three-digit code

[gmat740]

A three-digit code for certain logs 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

[spoiler]OA-B
IMO-A[/spoiler]

[prindaroy]

Case 1:

8 possibilities for 1st number, number 2 is 1, 10 possibilities for 3rd number

so 8 * 10 = 80

Case 2:

8 possibilities for 1st number, number 2 is 0, 9 possibilities for 3rd number

so 8 * 9 = 72

so total = 80 + 72 = 152 or B

[[email protected]]

Hi All,

We're told that a three-digit code for certain logs 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. We're asked for the total number of possible codes. This question is ultimately about a Permutation - but you'll also have to note that there are a few codes within that permutation that must be REMOVED.

The first digit can be 2, 3, 4, 5, 6, 7, 8 or 9 --> 8 possible digits
The second digit can be 0 or 1 --> 2 possible digits
The third digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 --> 10 possible digits.... but there is an additional rule (the code CANNOT end in '00').

Without that final 'restriction', there would be (8)(2)(10) = 160 codes
WITH that restriction though, we have to remove 8 of the codes: 200, 300, 400, 500, 600, 700, 800 and 900

Total possible codes = 160 - 8 = 152

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

A three-digit code for certain logs 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

Case 1: 2nd digit is 0
Number of options for the 3rd digit = 9. (Any digit but 0, since the 2nd and 3rd digits cannot both be 0.)
Number of options for the 1st digit = 8. (Any digit but 0 or 1.)
To combine these options, we multiply:
9*8 = 72.

Case 2: 2nd digit is 1
Number of options for the 3rd digit = 10.
Number of options for the 1st digit = 8. (Any digit but 0 or 1.)
To combine these options, we multiply:
10*8 = 80.

Total ways = Case 1 + Case 2 = 72 + 80 = 152.

The correct answer is B.

[Scott@TargetTestPrep]

A three-digit code for certain logs 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

[spoiler]OA-B
IMO-A[/spoiler]

Solution:

We see that the first digit has 8 choices, the second has 2, and the third has 10. Therefore, there are 8 x 2 x 10 = 160 possible 3-digit codes. However, this includes codes such as 200, 300, …, and 900, which are not allowed. Therefore, there are actually 160 - 8 = 152 possible codes.

Answer: B