Question

$6$ boys and $6$ girls sit in a row randomly. The probability that all $6$ girls sit together , is

• A. $\frac{1}{64}$
• B. $\frac{1}{8}$
• C. $\frac{1}{132}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{132}$

Total no. of possible ways of arrangement $=12!$

No. of ways of rearrangement in which $6$ girls are always together$=6!\times P_6^7$

$\therefore$ P($6$ girls sitting together)$=\frac{6!\times P_6^7}{12!}$

$=\frac{6!\times 7!}{1!\times 12!}=\frac{6!\times 7!}{12\times 11\times 10\times 9\times 8\times 7!}=\frac{6\times 5\times 4\times 3\times 2}{12\times 11\times 10\times 9\times 8}=\frac{1}{132}$