If the coefficients of $x$ and $x^{2}$ in $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$ are 4 and $-$5 respectively, then $2 p+3 q$ is equal to :

We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and <br/><br/>coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$ <br/><br/>$$ \begin{aligned} & (1+x)^p(1-x)^q \\\\ & =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\ & =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\right) x^2+\ldots \ldots \end{aligned} $$ <br/><br/>Coefficient of $x$ in $(1+x)^p(1-x)^q=-q+p$ <br/><br/>$\Rightarrow p-q=4$ ...........(i) <br/><br/>$$ \text { Coefficient of } x^2 \text { in }(1+x)^p(1-x)^q=\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{q^2-q-2 p q+p^2-p}{2} =-5 \\\\ & \Rightarrow \frac{p^2+q^2-2 p q-(p+q)}{2} =-5 \\\\ & \Rightarrow (p-q)^2-(p+q) =-10 \\\\ & \Rightarrow (4)^2-(p+q) =-10 \quad[\because \text { From Eq. (i) }] \\\\ & \Rightarrow p+q =26 ...........(ii) \end{aligned} $$ <br/><br/>Form Eqs. (i) and (ii), we get $p=15, q=11$ <br/><br/>$\therefore 2 p+3 q=2 \times 15+3 \times 11=30+33=63$