$(r+1{)}^{th}$ term in the expansion of $(1-x{)}^{-4}$ will be

The correct option is B

$(1-x{)}^{-4}$ = [$\frac{\mathrm{1.2.3}}{6}{x}^0+\frac{\mathrm{2.3.4}}{6}x+\frac{\mathrm{3.4.5}}{6}{x}^2+\frac{\mathrm{4.5.6}}{6}{x}^3+.......+\frac{(r+1)(r+2)(r+3)}{6}{x}^r+.........]$

Therefore $T_{r+1}$ = $\frac{(r+1)(r+2)(r+3)}{6}{x}^r$.