Question

Let $k$ be the number of three digit numbers, which are of the form $xyz$, with $x<y,z<y$ and $x\ne 0$. Find sum of digits of $k$.

Answer 1

Since $x\ge 1$, then $y\ge 2$
$\because$ If $y=n$, then $x$ takes the values from $1$ to $n-1$ and $z$ takes the values from $0$ to $n-1$ for $n>1$.
Thus for each values of $y,(2\le y\le 9),$ then total combinations of $x$ & $z$ are $n(n-1)$.
Hence the three digit numbers are of the form $xyz$
$\sum _{n=2}^{9}n(n-1)=\sum _{n=1}^{9}n(n-1)=\sum _{n=1}^9{n}^2-\sum _{n=1}^{9}n$
$=\frac{9(9+1)(18+1)}{6}-\frac{9.(9+1)}{2}=285-45=240$
Sum of digit is $2+4+0=6$