Question
If ${^\text{16}}\text{C}_{\text{r}}={^\text{16}}\text{C}_{\text{r+2}},$ find ${^\text{7}}\text{C}_{4}.$

Answer

We have, If ${^\text{n}}\text{C}_{\text{r}}={^\text{n}}\text{C}_{\text{p}}$ then r + p = n 16 = r + r + 2 r = 7 then ${^\text{n}}\text{C}_{\text{4}}={^\text{7}}\text{C}_{\text{4}}$ $\Rightarrow \frac{7!}{4!(7-4)!}$ $\Rightarrow \frac{7\times5\times6}{3\times2}$ $\Rightarrow 35$

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