Last Updated on May 10, 2023
GMAT OFFICIAL GUIDE PS
Solution:
We are given the following constraints for a 3-digit code:
The first number 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. Thus, we must create two different scenarios for our digits.
Scenario One (second digit is zero)
1st digit: 8 values (since it can be any digit from 2 to 9, inclusive)
2nd digit: 0
3rd digit: 9 values (since it can’t be 0 again but it can be any digit from 1 to 9, inclusive)
Thus, in scenario one there are 8 x 1 x 9 = 72 ways to create the 3-digit code.
Scenario Two (second digit is 1)
1st digit: 8 values (since it can be any digit from 2 to 9, inclusive)
2nd digit: 1
3rd digit: 10 values (since it can be any digit from 0 to 9, inclusive)
Thus in scenario two there are 8 x 1 x 10 = 80 ways to create the 3-digit code
Combining our scenarios there are 72 + 80 = 152 ways to create the 3-digit code.
Answer: B
