Question

If ${}^n{C}_{12}{=}^n{C}_8$, then find the value of ${}^n{C}_{17}$.

Answer 1

We have for ${}^n{C}_r{=}^n{C}_{n-r}$.

Now given

${}^n{C}_{12}{=}^n{C}_8$

or, ${}^n{C}_{n-12}{=}^n{C}_8$

$\Rightarrow n-12=8$

or, $n=20$.

Then ${}^n{C}_{17}$

${=}^{20}{C}_{17}$

${=}^{20}{C}_{20-17}$

${=}^{20}{C}_3$

$=\frac{\mathrm{20.19.18}}{\mathrm{3.2.1}}$

$=20.57$

$=1140$