The sum of the binomial coefficients of the $3^{\mathrm{rd}}, 4^{\mathrm{th}}$ terms from the beginning and from the end of ${(\mathrm{a}+\mathrm{x})}^{\mathrm{n}}$ is 440 then $\mathrm{n}=$ 
 

${ }^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_3{+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-2}{+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-3}=440$

$2\left({ }^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_3\right)=440$

${ }^{\mathrm{n}}{C}_2{+}^{\mathrm{n}}{C}_3=220$

$\frac{\mathrm{n}(\mathrm{n}-1)}{2!}+\frac{\mathrm{n}(\mathrm{n}-2)(\mathrm{n}-1)}{3!}=220$

$\frac{\mathrm{n}(\mathrm{n}-1)}{2}\left(1+\frac{\mathrm{n}-2}{3}\right)=220$

$\frac{\mathrm{n}(\mathrm{n}-1)}{2}\left(\frac{\mathrm{n}+1}{3}\right)=220$

$\frac{\mathrm{n}\left({\mathrm{n}}^2-1\right)}{6}=220$

${\mathrm{n}}^3-\mathrm{n}-1320=0$

$\text{∴ }\mathrm{n}=11$