M13-35

Question

How many distinct positive four-digit integers can be formed using the digits 0, 2, and 3?

A. 48
B. 54
C. 64
D. 72
E. 81

Bunuel · Answer

Official Solution:

How many distinct positive four-digit integers can be formed using the digits 0, 2, and 3?

A. 48
B. 54
C. 64
D. 72
E. 81

For the first digit, there are two options: 2 or 3. For each of the remaining three digits there are three options: 0, 2, or 3. Therefore, the total number of combinations is  2*3*3*3 = 54 .

Answer: B

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sb995 · Answer

I did not quite understand the solution. Hi, could you please explain why the following does not provide the correct answer. Want to make sure I am clear on the concept. Thank you:

Each number except 0 can be placed into any of the 4 slots of a 4 digit integer so the following combination for the three digits would be: 3 * 4 * 4.

0 can be placed into any of the 3 slots, 2 can be placed into any of the 4 slots, and 3 can be placed into any of the 4 slots.

Bunuel · Answer

Your method doesn’t make any sense. You’re treating it like we are “placing” the digits 0, 2, and 3 into slots one by one, but that’s not the problem. If you say, for example, that 0 goes into the second digit, then that position is fixed, 2 doesn’t suddenly still have four options, because one slot is already occupied. Digits don’t get assigned to slots like that. The correct approach is to count choices per position: first digit has 2 options (2 or 3), and each of the next three digits has 3 options (0, 2, 3). That gives 2*3*3*3 = 54.

M13-35

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