 QUESTION

# 6 boys and 6 girls sit in a row at random. What is the probability that all the girls sit together?(a) $\dfrac{1}{432}$(b) $\dfrac{12}{431}$(c) $\dfrac{1}{132}$(d) None of these

Hint: First arrange the 6 boys with respect to each other in $6!$ ways. Now, draw a notation to understand things better. You have 7 places between the boys where all the girls can be placed. You have to choose 1 out of these 7 places and then arrange the 6 girls there. This can be done in $7\times 6!$ ways. Now, multiply $6!$ by $7\times 6!$ to find the number of favourable outcomes and divide this by $12!$ which is the total number of outcomes. The resultant will be your final answer.

In this question, we are given that 6 boys and 6 girls sit in a row at random.
We need to find the probability that all the girls sit together.
There are 6 boys and 6 girls. So, in total we have 12 children.
Total number of possible ways of arrangement of 12 children = $12!$ ways.
Now, let us first arrange the 6 boys with respect to each other. The number of ways to arrange 6 boys = $6!$ ways.
Now, once the boys are arranged with respect to each other, we have to find a place where all the 6 girls can sit together. To understand this better, see the notation below:
__ B __ B __ B __ B __ B __ B __
In the above notation, B represents the boys whose positions have been fixed with respect to each other and __ represents an empty place where the girls can sit. So, after arranging the boys, we see that there are 7 places between the boys where all the girls can be placed. We have to choose 1 out of these 7 places and then arrange the 6 girls there.
So, the number of ways to choose 1 out of 7 places = ${}_{1}^{7}C=7$ ways.
The number of ways to arrange 6 girls = $6!$ ways.
So, the number of ways in which 6 boys and 6 girls can sit in a row such that all the girls sit together are = Number of favourable outcomes = Number of ways to arrange 6 boys $\times$ number of ways to choose 1 out of 7 places $\times$ number of ways to arrange 6 girls
Number of favourable outcomes = $6!\times 7\times 6!=7!\times 6!$
Total number of outcomes = $12!$
Now, we know that $\text{Probability =}\dfrac{\text{No}\text{. of favourable outcomes}}{\text{Total no}\text{. of outcomes}}$
Using this formula, we get the following:
Probability that all the girls sit together = $\dfrac{7!\times 6!}{12!}$
$\dfrac{7!\times 6!}{12\times 11\times 10\times 9\times 8\times 7!}=\dfrac{6\times 5\times 4\times 3\times 2}{12\times 11\times 10\times 9\times 8}=\dfrac{1}{132}$
So, the probability that all the girls sit together = $\dfrac{1}{132}$
Hence, option (c) is correct.

Note: In this question, it is very important to arrange the boys first and find the number of ways to do so and then find the number of ways the girls can be arranged. Also, draw such a notation as it will make it easier for you to understand the situation and approach accordingly.