Question

The number of three digit numbers of the form xyz such that x < y and $z\le y$ is
If zero is included it will be at ${}^9{C}_2$ no. s
If zero is excluded \( \left\{$$\begin{array}{ccc}x,y,zalldiff.& \Rightarrow & {}^9{C}_3\times 2!\\ x=z<y& \Rightarrow & {}^9{C}_2\\ x<y=z& \Rightarrow & {}^9{C}_{2}N{o}^{\prime }s\end{array}$$\right\)
Total number of ways = 276
Alternative
y can be from 2 to 9 so total number of ways =${\sum }_{r=2}^9(r^2-1)=276$

• A. 276
• B. 285
• C. 240
• D. 244

Answer 1

The correct option is A 276

If zero is included it will be at ${}^9{C}_2$ no. s
If zero is excluded \{begin{matrix}
x,y,z ~all~diff. & \Rightarrow & ^9C_3 \times 2!\\
x=z < y & \Rightarrow & ^9C_2\\
x < y =z & \Rightarrow & ^9C_2 No's
\end{matrix}\
Total number of ways = 276
Alternative
y can be from 2 to 9 so total number of ways =${\sum }_{r=2}^9(r^2-1)=276$