The coefficient of ${\mathrm{x}}^4$ ln the expansion of ${\left(1-\mathrm{x}+{\mathrm{x}}^2-{\mathrm{x}}^3\right)}^4$ is

$\begin{array}{l}{\left(1-\mathrm{x}+{\mathrm{x}}^2-{\mathrm{x}}^3\right)}^4={\left(1+{\mathrm{x}}^2\right)}^4(1-\mathrm{x}{)}^4\\ =\left({ }^4{\mathrm{C}}_0{+}^4{\mathrm{C}}_1{\mathrm{x}}^2{+}^4{\mathrm{C}}_2{\mathrm{x}}^4{+}^4{\mathrm{C}}_3{\mathrm{x}}^6{+}^4{\mathrm{C}}_4{\mathrm{x}}^8\right)\left({ }^4{\mathrm{C}}_0{-}^4{\mathrm{C}}_1\mathrm{x}{+}^4{\mathrm{C}}_2{\mathrm{x}}^2{-}^4{\mathrm{C}}_3{\mathrm{x}}^3{+}^4{\mathrm{C}}_4{\mathrm{x}}^4\right)\end{array}$

$\therefore$ Coefficient ${\mathrm{x}}^4$ is ${ }^4{{\mathrm{C}}_0}^4{\mathrm{C}}_4{+}^4{{\mathrm{C}}_1}^4{\mathrm{C}}_2{+}^4{{\mathrm{C}}_2}^4{\mathrm{C}}_0=1+\left(4\right)\left(6\right)+6=1+24+6=31$