Question

How many numbers between 100 and 1000 are such that exactly one of the digits is 6?

Answer 1

fixing $6$ in hundreds place

$\frac{6}{\overline{\downarrow 1}}\frac{}{\overline{\downarrow 9}}\frac{}{\overline{\downarrow 9}}\Rightarrow 1\times 9\times 9=81$

$9$ numbers $(0,1,2,3,4,5,6,7,8,9)$ are possible in units and tens place

fixing $6$ in tens place

$\frac{}{\overline{\downarrow 8}}\frac{6}{\overline{\downarrow 1}}\frac{}{\overline{\downarrow 9}}\Rightarrow 8\times 1\times 9=72(1,2,3,4,5,7,8,9)(0,1,2,3,4,5,7,8,9)$

fixing $6$ in units place

$\frac{}{\overline{\downarrow 8}}\frac{}{\overline{\downarrow 9}}\frac{}{\overline{\downarrow 1}}\Rightarrow 8\times 1\times 9=72(1,2,3,4,5,7,8,9)(0,1,2,3,4,5,7,8,9)\Rightarrow 8\times 9\times 1=72$

Hence, total no. of possible selection

$=81+72+72$

$=225$