Question

The number of ways of choosing triplet $(x,y,z)$ such that $z>max\text{of}(x,y)$ and $x,y,z\in \{1,2,...,n,n+1\}$ is

• A. ${}^{n+1}{C}_3+{}^{n+2}{C}_3$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. ${}^{n+3}{C}_3$
• D. $2{(}^{n+2}{C}_3)-{}^{n+1}{C}_2$

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{n(n+1)(2n+1)}{6}$
D $2{(}^{n+2}{C}_3)-{}^{n+1}{C}_2$

When $z=n+1$, we can choose $x,y$ from $\{1,2,...,n\}$.
When $z=n+1,x,y$ can be chosen in $n^2$ ways and when $z=n,x,y$ can be chosen in $(n-1{)}^2$ ways and so on. Therefore, the number of ways of choosing triplets is

$n^2+(n-1{)}^2+....+1^2=\frac{1}{6}n(n+1\left)\right(2n+1)$

Alternate Solution:
Possiblilities of the triplets are
$x=y<z,x<y<z,y<x<z$ can be chosen in ${}^{n+1}{C}_2,{}^{n+1}{C}_3,{}^{n+1}{C}_3$ ways. Therefore number of ways
$={}^{n+1}{C}_2+2{(}^{n+1}{C}_3)={}^{n+2}{C}_3+{}^{n+1}{C}_3=2{(}^{n+2}{C}_3)-{}^{n+1}{C}_2$