If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$    the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

Solution :
We have $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n}$$
$$\eqalign{ & \left[ {1 + mx + \frac{{m\left( {m - 1} \right)}}{{2!}}{x^2} + .....} \right]\left[ {1 - nx + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^2} - .....} \right] \cr & = 1 + \left( {m - n} \right)x + \left[ {\frac{{m\left( {m - 1} \right)}}{2} + \frac{{n\left( {n - 1} \right)}}{2} - mn} \right]{x^2} + ..... \cr & {\text{Given, }}m - n = 3\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}\frac{1}{2}m\left( {m - 1} \right) + \frac{1}{2}n\left( {n - 1} \right) - mn = - 6 \cr & \Rightarrow \,\,{m^2} + {n^2} - 2mn - \left( {m + n} \right) = - 12 \cr & \Rightarrow \,\,{\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12 \cr & \Rightarrow \,\,m + n = 9 + 12 = 21\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2), we get
$$m = 12$$