A) \[\frac{1}{77}\]
B) \[\frac{1}{132}\]
C) \[\frac{1}{231}\]
D) None of these
Correct Answer: B
Solution :
6 boys and 6 girls can be arranged in a row in \[12\,\,!\] ways. If all the 6 girls are together, then the number of arrangement are \[7\,\,!\,\times \,6\,\,!\]. Hence required probability\[=\frac{7\,\,!\,.\,6\,\,!}{12\,\,!}\] \[=\frac{6\times 5\times 4\times 3\times 2}{12\times 11\times 10\times 9\times 8}=\frac{1}{132}\].
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