$1^{st}$ solution : $^{n+1}C_{3} -^{n}C_{3} = 10 $
$\Rightarrow \frac{\left(n+1\right)n\left(n-1\right)}{6} -\frac{n\left(n-1\right)\left(n-2\right)}{6}=10$
$ \Rightarrow 3n\left(n-1\right)=60 \Rightarrow n\left(n-1\right)=20$
$ \Rightarrow n^{2} -n -20 =0 $
$\Rightarrow \left(n-5\right)\left(n+4\right)= 0 \:\:\: \therefore \:n=5$
$2^{nd}$ solution : $^{n+1}C_{3} - \,^{n}C_{3} =10$
$ \Rightarrow \,^{n}C_{2} =10 \Rightarrow \frac{n\left(n-1\right)}{2}=10 $
$\Rightarrow n^{2} -n -20 =0.\, \therefore \: n=5$
Here we have used $^{n}C_{r}+\,^{n}C_{r+1}=\,^{n+1}C_{r+1}$