Suppose $x^2$ occurs in the (r + 1)th term of the expansion $(1 + x)^m$. Now $T_{r+1}$ = $\,\,{}^{m}C_r x^r$ On comparing power of x in $x^2$ and $T_{r+1}$, we get r = 2. Thus coefficient of $x^2$ = $\,\,{}^{m}C_2$ ⇒ $\,\,{}^{m}C_2$ = 6 (given) ⇒ $\frac{m!}{2!(m-2)!}$ = 6 ⇒ m (m – 1) = 12 ⇒ $m^2$ – m – 12 = 0 ⇒ (m + 3) (m – 4) = 0 ⇒ m = – 3, 4 ⇒ m = 4 (Since m cannot be negative)