If $^{n} C _{9}=\,\,^{n} C _{8},$ find $^{n} C _{17}$

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We have $^{n} C _{9}=\,^{n} C _{8}$

i.e.,   $\frac{n !}{9 !(n-9) !}=\frac{n !}{(n-8) ! 8 !}$

or   $\frac{1}{9}=\frac{1}{n-8}$ or $n-8=9$ or $n=17$

Therefore   $^{n} C_{17}=\,^{17} C_{17}=1$

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