Question:
If nC10 = nC12, find 23Cn.
Solution:
Given: ${ }^{n} C_{10}={ }^{n} C_{12}$
We have:
${ }^{n} C_{10}={ }^{n} C_{12}$
$\Rightarrow n=12+10=22 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x=y\right.$ or, $\left.n=x+y\right]$
Now, ${ }^{23} C_{22}={ }^{23} C_{1} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$
$\Rightarrow{ }^{23} C_{22}={ }^{23} C_{1}=\frac{23}{1} \times{ }^{22} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$
$\Rightarrow{ }^{23} C_{22}=23 \quad\left[\because{ }^{n} C_{0}=1\right]$