Question
If $^nC_9 =\ ^nC_8,$ find $^nC_{17}$

Answer

We have, $^nC_9 =\ ^nC_8$
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 $n = 17$
Thus, $^nC_{17} =\ ^{17}C_{17} = 1$

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