Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have to form a number of n digits having digits from 1 to 9. Constraint is that first and last digit must be same and no two consecutive digits must be same. How many such number of n digits can be there?

share|cite|improve this question
$n$ has to be larger than $2$ then? Try some examples...! Is this homework? – draks ... Oct 5 '12 at 6:48
Not a homework. Came through it while practicing a problem. – Shashwat Kumar Oct 5 '12 at 6:52
I guess $n=1$ would fit the constraints given here. – user22805 Oct 5 '12 at 6:57
up vote 1 down vote accepted

Since the first and last digits have to be the same, this is the same as asking how many ways there are to colour $n-1$ points on a circle with $9$ colours such that any two adjacent points have different colours. The answer is given here: $(-1)^{n-1}(9-1)+(9-1)^{n-1}=(-1)^{n-1}\cdot8+8^{n-1}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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