Question

A three-digit code for certain locks uses the digits $0,1,2,3,4,5,6,7,8,9$ according to the following constraints. The first digit cannot be $0$ or $1$, the second digit must be $0$ or $1$, and the second and third digits cannot both be $0$ in the same code. How many different codes are possible?

• A. $144$
• B. $152$
• C. $160$
• D. $168$
• E. $176$

Answer 1

Answer: (B)

$152$

The first digit can be fill in $8$ ways. For $2nd$ digit it can be one of $(0,1)$

Now, assume second digit as $1$ the $3rd$ digit can take $10$ values.

$\therefore$ Number of codes $=8\times 1\times 9=72$

$\therefore$ total $80+72=152$

Hence, the answer is $152.$