Question

A three-digit number for certain locks uses the digits $0,1,2,3,4,5,6,7,8,9$according to the following constraints. The first digits cannot be $0$or $1$, the second digit must be $0$ or $1$, and the second and third digits cannot be $0$in the same code. How many different codes are possible?
A) $144$
B) $152$
C) $160$
D) $168$
E) $176$

Answer 1

Hint: In this question evaluate the possible combination for first digit, second digit and third digit according to the given condition. There will be two cases for the second and third digits.

Complete step-by-step answer:
We are given that a three-digit number for certain locks uses the digits $0,1,2,3,4,5,6,7,8,9$.
There are some constraints in the question.
We solve our problem by making the cases.
Case-1
We are given that first digits cannot be $0$or $1$ it means the possible cases for first digit will be $8$
The second digit must be $0$ or $1$
In this case we consider the second digit will be $0$.
For the third digit the condition is that second and third digits cannot be $0$in the same code.
It means in this case no $0$ for the third digit.
Therefore, for the third digit there will be $9$ possibilities.
Hence, in this case the total codes will be $8 \times 9 = 72$
Case-2
The number of cases for the first digit will remain the same.
In this case we consider the second digit will be $1$.
If the second digit will be $1$ then there is no constraint for the third digit.
It means for the third digit there will be $10$ possibilities.
Hence, in this case the total codes will be $8 \times 10 = 80$
There will be no further cases.
Therefore, the total number possibilities will be the sum of both the cases.

Hence, $80 + 72 = 152$ different codes are possible.
Therefore, option (B) is correct.

Note:
In these types of questions the constraints define the possibilities. Analyse the constraint given in the question and then make cases according to them and don’t forget to add the total number of each case.