Question

Six boys and six girls are to sit in a row at random. The probability that all the six girls and all the six boys sit together is
A) \[\dfrac{{6!6!}}{{12!}}\]
B) \[\dfrac{{6!7!}}{{12!}}\]
C) \[\dfrac{{2 \times 6!6!}}{{12!}}\]
D) \[\dfrac{{2 \times 6!7!}}{{12!}}\]

Answer 1

Hint- Here we have to consider six boys and six girls to be different from each other. Six boys to be considered different from each other and six girls to be considered different from each other. We know that the set of all possible outcomes is called sample space in probability. For example, tossing a coin gives either head or tail. So, it has two outcomes which is its sample space. And an event is a set of outcomes of an experiment. Probability is the ratio of favorable outcomes to the total number of outcomes. That is $\text{Probability} = \dfrac{\text{ Favourable outcomes}}{\text{Total number of outcomes}}$.

Complete step-by-step answer:
According to question,
6 boys and 6 girls are there to be seated in a row,
$\therefore $ Number of ways in which 6 boys and 6 girls can be seated together in a row =2
There are 6 boys, no. of ways boys can be seated together in a row$ = 6!$
There are 6 girls, no. of ways girls can be seated together in a row $ = 6!$
$\therefore $ Total number of ways all the six girls and all the six boys can sit together in a row $ = 2 \times 6!6!$
Total number of possible ways 12 persons comprising 6 boys and 6 girls can be seated together in a row $ = 12!$
$\therefore $ Probability that all the six girls and all the six boys sit together $\dfrac{{2 \times 6!6!}}{{12!}}$
Hence, option (C) is the correct answer.

Note- Here, it can be thought that all the girls are tied together in a rope and all the boys are tied together in a rope separately. Also, all boys can change their position and all girls can change position with each other. So, there will be two sets, one consisting of boys and another of girls. By using this approach, we can easily solve such problems.