The number of ways of choosing triplet $(x,y,z)$ such that $z>max\left\{\begin{array}{c}x,y\end{array}\right\}andx,y,z\in \left\{\begin{array}{c}1,2,\dots ,n,n+1\end{array}\right\}$ is

The correct options are
A $1^2+2^2+...+n^2$
C $2^{\left(\begin{array}{c}{n+2}_{C_3}\end{array}\right)}{-}^{{n+1}_{C_2}}$
D $\frac{1}{6}n(n+1)(2n+1)$
When $z=n+1$ we can choose $x,y$ from $\left\{\begin{array}{c}1,2,\dots ,n\end{array}\right\}$
$\therefore$ when $z=n+1,x,y$ can be chosen in $n^2$ ways and $z=n,y$ can be chosen in $(n-1{)}^2$ ways and so on
$\therefore n^2+(n-1{)}^2+\cdots +1^2=\frac{1}{6}n(n+1)(2n+1)$ ways of choosing triplets
Alternatively triplets with $x=y<z,x<y<z,y<x<z$ can be chosen in ${{}^{n+1}}_{C_2},{{}^{n+1}}_{C_3},{{}^{n+1}}_{C_3}$ ways.
There are ${{}^{n+1}}_{C_2}+2\left(\begin{array}{c}{{}^{n+1}}_{C_3}\end{array}\right)={{}^{n+2}}_{C_2}+{{}^{n+1}}_{C_3}=2\left(\begin{array}{c}{{}^{n+2}}_{C_3}\end{array}\right)-{{}^{n+1}}_{C_2}$.