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).
