Given that $(1 + 2x - x^{2})^{4}$
Let $y=2x-x^{2}$
So we get $(1+y)^{4}$
Now we can apply Binomial Theorem
$(a+b)^{n} =\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}\cdot (b)^{k}$
$(1+y)^{4} = \sum_{k=0}^{4}\binom{4}{k}(1)^{4-k} \cdot(y)^{k}$
$(1+y)^{4} = \sum_{k=0}^{4}\binom{4}{k}y^{k}$
we can expend all the term
$(1+y)^{4} = \binom{4}{0}y^{0}+\binom{4}{1}y^{1}+\binom{4}{2}y^{2}+\binom{4}{3}y^{3}+\binom{4}{4}y^{4}$
$(1+y)^{4}=1+4y+6y^{2}+4y^{3}+y^{4}$
Now we can put the valye of $y=2x-x^{2},$and we get
$(1+2x-x^{2})^{4}=1+4(2x-x^{2})+6(2x-x^{2})^{2}+4(2x-x^{2})^{3}+(2x-x^{2})^{4}$
Only one term can be give the $x^{7}$ and term is $(2x-x^{2})^{4}$
So,we can apply again Binomial theorem,
$(2x-x^{2})^{4} =\sum_{r=0}^{4}\binom{4}{r}(2x)^{4-r}\cdot(-x^{2})^{r}$
We can expend all the term
$(2x-x^{2})^{4}=\binom{4}{0}(2x)^{4}\cdot(-x^{2})^{0}+\binom{4}{1}(2x)^{3}\cdot(-x^{2})^{1}+\binom{4}{2}(2x)^{2}\cdot(-x^{2})^{2}+\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}+\binom{4}{4}(2x)^{0}\cdot(-x^{2})^{4}$
Only one tern can give $x^{7}$ and the term is $\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}$
$\Rightarrow\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}$
$\Rightarrow 4(2x)\cdot(-x^{6})$
$\Rightarrow -8x^{7}$
So,coefficient of $x^{7}$ is$: -8$