Solve the following


If nC10 = nC12, find 23Cn.


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, ${ }^{23} C_{22}={ }^{23} C_{1} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow{ }^{23} C_{22}={ }^{23} C_{1}=\frac{23}{1} \times{ }^{22} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$

$\Rightarrow{ }^{23} C_{22}=23 \quad\left[\because{ }^{n} C_{0}=1\right]$

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