The coefficient of x⁴ in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^6$in powers of x is ....... 

The coefficient of x⁴ in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^6$

= coefficient of x⁴ in ${\left(\frac{1-{\mathrm{x}}^4}{1-\mathrm{x}}\right)}^6$

$\begin{array}{l}=\text{ coefticient of }{\mathrm{x}}^4\text{ in }\left(1-6{\mathrm{x}}^4\right)(1-\mathrm{x}{)}^{-6}\\ =\text{ coefficient of }{\mathrm{x}}^4\text{ in }\left(1-6{\mathrm{x}}^4\right)\left[1{+}^6{\mathrm{C}}_1\mathrm{x}{+}^7{\mathrm{C}}_2{\mathrm{x}}^2+\dots \right]\end{array}$

${=}^9{\mathrm{C}}_4-6=\frac{9\times 8\times 7\times 6}{4\times 3\times 2}-6=126-6=120$