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If 6 boys and 6 girls sit in a row at random, then the probability that all the girls sit together is
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The correct answer is:
$\frac{1}{132}
$\frac{1}{132}
When all the girls sit together, then we may consider it as one group
So, number of arrangements $=\lfloor\underline{6} \times\lfloor\underline{7}$
Now, total number of arrangements of 12 persons $=12$
$$
\text { Hence, required probability }=\frac{6 ! 7 !}{12 !}=\frac{1}{132}
$$
So, number of arrangements $=\lfloor\underline{6} \times\lfloor\underline{7}$
Now, total number of arrangements of 12 persons $=12$
$$
\text { Hence, required probability }=\frac{6 ! 7 !}{12 !}=\frac{1}{132}
$$
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