A warehouse manager plans to assign each county a code consisting of

Method 1:

Writing out the single digits and pairs until one reaches a value that is greater than or equal to 16, one finds that 6 digits are needed.

Answer C

Method 2:

The codes comprise of two-digit codes and single-digit codes. The two-digit codes are created by seeing how many pairs can be made from \(n\) amount of digits and the single-digit codes are just the amount of digits, \(n\). The sum of these two amounts is the amount of possible codes that can be generated from \(n\) digits.

As a formula that looks like this: \(\frac{n!}{2!(n-2)!}+n\). As the question asks for the least amount of digits and we do not know if \(n\) amount of digits will yield exactly 16 possibilities or more, it will be easiest to see the number of digits required for \(>15\) possibilities.

\(\frac{n!}{2!(n-2)!}+n > 15\)

\(\frac{n!}{(n-2)!} > 30 - 2n\)

The factorials will make this a messy affair, so to get rid of them, \(\frac{n!}{(n-2)!} \) needs to be rewritten.
Looking at n=5: \(\frac{5!}{3!}= 5*4\) which in terms of n is: \(n*(n-1) = n^2 - n\). Plugging that in for \(\frac{n!}{(n-2)!}\):

\(n^2 - n > 30 - 2n\)

\(n^2 + n > 30\)

\(n(n+1) > 30\)

We know that \(5*6 = 30\), so to get 15 possibilities \(n=5\). This means that to cover 16 counties one will need \(n = 6\).

Answer C