The letters of the Latin Alphabet

The sum of the weights is $(n+1)^{n-1}$ . We prove this via a standard double counting in and out argument, my teammate solved it like this in the test. Suppose there are $\binom{n}{k}a_k$ words of length $n$ that use $k$ letters. Then the sum of the weights is $\sum_{k=1}^n\frac{\binom{n}{k}a_k}{n-k+1}=\sum_{k=1}^n\frac{\binom{n+1}{k}a_k}{n+1} =\frac{1}{n+1}\sum_{k=1}^n \binom{n+1}{k}a_k$ . The sum on the right gives us the number of words of length $n$ from an alphabet of size $n+1$ and is therefore $(n+1)^n$ . dividing by $n+1$ yields the desired $n+1^{n-1}$ . Thus the sum of weights is $27^{25}=3^{75}$ and therefore the answer is $225$