If ${^{n}}{C}_{10} = {^{n}}{C}_{12}$, find ${^{23}}{C}_{n}$
Given: $\;$ ${^{n}}{C}_{10} = {^{n}}{C}_{12}$
i.e. $\;$ $\dfrac{n!}{\left(n - 10\right)! \times 10!} = \dfrac{n!}{\left(n - 12\right)! \times 12!}$
i.e. $\;$ $\left(n - 12\right)! \times 12! = \left(n - 10\right)! \times 10!$
i.e. $\;$ $\left(n - 12\right)! \times 12 \times 11 \times 10! = \left(n - 10\right) \times \left(n - 11\right) \times\ \left(n - 12\right)! \times 10!$
i.e. $\;$ $12 \times 11 = \left(n - 10\right) \times \left(n - 11\right)$
$\implies$ $n - 10 = 12$ $\;$ OR $\;$ $n - 11 = 11$
i.e. $\;$ $n = 22$
$\begin{aligned}
\therefore \; {^{23}}{C}_{n} & = {^{23}}{C}_{22} \\\\
& = \dfrac{23!}{\left(23 - 22\right)! \times 22!} \\\\
& = \dfrac{23!}{22!} = 23
\end{aligned}$