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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