6 boys and 6 girls sit in a row. find the probability that all the 6 girls sit together.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{boys} \mathrm{and} \mathrm{girls}=12 \\ \mathrm{When} \mathrm{all} \mathrm{girls} \mathrm{sit} \mathrm{together}, \mathrm{we} \mathrm{can} \mathrm{group} \mathrm{them} \mathrm{into} \mathrm{one} \\ \mathrm{Hence} \mathrm{we} \mathrm{have} \mathrm{total}=7 \mathrm{people} \\ \mathrm{Ways} \mathrm{of} \mathrm{arranging}=7!\times \left(\mathrm{ways} 6 \mathrm{girls} \mathrm{arrange} \mathrm{among} \mathrm{themselves}\right)=7!.6! \\ \mathrm{Total} \mathrm{arrangements}=12! \\ \boxed{\mathrm{Probability}=\frac{7!.6!}{12!}=\frac{1}{132}} \end{gathered}$$

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