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Q. If $^nC_9 = {^nC_8}$, what is the value of $^nC_{17}$ ?

Permutations and Combinations

Solution:

$\frac{n!}{9!\left(n-9\right)!} = \frac{n!}{8!\left(n-8\right)!}$
$ \Rightarrow \frac{1!}{9\times8! \left(n-9\right)!} = \frac{1!}{8!\left(n-8\right)\left(n-9\right)!} $
$\Rightarrow \frac{1}{9} = \frac{1}{\left(n-8\right) } \Rightarrow 9 = n - 8$
$ \Rightarrow 9 + 8 = n \Rightarrow n = 17$
$ \therefore ^{n}C_{17} = ^{17}C_{17} = 1 $ $[ \because \, {^{n}C_{n}} = 1 ] $