Question
If ${ }^n C_9={ }^n C_8$, find ${ }^n C_{17}$

Answer

We have, ${ }^n \mathrm{C}_9={ }^{\mathrm{n}} \mathrm{C}_8$
i.e., $\frac{n!}{9!(n-9)!}=\frac{n!}{(n-8)!8!}$
or $\frac{1}{9}=\frac{1}{n-8}$ or $\mathrm{n}-8=9$ or $\mathrm{n}=17$
Thus, ${ }^n \mathrm{C}_{17}={ }^{17} \mathrm{C}_{17}=1$

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