Question

A small combination lock on a suitcase has 55 ​wheels, each labeled with the 10 digits 0 to 9. How many 55 digit combinations are possible if successive digits must be​ different?

Answer 1

Suppose $a_n$ is the number of possible combinations for a suitcase with a lock consisting of $n$ wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for $n+1$ wheels, or $a_{n+1}$ is given recursively by the formula

$a_{n+1}=9a_n$

starting with $a_1=10$ (because you can start the combination with any one of the ten available digits 0 through 9).

For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).

By substitution, you have

$a_{n+1}=9a_n=9^2a_{n-1}=9^3a_{n-2}=\cdots=9^na_1=10\times9^n$

This means a lock with 55 wheels will have

$a_{55}=10\times9^{54}$

possible combinations (a number with 53 digits).