Photo line-up with the top three in the middle

An $11$-member chess club lines up for a photo. The three highest-rated members stand in the middle three spots in any order. The other eight split four to each side, and on each side the ratings must be monotone (each side may independently be increasing or decreasing). All ratings differ. How many line-ups are possible?

$1680$. The middle three can be arranged in $3!=6$ ways. Choose which four of the other eight go on the left, $\binom84=70$; each side then has $2$ admissible orderings. Total $6\cdot70\cdot2\cdot2=1680$.