AMC 12 Mock Exam 2 Problem 22

If the roots are $r,s,t$, then $P(x)=(x-r)(x-s)(x-t)$, so $$(1-r)(1-s)(1-t)=P(1)=1+a+b+c=24.$$ Conversely, any multiset $\{r,s,t\}$ of integers with $(1-r)(1-s)(1-t)=24$ gives a valid polynomial, and different multisets give different polynomials. With $u=1-r$, $v=1-s$, $w=1-t$, we count multisets $\{u,v,w\}$ of nonzero integers with product $24$. Either all three are positive or exactly two are negative.

All positive: $\{1,1,24\},\{1,2,12\},\{1,3,8\},\{1,4,6\},\{2,2,6\},\{2,3,4\}$: $6$ multisets.

Two negative: start from one of these six and choose which value stays positive. A multiset with three different values gives $3$ choices; $\{1,1,24\}$ and $\{2,2,6\}$ give $2$ each. That is $4\cdot3+2\cdot2=16$.

The total is $6+16=\boxed{22}$.