Six boys and six girls sit in a row randomly. Find the probability that six girls sit together?

(a)$\frac{1}{32} (b)\frac{2}{7}$

(c)$\frac{5}{12}$ (d) None of these

what i have tried

Since six girls need to sit together so the number of combination of girls sitting next to each can be formed =$12 \choose 6$=924
The number arrangement that can be done to make boys and girls sit on $12$ seats= $2^{12}$

Therefore the probability of girls sitting next to each other=$\frac{12 \choose 6}{2^{12}}=\frac{231}{1024}$

But i think so this is not one of the option and so please help me with the problem.

  • 2
    $\begingroup$ There are $\binom{12}{6}$ equally likely ways to choose $6$ seats and put "Reserved for Girls" signs on them. How many choices are favourable (all the chosen seats are together)? $\endgroup$ Feb 5, 2015 at 16:52

2 Answers 2


Consider the $6$ girls as one person, then the total number of situations when $6$ girls sitting together is $7!$ multiplied by the permutations of girls $6!$, the total number of all permutations is $12!$, hence the probability is $\frac{6!7!}{12!}=\frac{1}{132}$.

  • $\begingroup$ but the answer given in solution is $\frac{1}{32}$ $\endgroup$
    – CY5
    Feb 5, 2015 at 17:20
  • 1
    $\begingroup$ @CY5 This is probably a typo. If you're not convinced, here is a solution on a random book I found on google books.google.de/… $\endgroup$
    – Qidi
    Feb 6, 2015 at 0:30

The total cases will be $n(s)=12!$.
To find the favourable cases, we assume the group of girls to be a single object and find their permutations. This is given by $7!\cdot 6!$. The first term represents their permutations with the boys and the second represents the permutations among themselves.
Hence, total answer will be given by $$\frac{7!\cdot 6!}{12!}=\frac{1}{132}$$
Hence the answer is (D)


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