The coefficient of $x^7$ in the polynomial expansion of 
\[(1+2x-x^2)^4\]
is
$\mathrm{(A)\ } -8 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } -12 \qquad \mathrm{(E) \  }\text{none of these}$
Let's write out the multiplication, so that it becomes easier to see.
\[(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)\]
We can now see that the only way to get an $x^7$ is by taking three $-x^2$ and one $2x$. There are $\binom{4}{1}=4$ ways to pick which term the $2x$ comes from, and the coefficient of each one is $(-1)^3(2)=-2$. Therefore, the coefficient of $x^7$ is $(4)(-2)=-8, \boxed{\text{A}}$.