You start with a zero and try to reach exactly 100 in 4 stages, each stage consisting of either increasing or decreasing your current number by a factorial n!( number like 1,2,6,24,120,720 etc) .

How many sequences exist to accomplish this task?

I am assuming that selection of the first addend is the "first" STAGE (adding it to zero, the starting point). Hence the combination 2! + 2! + 5! - 4! is the only SET, since any higher factorial is more than 100 greater than any three lower factorials.

This seems too simple. Perhaps by SEQUENCE each ordering of these operations is considered a new sequence. There would be 24 orders if the four bases were all different, but since 2 is used twice, there are only 12 distinct sequences:

2245 2254 2425 2452 2524 2542 4225 4252 4522 5224 5242 5422

This assumes that the current number at some point(s) may be negative. Other interpretations?