Question

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

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

Answer 1

Answer: (B), (C), (D)

The correct options are
B $n(n+1)(2n+1)/6$
C $1^2+2^2+...+n^2$
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
$n^2+(n-1{)}^2+...+1^2=\frac{1}{6}n(n+1\left)\right(2n+1)$
Alternatively triplets with $x=y<z,x<y<z,y<z<x$ can be chosen in ${}^{n-1}{C}_2{,}^{n+1}{C}_3{,}^{n+1}{C}_3$ ways. Therefore,
${}^{n+1}{C}_2+2{(}^{n+1}{C}_3){=}^{n+2}{C}_2{+}^{n+1}{C}_3=2{(}^{n+2}{C}_3){-}^{n+1}{C}_2$.