Here is the problem:
A group of $20$ students, including $3$ particular girls and $4$ particular boys, are to be lined up in two rows with $10$ students each. In how many ways can this be done if the $3$ particular girls must be in the front row while the $4$ boys must be in the back row?
Below is my solution, which is wrong according to the answer in the book. The question is where is the mistake?
First, choose $7$ students to be with the girls in the first row. There are $\binom{13}{7}$ ways to do this. The remaining six students are with the boys. There are $3!$ ways to arrange the girls and $4!$ ways to arrange the boys. If you treat the groups of boys and girls as single objects, you have $8$ objects in the first row and $7$ objects in the second row. Then the total number of ways to arrange the students as required is $$\binom{13}{7} \cdot 3! \cdot 7! \cdot 4! \cdot 8!$$
Best Answer
It seems to me that you've accounted for all of the ways the particulars can be ordered within the set of particulars, but not for how they might be interspersed with the row at large. My answer would have been ${13 \choose 7} \times 10! \times 10!$ -- that is, I worry about the ordering of the rows at the end rather than ordering the particulars and non-particulars separately. Is that the answer they were looking for?