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Question Number 129642 by SOMEDAVONG last updated on 17/Jan/21

$$\mathrm{show}\mathrm{that}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}={\mathrm{C}}_{\mathrm{n}-1}^{\mathrm{k}-1}+{\mathrm{C}}_{\mathrm{n}-2}^{\mathrm{k}-1}+....+{\mathrm{C}}_{\mathrm{k}}^{\mathrm{k}-1}+{\mathrm{C}}_{\mathrm{k}-1}^{\mathrm{k}-1}.$$

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