Question

6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.

Answer 1

$$\begin{gathered} \mathrm{Here},6\text{boys and 6 girls can be arranged in a line in 12! ways.} \\ \text{Total possible outcomes}=12! \\ \text{Consider 6 girls as a single element X.} \\ \text{Now, 6 boys and X can be arranged in a line in 7! ways and girls can be arranged in 6! ways among them.} \\ P\left(\text{all girls are together}\right)=\frac{7!\times 6!}{12!} \\ =\frac{7\times 6\times 5\times 4\times 3\times 2\times 1\times 6\times 5\times 4\times 3\times 2\times 1}{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \\ =\frac{1}{11\times 12} \\ =\frac{1}{132} \end{gathered}$$