Orders for breaking stacked targets column by column

Twelve targets hang in four vertical stacks containing $2$, $3$, $3$ and $4$ targets. On each shot an archer picks a stack that still has targets and breaks the lowest remaining target in it. Counting two sequences as different whenever the order in which targets break differs, how many ways can all twelve targets be broken?

$277{,}200$. Within a stack the order is forced, so a full sequence is determined by the list of stack labels chosen at each of the 12 shots. That is an arrangement of the multiset with 2, 3, 3 and 4 copies, counted by $\tfrac{12!}{2!\,3!\,3!\,4!}=277{,}200$.