Prove this binomial sum

Question

Following problem is interesting

Show that: $$\sum_{i=1}^{n-1}\binom{n-1}{i} i^{i-1}(n-i)^{n-i-1}=n^{n-1}-n^{n-2}$$

Answer 1

This computation is very similar to the one at this MSE link.

We can prove this using the labelled tree function that is known from combinatorics.

This will provide a closed form of the exponential generating function of the two terms that are involved.

We seek to show that $$\sum_{k=1}^n {n\choose k} k^{k-1} (n+1-k)^{n-k} = (n+1)^n - (n+1)^{n-1} = n \times (n+1)^{n-1}.$$

The species of labelled trees has the specification $$\mathcal{T} = \mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$ which gives the functional equation $$T(z) = z \exp T(z).$$

We have $n! [z^n] T(z) = n^{n-1}$, for a proof consult the link from the introduction. This implies that $$T'(z) = \sum_{n\ge 1} n^{n-1} \frac{z^{n-1}}{(n-1)!} = \sum_{n\ge 0} (n+1)^{n} \frac{z^{n}}{n!}.$$

Simplify using $$z T'(z) = z \left(\exp T(z) + z \exp T(z) T'(z) \right) = T(z) + z T(z) T'(z)$$ which implies that $$T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$

Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that

$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$

i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

In the present case we have $$A(z) = T(z) \quad\text{and}\quad B(z) = T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$

The equality that we seek to prove is the convolution of the two exponential generating functions $A(z)$ and $B(z)$ and to verify it we must show that $$n! [z^n] A(z) B(z) = n \times (n+1)^{n-1}$$ But we have $$A(z) B(z) = \frac{1}{z} \frac{T(z)^2}{1- T(z)}.$$

It follows that $$n! [z^n] A(z) B(z) = n! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{z} \frac{T(z)^2}{1- T(z)} dz.$$

Using the same substitution as before (consult link) this becomes $$n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+2))}{w^{n+2}} \times \frac{w^2}{1-w} \times (\exp(-w) - w\exp(-w)) dw \\ = n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n}} dw \\ = n! \frac{(n+1)^{n-1}}{(n-1)!} = n \times (n+1)^{n-1}.$$

The labelled tree function recently appeared at this MSE link.