For the natural numbers $m, n,$ if $(1-y{)}^m(1+y{)}^n=1+a_{1}y+a_2{y}^2+\dots .+a_{m+n}{y}^{m+n}$ and $a_1=a_2=10,$ then the value of $m+ n,$ is equal to:

We have,

$(1-y{)}^m(1+y{)}^n$

$=\left({}^{m}C_0-{}^{m}C_{1}y+{}^{m}C_2{y}^2+...+{}^{m}C_m{\left(-1\right)}^m{y}^m\right)\left({}^{n}C_0+{}^{n}C_{1}y+{}^{n}C_2{y}^2+...+{}^{n}C_n{y}^n\right)$

Coefficient of $y\left(a_1\right)=1·{}^{n}C_1+{}^{m}C_1(-1)$

$=n-m=10 \dots \left(1\right)$

Coefficient of $y^2\left(a_2\right)$

$=1{}^{n}C_2-{}^{m}C_1{}^{n}C_1+1^m{C}_2=10$

$\Rightarrow \frac{n(n-1)}{2}-m·n+\frac{m(m-1)}{2}=10$

$\Rightarrow m^2+n^2-2mn-(n+m)=20$

$\Rightarrow (n-m{)}^2-(n+m)=20$

$\Rightarrow n+m=80 \dots \left(2\right)$

By equation $\left(1\right)$ and $\left(2\right)$

$m=35, n=45$

Hence, $m+n=80$