If in the binomial expansion of $(1-x{)}^m(1+x{)}^n$, the coefficients of $x\text{ and }{x}^2$ are respectively 3 and -4, then the ratio $m:n$ is equal to 

We have 

$\begin{array}{l}(1-x{)}^m(1+x{)}^n\\ =\left(1-mx{+}^m{C}_2{x}^2-\dots \right)\left(1+nx{+}^n{C}_2{x}^2+\dots \right)\\ =1+(n-m)x+\left({ }^m{C}_2-mn{+}^n{C}_2\right)x^2+\dots \end{array}$

we are given 

$\begin{array}{l}n-m=3{,}^m{C}_2-mn{+}^n{C}_2=-4\\ \Rightarrow n-m=3,\frac{1}{2}\left(m^2-2mn+n^2-(m+n)\right)=-4\\ \Rightarrow n-m=3,(m-n{)}^2-(m+n)=-8\end{array}$

$\begin{array}{l}\Rightarrow n-m=3,m+n=17\\ \therefore n=10,m=7\\ \Rightarrow m:n=7:10\end{array}$