Question

The number of three digit numbers $N=abc$ such that $a\le b>c$, is equal to

• A. 258
• B. 265
• C. 285
• D. 295

Answer 1

Answer: (C)

(i) $a\ne b\ne c($ all distinct and $c\ne 0)\Rightarrow$ Total cases $=2\left({}^9{C}_3\right)$ e.g. $1,2,3$

(ii) $a\ne b,c=0(a<b>0)\Rightarrow$ Total cases $={}^9{C}_2$
(iii) $a=b,c\ne 0(a=b>c)\Rightarrow$ Total cases $={}^9{C}_2$
(iv) $a=b,c=0(a=b>0)\Rightarrow$ Total cases $={}^9{C}_1$
(v) $a=c\ne 0(a<b>c)\Rightarrow$ Total cases $={}^9{C}_2$ e.g. $1,4\Rightarrow 141$ (formed)
$\therefore$ Number of such 3 digit numbers $=2\cdot \left({}^9{C}_3\right)+3\cdot \left({}^9{C}_2\right)+{}^9{C}_1=285$.