Let T(n) be the number of ternary strings of length n in which the number of 0's and the number of 1's is odd.
Consider the last digit of a ternary string of length n:
- If the last digit is 0, then the remaining n-1 digits must have an even number of 0's and an odd number of 1's.
- If the last digit is 1, then the remaining n-1 digits must have an odd number of 0's and an even number of 1's.
- If the last digit is 2, then the remaining n-1 digits can have any combination of 0's and 1's.
Therefore, we can write the recurrence relation as follows: T(n) = 2T(n-2) + 3^(n-1)
The first term in the recurrence relation corresponds to the cases where the last digit is either 0 or 1, and the remaining n-1 digits satisfy the conditions of the problem. There are 2 ways to choose whether the last digit is 0 or 1, and the number of valid strings of length n-2 is T(n-2).
The second term corresponds to the case where the last digit is 2, and the remaining n-1 digits can have any combination of 0's and 1's. There are 3 possible digits that can be used for each of the n-1 positions, so the total number of valid strings of length n with a last digit of 2 is 3^(n-1).
The initial conditions for the recurrence relation are: T(0) = 0 T(1) = 0 T(2) = 2 (the two valid strings are "01" and "10")
Using the recurrence relation and the initial conditions, we can compute T(n) for any positive integer n.