7}{{C}_{3}} \times 8!}{12!}$
(d) None of these

Answer 1

Hint: We start solving the problem by finding the total number of ways that girls and boys can be seated randomly using the fact that the total no. of ways of arranging n-objects randomly is $n!$ ways. We then find the total no. of ways of arranging 8 girls by leaving a gap for 4 boys to sit. We then find the total no. of ways that the boys can sit in those gaps left by girls. We then find the probability using $P=\dfrac{\text{total number of ways that each boy sits between two girls}}{\text{total number of ways of arranging 8boys and 4girls}}$ and making necessary calculations to get the required result.

Complete step-by-step solution
According to the problem, we are given that four boys and eight girls sit at random in a row and we need to find the probability that each boy sits between two girls.
Let us find the total no. of ways that all boys and girls can be seated randomly. We know that the total no. of ways of arranging n-objects randomly is $n!$ ways. So, we can arrange 8 girls and 4 boys in a total of $12!$ ways ---(1).
Now, let us find the total number of ways such that each boy sits in between two girls.
Let us assume that the girls have already sat in their respective places leaving the space between each other for boys to the seat. So, we get a total of $8!$ ways for sitting in such a condition. Let us represent this situation by mentioning the girls as G and the gap between them as _.
So, the current situation is $\begin{array}{*{35}{l}}
   G & \_ & G & \_ & G & \_ & G & \_ & G & \_ & G & \_ & G & \_ & G \\
\end{array}$.
We can see that there are 7 places in which boy 4 can sit. Let us first the no. of ways of choosing 4 places out of 7 places for boys to sit. We know that the total number of ways choosing ‘r’ objects out of ‘n’ objects is ${}^{n}{{C}_{r}}$ ways. So, we can choose 4 places in ${}^{7}{{C}_{4}}$ ways and we can arrange 4 boys in 4 chosen places in $4!$ ways.
So, we get the total number of ways such that each boy sits in between two girls by multiplying all the obtained values. So, we get ${}^{7}{{C}_{4}}\times 4!\times 8!$ ways.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, we get
$\Rightarrow {}^{7}{{C}_{4}}\times 4!\times 8!=\dfrac{7!}{4!3!}\times 4!\times 8!$.
$\Rightarrow {}^{7}{{C}_{4}}\times 4!\times 8!=\dfrac{7!\times 8!}{3!}$.
Now, let us find the probability that each boy sits between two girls. Let us assume it as P.
So, we have $P=\dfrac{\text{total number of ways that each boy sits between two girls}}{\text{total number of ways of arranging 8boys and 4girls}}$.
\[\Rightarrow P=\dfrac{\dfrac{7!\times 8!}{3!}}{12!}\].
\[\Rightarrow P=\dfrac{7!\times 8!}{3!\times 12!}\].
So, we have found the required probability as \[\dfrac{7!\times 8!}{3!\times 12!}\].
∴ The correct option for the given problem is (a).

Note: We should not stop finding the number of ways for boys to sit between the girls as the arrangement is also very important in this type of problem. We should not confuse or make mistakes while finding the required probability. Similarly, we can expect problems to find the total number of ways that all boys sit in between 8 girls such that 4 girls are present on either side of the group of boys.