Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1. Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1.
My approach:
Total number of cases=$9^3$
Considering that a,b,c are 0 to 9 and and for the HCF to be 1 there should be no common factor of 2 or 3 .
So total number of cases is equal to
$9^3-3^3-2^3=694 $
Now i might have made a total mess here but i what should be the shortest possible method and obviously the correct answer?

For each leading digit from $1$ to $9$ there are $100$ possible integers. If the second and third digits both share a divisor greater than $1$ with the leading digit then we exclude that case.
So if $n$ digits share a divisor with the leading digit, we exclude $n^2$ cases.
$$
\begin{array}{c|crrr}
\text{Leading digit} & \text{Excluded digits} & \text{Size} & \text{Size}^2 & 100-\text{Size}^2 \\
\hline
1 & \text{None} & 0 & 0 & 100 \\
2 & 0,2,4,6,8 & 5 & 25 & 75 \\
3 & 0,3,6,9 & 4 & 16 & 84 \\
4 & 0,2,4,6,8 & 5 & 25 & 75 \\
5 & 0,5 & 2 & 4 & 96 \\
6 & \text{See below} & & & 63 \\
7 & 0,7 & 2 & 4 & 96 \\
8 & 0,2,4,6,8 & 5 & 25 & 75 \\
9 & 0,3,6,9 & 4 & 16 & 84 \\
Total & & & & 748 \\
\end{array}
$$
The leading digit of $6$ is a special case because both $2$ and $3$ are divisors. We can use Inclusion-Exclusion.
Exclude $\{0,2,4,6,8\}$ for $25$ cases. Exclude $\{0,3,6,9\}$ for $16$ cases. But $\{0,6\}$ has been double-counted so add back $4$ cases. $100-25-16+4=63$
