If ${ }^{n+1}{C}_3-{}^n{C}_3=21$, then $\text{'}n\text{'}=$

$\begin{array}{l}{ }^{n+1}{C}_3-{}^n{C}_3=21\\ { }^n{C}_2+{}^n{C}_3-{}^n{C}_3=21(\because {}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r)\\ { }^n{C}_2=21\\ \frac{n(n-1)}{2!}=21\Rightarrow n^2-n-42=0\\ (n-7)(n+6)=0\end{array}$

$n=7$ (or) $n=-6$