If ${ }^{2n}{C}_3{:}^n{C}_2=44:3,$ hen for which of the following values of r, the value of ${ }^n{C}_r$ will be 15 ?

$$\begin{gathered} \frac{\left(2n\right)!}{(2n-3)!\cdot 3!}\times \frac{2!\times (n-2)!}{n!}=\frac{44}{3} \\ \begin{array}{l}\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{3n(n-1)}=\frac{44}{3}\\ \Rightarrow 4(2n-1)=44\\ \Rightarrow 2n=12\\ \Rightarrow n=6\end{array} \end{gathered}$$

Now, $$\begin{gathered} { }^6{C}_r=15 \\ {\Rightarrow }^6{\mathrm{C}}_r{=}^6{\mathrm{C}}_2 \end{gathered}$$

or ${ }^6{\mathrm{C}}_4$

$\Rightarrow r=2,4$