Question
If $^nC_{10} = ^nC_{12}$, Find $^{23}C_n$.

Answer

We have, If ${ }^n \mathrm{C}_{\mathrm{p}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{q}}=\mathrm{n}$
Then $\mathrm{p}+\mathrm{q}=\mathrm{n}$ Also,
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{10}={ }^{\mathrm{n}} \mathrm{C}_{12} 10+12=\mathrm{n} \Rightarrow \mathrm{n}=22$
Applying (i), ${ }^{23} \mathrm{C}_{22}=\frac{23!}{22!1!}$ $=\frac{23 \times 22!}{22!}=23$

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