Question

The number of three-digit numbers of the form $xyz$ such that $x<y$ and $z\le y$ is

• A. 279
• B. 285
• C. 240
• D. 244

Answer 1

Answer: (A)

If zero is included it will be at $z\Rightarrow {}^9{C}_2$
If zero is excluded $\{\begin{array}{ll}x,y,z\text{ all diff. }& \Rightarrow {}^9{C}_3\times 2!\\ x=z<y& \Rightarrow {}^9{C}_2\\ x<y=z& \Rightarrow {}^9{C}_2\end{array}$
The total number of ways is $276$ .
Alternate method :
$y$ can be from $2$ to $9$ ; so total number of ways is
$\sum _{r-2}^9\left(r^2-1\right)=276$