With Greedy algorithm, the order of multiplication is:
$A_{1}(A_{2}A_{3})$ , therefore the number of multiplications is $400 + 600 = 1000$
With dynamica programming which always picks the optimal order, the order is:
$(A_{1}A_{2})A_{3}$ , therefore the number of multiplications is $600 + 12 = 612$
Hence, Person Y would save $388$ multiplications.