smc

Suppose that $P(x)=a{x}^3+b{x}^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $$P(-1)=-a+b-c+d=-9.$$ Let $a^{\prime }=9-a$ and $c^{\prime }=9-c.$ Note that both of $a^{\prime }$ and $c^{\prime }$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a^{\prime },b,c^{\prime },d)$ have one-to-one correspondence. We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or $$a^{\prime }+b+c^{\prime }+d=9.$$ By the stars and bars argument, there are $\left(\genfrac{}{}{0ex}{}{9+4-1}{4-1}\right)=\boxed{220}$ ordered quadruples $(a^{\prime },b,c^{\prime },d).$