No Face-to-face!

We solve using casework. We see that there are 210 ways to arrange them without restrictions. Now we must subtract ways that have the girls facing each other. Note that there are 3 pairs that could have girls facing each other. Just using simple casework we find there are 166 ways to arrange the kids so that girls that face each other. Subtracting, we see that the answer is 44 ways.

Basically we have four seats not facing any other seat (call them "single" seats) and three pairs of seats that face each other (call them "double seats"). Only one girl can occupy each pair, so we only have two cases:

• Case 1: Four girls occupy the single seats, two girls occupy the double seats

Then there's clearly only one way for the single seats to be taken (girls aren't distinguished). For the double seats, there are $\binom{3}{2}$ ways to choose the pairs that get the girls, and $2^2$ ways to choose which of the seats in the respective pairs are taken by the girls. This gives a total of $1 \cdot \binom{3}{2} \cdot 2^2 = 12$ ways.

• Case 2: Three girls occupy the single seats, three girls occupy the double seats.

Following the same logic, there are $\binom{4}{3}$ ways to choose the occupied single seats. Each pair of double seats gains one girl, and there are $2^3$ ways to choose the seats in the respective pairs that are taken by the girls. So the number of ways is $\binom{4}{3} \cdot 2^3 = 32$ .

Adding the two together gives $12+32 = \boxed{44}$ .

same as ivan