Ten-digit codes that fall then rise

A ten-digit code $ABCDEFGHIJ$ uses each digit $0$ through $9$ exactly once. The digits must strictly decrease from $A$ to $E$ ($A>B>C>D>E$) and then strictly increase from $E$ to $J$ ($E<F<G<H<I<J$). How many such codes are there?

$126$. $E$ is below every other digit, so $E=0$. Choosing which 4 of the remaining 9 digits go before it fixes the whole code, since each side has only one valid order. That gives $\binom94=126$.