Question

Number of natural numbers between $100$ and $1000$ such that at least one of their digits is $7$, is

• A. $243$
• B. $252$
• C. $268$
• D. $648$

Answer 1

The correct option is B $252$

Case$I$ When only one digit is $7$

$\left(i\right)$ When the ones place is $7$

it can be done in $8\times 9\times 1$ways$=72$ways

When only one digit is $7$

$\left(ii\right)$ When the tens place is $7$

it can be done in $8\times 9\times 1$ways$=72$ways

$\left(ii\right)$ When the hundreds place is $7$

it can be done in $9\times 9$ways$=81$ways

Thus, when only one digit is $7$, it can be done in $72+72+81=225$ways

Case$II$ When two digits are $7$

$\left(i\right)$when ones and tens place is $7$, it can be done in $8$ ways.

$\left(ii\right)$when hundreds and tens place is $7$, it can be done in $9$ ways.

$\left(iii\right)$when ones and hundreds place is $7$, it can be done in $9$ ways.

Total in $8+9+9=26$ ways

Case$III$ When all three digits are $7$ only $1$ is possible

Thus, total number of ways$=225+26+1=252$ways.