Given $C(n, 12)=C(n, 8)$ $\begin{array}{l}\Rightarrow{ }^{n} C_{12}={ }^{n} C_{8} \\ \Rightarrow \frac{n !}{(n-12) ! 12 !}=\frac{n !}{(n-8) ! 8 !} \\ \Rightarrow \frac{1}{(n-12) !(12 \times 11 \times 10 \times 9 \times 8 !)} \\ = & \frac{1}{(n-8)(n-9)(n-10)(n-11)(n-12) ! 8 !} \\ \Rightarrow \frac{1}{12 \times 11 \times 10 \times 9}=\frac{1}{(\mathrm{n}-8)(\mathrm{n}-9)(\mathrm{n}-10)(\mathrm{n}-11)} \\ \Rightarrow(n-8)(n-9)(n-10)(n-11) \\ =12 \times 11 \times 10 \times 9 \\ \Rightarrow n-8=12, n-9=11, n-10=10 \text { and } n-11=9 \\ \Rightarrow \mathrm{n}=20 \\ \Rightarrow C(22, n)={ }^{22} C_{20} \\ \quad=\frac{22 !}{2 ! 20 !}=\frac{22 \times 21}{2}=231\end{array}$