Question

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

• A. $\frac{6!6!}{12!}$
• B. $\frac{6!7!}{12!}$
• C. $\frac{2\times 6!6!}{12!}$
• D. $\frac{2\times 6!7!}{12!}$

Answer 1

Answer: (B)

$\frac{6!7!}{12!}$

Total ways in which all the $12$ people sit : $12!$
Now let's say that all the $6$ girls are sitting together. Hence, they can be treated as an individual group. So, now we have $7$ individual groups ($6$ individual boys and $1$ group of girls). Arranging them is $7!$ ways. Also, the internal permutation among the $6$ girls is $6!$.