Prove that ${^{15}}{C}_{3} + 2 \times {^{15}}{C}_{4} + {^{15}}{C}_{5} = {^{17}}{C}_{5}$
$\begin{aligned} LHS = {^{15}}{C}_{3} + 2 \times {^{15}}{C}_{4} + {^{15}}{C}_{5} & = {^{15}}{C}_{3} + {^{15}}{C}_{4} + {^{15}}{C}_{4} + {^{15}}{C}_{5} \\\\ & = \left({^{15}}{C}_{4} + {^{15}}{C}_{3}\right) + \left({^{15}}{C}_{5} + {^{15}}{C}_{4}\right) \\\\ & = {^{16}}{C}_{4} + {^{16}}{C}_{5} \;\;\;\;\; \left[\because \; {^{n}}{C}_{r} + {^{n}}{C}_{\left(r - 1\right)} = {^{\left(n + 1\right)}}{C}_{r}\right] \\\\ & = {^{17}}{C}_{5} = RHS \end{aligned}$