# Find the no. of ways in which 6 boys and 6 girls can be seated in a row such that no two girls are seated together

My approach to this problem was as follows - Seat all the boys in a row before the girls. Then arrange the 6 girls in the 7 places between the boys(5 places between the boys and 2 places at the extremes). Therefore, the answer will come out to be $$6! * 7!$$ .

What is written in the solutions - Seat all the 6 girls before the boys in a row. Now find the number of ways of seating 6 boys in the 5 places between the girls. For the one boy left, the number of places to sit in is 12. Therefore, the answer comes out to be $$6! * 6! * 12$$. What is the cause of these two conflicting answers? Is my approach wrong or is the approach given in the book wrong?

• @TahaDirek But in my approach, there is one place left as there are 7 places and only 6 girls. That one place can fall in between 2 boys and hence my answer includes the cases where 2 boys are sitting together. – Pratham Yadav Mar 16 '20 at 12:10
• Sorry, I've misread. – Taha Direk Mar 16 '20 at 12:20
• The answer of $6!\cdot 7!$ is correct. The answer of $6!\cdot 6!\cdot 12$ is incorrect. To see why, look at what happens for if two boys are between two girls. Using lowercase for girls and uppercase for boys, you might have had the scenario $aABb\cdots$ where $A$ was placed between $a$ and $b$ in the first step and then $B$ was placed after $A$, or you might have had the situation $aABb\cdots$ where $B$ was placed between $a$ and $b$ and then $A$ was placed before $B$ in the second step. – JMoravitz Mar 16 '20 at 12:31
• @JMoravitz Thanks. I've got it now. Such types of cases are difficult to visualize and it becomes difficult to find out which approach is the correct one. – Pratham Yadav Mar 16 '20 at 12:35