Find the number of three-digit numbers (integers from 100 to 999) that contain no two consecutive equal digits.

My thoughts were, there a 1-9 choices for position one; so 9 choices. For the second position there are 0-9 choices but it cannot be the same as the first, so there are 9 choices, and for the last position there are 0-9 choices, but it can't be the same as the second so there are again 9 choices. So my answer is $9^3 = 729$.

I have checked on this website and found a similar problem, but their reasoning and answer did not match with my answer. Please tell me what I am doing wrong and what the correct answer is.

  • $\begingroup$ Your reasoning seems correct. $\endgroup$ – Aniruddha Deshmukh Jun 25 '17 at 6:04

You reasoning is correct: there are 9 options for every digit, resulting in $9^3 = 729$ valid three-digit numbers. You can verify this result with the following Python script:

i = 0
for a in range(1, 10):
  for b in range(0, 10):
    for c in range(0, 10):
      if a != b != c:
        i += 1

In general, the number of $n$-digit numbers with no two consecutive equal digits equals $9^n$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.