Question

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

(a) $\frac{1}{432}$ (b) $\frac{12}{431}$ (c) $\frac{1}{132}$ (d) None of these

Answer 1

Since 6 boys and 6 girls can sit in a row at random
Then, the total number of arrangement = 12!
Suppose all girls sit together
∴ all girls can be grouped together and regarded as a single group.
Hence, we have 6 boys and 1 group of girls = 7 people
Hence, Number of arrangements = 7!
Also, these 6 girls can be arranged among themselves in 6! ways

$\begin{array}{rcl}\therefore \mathrm{Required}\mathrm{probability}& =& \frac{6!\times 7!}{12!}\\ & =& \frac{6\times 5\times 4\times 3\times 2}{12\times 11\times 10\times 9\times 8}\\ \mathrm{i}.\mathrm{e}\mathrm{required}\mathrm{probability}& =& \frac{1}{132}\end{array}$

Hence, the correct answer is option C.